[{"Name":"Introduction to Word Problems","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Tutorial","Duration":"5m 1s","ChapterTopicVideoID":7786,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/7786.jpeg","UploadDate":"2020-09-21T14:52:27.2600000","DurationForVideoObject":"PT5M1S","Description":null,"MetaTitle":"Tutorial: Video + Workbook | Proprep","MetaDescription":"Word Problems - Introduction to Word Problems. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/ordinary-differential-equations/word-problems/introduction-to-word-problems/vid7830","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.570","Text":"Ordinary differential equations appear often as part of a word problem."},{"Start":"00:06.570 ","End":"00:13.319","Text":"Now they are so varied in nature that it\u0027s really hard to say something very general,"},{"Start":"00:13.319 ","End":"00:16.155","Text":"but choose an example."},{"Start":"00:16.155 ","End":"00:20.010","Text":"Often it relates to questions about tangents,"},{"Start":"00:20.010 ","End":"00:22.936","Text":"and normals to curves,"},{"Start":"00:22.936 ","End":"00:24.645","Text":"and I\u0027ll just pick one example."},{"Start":"00:24.645 ","End":"00:26.040","Text":"This is really more included for"},{"Start":"00:26.040 ","End":"00:29.745","Text":"completeness sake because each word problem is different."},{"Start":"00:29.745 ","End":"00:31.620","Text":"But let\u0027s take an example,"},{"Start":"00:31.620 ","End":"00:35.790","Text":"which will say find the curve"},{"Start":"00:35.790 ","End":"00:40.260","Text":"and I\u0027ll give some properties of that curve with tangents or normal,"},{"Start":"00:40.260 ","End":"00:41.645","Text":"I\u0027ll choose the normal."},{"Start":"00:41.645 ","End":"00:49.310","Text":"Who\u0027s normal, the line perpendicular to the tangent, at each point."},{"Start":"00:49.310 ","End":"00:51.650","Text":"Let\u0027s call that point x,"},{"Start":"00:51.650 ","End":"01:02.060","Text":"y on the curve on it has a slope equal to y over x."},{"Start":"01:02.060 ","End":"01:03.560","Text":"That\u0027s 1 condition,"},{"Start":"01:03.560 ","End":"01:10.400","Text":"and I\u0027ll also say about it that it passes through a given point,"},{"Start":"01:10.400 ","End":"01:16.455","Text":"passes through, let\u0027s say the point 3, 4."},{"Start":"01:16.455 ","End":"01:21.710","Text":"Now, we don\u0027t see anywhere a differential equation,"},{"Start":"01:21.710 ","End":"01:23.785","Text":"so we have to set it up."},{"Start":"01:23.785 ","End":"01:26.090","Text":"Now let\u0027s see, we\u0027re talking about a normal,"},{"Start":"01:26.090 ","End":"01:28.460","Text":"normal is perpendicular to the tangent."},{"Start":"01:28.460 ","End":"01:30.590","Text":"The tangent is related to the slope,"},{"Start":"01:30.590 ","End":"01:32.630","Text":"slope is related to the derivative."},{"Start":"01:32.630 ","End":"01:39.710","Text":"Let\u0027s start off with the slope of the tangent because that\u0027s closer to the derivatives."},{"Start":"01:39.710 ","End":"01:44.300","Text":"Slope with a tangent in fact is equal to y prime."},{"Start":"01:44.300 ","End":"01:49.865","Text":"If you\u0027re using the other notation, it\u0027s dy/dx."},{"Start":"01:49.865 ","End":"01:54.940","Text":"Now, the slope of the normal is,"},{"Start":"01:54.940 ","End":"01:59.075","Text":"because it\u0027s perpendicular, it\u0027s the negative reciprocal."},{"Start":"01:59.075 ","End":"02:03.815","Text":"It\u0027s minus 1 over y prime."},{"Start":"02:03.815 ","End":"02:12.305","Text":"Or if you\u0027re using the other notation that would come out to be minus dx over dy."},{"Start":"02:12.305 ","End":"02:15.830","Text":"Now, if we compare this to y over x,"},{"Start":"02:15.830 ","End":"02:18.950","Text":"we have a differential equation that,"},{"Start":"02:18.950 ","End":"02:20.863","Text":"I\u0027ll use this form,"},{"Start":"02:20.863 ","End":"02:31.010","Text":"that minus dx/dy is equal to y/x."},{"Start":"02:31.010 ","End":"02:37.490","Text":"The reason I chose this form is that this now gives me a separable equation,"},{"Start":"02:37.490 ","End":"02:40.085","Text":"meaning I can use separation of variables."},{"Start":"02:40.085 ","End":"02:43.670","Text":"Because what I can do is cross multiply,"},{"Start":"02:43.670 ","End":"02:48.760","Text":"so this will give me that minus"},{"Start":"02:48.920 ","End":"02:56.510","Text":"xdx is equal to ydy so the variables are separated."},{"Start":"02:56.510 ","End":"02:58.040","Text":"X is on the left,"},{"Start":"02:58.040 ","End":"02:59.165","Text":"y is on the right."},{"Start":"02:59.165 ","End":"03:01.250","Text":"You can refer it the other way round. You know what else?"},{"Start":"03:01.250 ","End":"03:04.700","Text":"Switch sides. We put an integral sign in front of each."},{"Start":"03:04.700 ","End":"03:11.125","Text":"I\u0027ve got the integral of ydy equals,"},{"Start":"03:11.125 ","End":"03:13.230","Text":"input the minus in front of the integral,"},{"Start":"03:13.230 ","End":"03:20.540","Text":"minus the integral of xdx and this comes out to be y"},{"Start":"03:20.540 ","End":"03:28.360","Text":"squared over 2 equals this is minus x squared over 2,"},{"Start":"03:28.360 ","End":"03:30.920","Text":"but I need a constant."},{"Start":"03:30.920 ","End":"03:33.980","Text":"Now, I can find the constant because I have"},{"Start":"03:33.980 ","End":"03:38.795","Text":"an extra condition I haven\u0027t used that it passes through 3, 4."},{"Start":"03:38.795 ","End":"03:40.565","Text":"I\u0027ll continue over here."},{"Start":"03:40.565 ","End":"03:43.670","Text":"If I plug in 3, 4,"},{"Start":"03:43.670 ","End":"03:49.935","Text":"then I\u0027ve got the 4 squared over 2 is equal"},{"Start":"03:49.935 ","End":"03:57.810","Text":"to minus 3 squared over 2 plus the constant."},{"Start":"03:57.810 ","End":"04:00.765","Text":"The constant equals,"},{"Start":"04:00.765 ","End":"04:03.185","Text":"bringing everything to the other side,"},{"Start":"04:03.185 ","End":"04:07.895","Text":"I\u0027ve got 4 squared plus 3 squared over 2,"},{"Start":"04:07.895 ","End":"04:09.923","Text":"which is 16,"},{"Start":"04:09.923 ","End":"04:13.110","Text":"and 9 is 25 over 2."},{"Start":"04:13.110 ","End":"04:19.815","Text":"Now my equation becomes y squared over 2"},{"Start":"04:19.815 ","End":"04:27.480","Text":"equals minus x squared over 2 plus 25 over 2."},{"Start":"04:27.480 ","End":"04:30.425","Text":"That is the answer, but we should tidy it up a bit."},{"Start":"04:30.425 ","End":"04:35.680","Text":"Let\u0027s multiply everything by 2 and bring the x squared to the other side,"},{"Start":"04:35.680 ","End":"04:43.005","Text":"and I\u0027ll get x squared plus y squared equals 25."},{"Start":"04:43.005 ","End":"04:45.875","Text":"This is the answer to the question."},{"Start":"04:45.875 ","End":"04:53.015","Text":"You might identify this as a circle of radius 5 centered at the origin."},{"Start":"04:53.015 ","End":"04:56.375","Text":"As I said, each problem is different,"},{"Start":"04:56.375 ","End":"04:59.245","Text":"and this is just a sample one."},{"Start":"04:59.245 ","End":"05:01.630","Text":"I\u0027m done."}],"ID":7830},{"Watched":false,"Name":"Exercise 1","Duration":"1m 24s","ChapterTopicVideoID":7757,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.530","Text":"Here we\u0027re given a word problem which will turn out to give a differential equation."},{"Start":"00:04.530 ","End":"00:07.500","Text":"For a given curve, the slope of the tangent at each point x,"},{"Start":"00:07.500 ","End":"00:10.560","Text":"y on the curve is equal to minus x over y."},{"Start":"00:10.560 ","End":"00:12.660","Text":"Find the equation of the curve."},{"Start":"00:12.660 ","End":"00:19.995","Text":"The first thing is to remember that the slope of the tangent is also given by dy over dx."},{"Start":"00:19.995 ","End":"00:23.120","Text":"From the one hand, it\u0027s dy over dx in general."},{"Start":"00:23.120 ","End":"00:25.430","Text":"Here it\u0027s equal to minus xy."},{"Start":"00:25.430 ","End":"00:30.935","Text":"Then we simply get the equation that dy over dx=minus x over y."},{"Start":"00:30.935 ","End":"00:32.920","Text":"Now this is separable."},{"Start":"00:32.920 ","End":"00:38.980","Text":"Easily, we can cross-multiply and get that ydy is minus xdx."},{"Start":"00:39.410 ","End":"00:45.095","Text":"From here we can put the integral sign in front of each one of them."},{"Start":"00:45.095 ","End":"00:46.820","Text":"Both of these are easy integrals."},{"Start":"00:46.820 ","End":"00:48.740","Text":"Here we get y^2 over 2,"},{"Start":"00:48.740 ","End":"00:52.955","Text":"and here we have minus x^2 over 2 with the constant of integration."},{"Start":"00:52.955 ","End":"00:55.555","Text":"If I multiply both sides by 2,"},{"Start":"00:55.555 ","End":"01:00.620","Text":"then I\u0027ll get this and bring the x^2 over to the other side."},{"Start":"01:00.620 ","End":"01:02.165","Text":"We get this."},{"Start":"01:02.165 ","End":"01:05.615","Text":"After all, what is 2c but a general constant"},{"Start":"01:05.615 ","End":"01:12.725","Text":"k. What in fact we have is a circle with center at the origin."},{"Start":"01:12.725 ","End":"01:17.105","Text":"We have to assume that k is bigger or equal to 0,"},{"Start":"01:17.105 ","End":"01:20.690","Text":"otherwise there is no solution and the radius is"},{"Start":"01:20.690 ","End":"01:25.680","Text":"just the square root of k. That\u0027s it. We\u0027re done."}],"ID":7831},{"Watched":false,"Name":"Exercise 2","Duration":"6m 28s","ChapterTopicVideoID":7758,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.564","Text":"Here we have another word problem."},{"Start":"00:02.564 ","End":"00:05.640","Text":"Given a curve in the first quadrant which goes through the point(1,"},{"Start":"00:05.640 ","End":"00:11.670","Text":"3) and it has the property that the slope of its tangent at the point x,"},{"Start":"00:11.670 ","End":"00:15.120","Text":"y in general is equal to this expression."},{"Start":"00:15.120 ","End":"00:17.625","Text":"We have to find the equation of the curve."},{"Start":"00:17.625 ","End":"00:21.795","Text":"Now the slope of the tangent is just dy over dx,"},{"Start":"00:21.795 ","End":"00:27.240","Text":"and what I get is that dy over dx equals this,"},{"Start":"00:27.240 ","End":"00:31.260","Text":"but I also wrote this over a common denominator."},{"Start":"00:31.260 ","End":"00:34.920","Text":"What\u0027s inside the brackets is x over x plus y over x."},{"Start":"00:34.920 ","End":"00:39.240","Text":"In other words, this and I also added the fact that we\u0027re in the first quadrant,"},{"Start":"00:39.240 ","End":"00:42.200","Text":"which means that both x and y are bigger than 0."},{"Start":"00:42.200 ","End":"00:45.395","Text":"For instance, x is not 0."},{"Start":"00:45.395 ","End":"00:54.785","Text":"Then we can multiply both sides by x and dx and bring things over to the left-hand side."},{"Start":"00:54.785 ","End":"00:57.260","Text":"Simple algebra brings us this."},{"Start":"00:57.260 ","End":"01:00.440","Text":"Our goal is to try and separate the variables."},{"Start":"01:00.440 ","End":"01:03.200","Text":"Well, first of all, we show it\u0027s homogeneous."},{"Start":"01:03.200 ","End":"01:06.815","Text":"If I let this bit be m and this bit be n,"},{"Start":"01:06.815 ","End":"01:08.495","Text":"like we usually do,"},{"Start":"01:08.495 ","End":"01:14.030","Text":"then we get that both of them will be homogeneous of degree 1."},{"Start":"01:14.030 ","End":"01:15.470","Text":"Just going to present it."},{"Start":"01:15.470 ","End":"01:16.490","Text":"I\u0027m not going to go into it."},{"Start":"01:16.490 ","End":"01:21.500","Text":"It\u0027s very easy to see that m and n are both homogeneous functions of degree 1."},{"Start":"01:21.500 ","End":"01:25.280","Text":"If that\u0027s the case, the equation is homogeneous,"},{"Start":"01:25.280 ","End":"01:28.820","Text":"which means that if we make the appropriate substitution,"},{"Start":"01:28.820 ","End":"01:30.920","Text":"we can convert it into separable."},{"Start":"01:30.920 ","End":"01:36.110","Text":"The usual substitution is that y equals vx,"},{"Start":"01:36.110 ","End":"01:41.960","Text":"where v is a new variable and then dy is what you get when you do a product rule."},{"Start":"01:41.960 ","End":"01:46.159","Text":"I usually like to remember that in case I want to substitute"},{"Start":"01:46.159 ","End":"01:52.519","Text":"back that v is equal to y over x."},{"Start":"01:52.519 ","End":"01:54.785","Text":"That\u0027s useful for later."},{"Start":"01:54.785 ","End":"02:03.665","Text":"Continuing, what I basically did was that wherever I had a y in the equation,"},{"Start":"02:03.665 ","End":"02:08.450","Text":"say here, then I replace it by vx."},{"Start":"02:08.450 ","End":"02:12.660","Text":"This only occurred once and wherever we see dy,"},{"Start":"02:12.660 ","End":"02:15.245","Text":"which is only in one place here,"},{"Start":"02:15.245 ","End":"02:20.250","Text":"then we replace it by the dy from here."},{"Start":"02:20.250 ","End":"02:29.360","Text":"I can spell it out that the dy is this and the y is the VX."},{"Start":"02:29.360 ","End":"02:32.030","Text":"We\u0027re all straight now."},{"Start":"02:32.030 ","End":"02:40.185","Text":"Next we can expand the brackets and what we get is this,"},{"Start":"02:40.185 ","End":"02:43.820","Text":"x squared dv and so on and so on."},{"Start":"02:43.820 ","End":"02:53.335","Text":"What we can do now is collect dv separately and dx separately is dv."},{"Start":"02:53.335 ","End":"02:59.265","Text":"Here is dx, dx, dx."},{"Start":"02:59.265 ","End":"03:02.180","Text":"What I get is,"},{"Start":"03:02.180 ","End":"03:04.565","Text":"here\u0027s the x squared dv,"},{"Start":"03:04.565 ","End":"03:06.695","Text":"here with the dx,"},{"Start":"03:06.695 ","End":"03:08.270","Text":"all of the terms have x,"},{"Start":"03:08.270 ","End":"03:11.240","Text":"so I\u0027ll take that out and what I\u0027m left with is v"},{"Start":"03:11.240 ","End":"03:14.720","Text":"plus 1 plus v instead of v plus 1 plus v,"},{"Start":"03:14.720 ","End":"03:16.550","Text":"I write 2v plus 1."},{"Start":"03:16.550 ","End":"03:18.635","Text":"I don\u0027t think anyone objects to that."},{"Start":"03:18.635 ","End":"03:23.330","Text":"Now, bring this to the other side and then do the division so that we"},{"Start":"03:23.330 ","End":"03:28.835","Text":"get all the v\u0027s on the left with dv and all the x\u0027s on the right width dx."},{"Start":"03:28.835 ","End":"03:30.695","Text":"I\u0027m going to move to the next page."},{"Start":"03:30.695 ","End":"03:34.015","Text":"I\u0027m just going to copy this and put an integral sign in front of it."},{"Start":"03:34.015 ","End":"03:36.545","Text":"Here, this is what we had before,"},{"Start":"03:36.545 ","End":"03:38.390","Text":"just with an integral sign."},{"Start":"03:38.390 ","End":"03:41.990","Text":"Now, I\u0027m going to produce a formula that we should use."},{"Start":"03:41.990 ","End":"03:44.585","Text":"This is a formula I use often,"},{"Start":"03:44.585 ","End":"03:47.840","Text":"and that if we have the integral of a fraction"},{"Start":"03:47.840 ","End":"03:51.395","Text":"where the numerator is the derivative of the denominator,"},{"Start":"03:51.395 ","End":"03:54.755","Text":"then we know the answer in terms of the natural log."},{"Start":"03:54.755 ","End":"03:58.810","Text":"Here I have a denominator 2_v plus 1,"},{"Start":"03:58.810 ","End":"04:04.010","Text":"but I don\u0027t have its derivative exactly on the numerator because derivative is 2,"},{"Start":"04:04.010 ","End":"04:05.990","Text":"but we know how to fix that."},{"Start":"04:05.990 ","End":"04:09.380","Text":"Essentially I force it to be 2 here."},{"Start":"04:09.380 ","End":"04:12.380","Text":"But to compensate, I also put a 1/2"},{"Start":"04:12.380 ","End":"04:16.070","Text":"here so I haven\u0027t cheated anyone and I haven\u0027t changed anything."},{"Start":"04:16.070 ","End":"04:21.620","Text":"That gives us that the integral here is as follows."},{"Start":"04:21.620 ","End":"04:26.195","Text":"It\u0027s 1/2 natural log of the denominator."},{"Start":"04:26.195 ","End":"04:30.530","Text":"I didn\u0027t put the absolute value because y and x are positive,"},{"Start":"04:30.530 ","End":"04:31.805","Text":"so v is positive,"},{"Start":"04:31.805 ","End":"04:36.170","Text":"this is positive, and here also x is positive."},{"Start":"04:36.170 ","End":"04:39.020","Text":"There is minus. The integral of 1 over x is"},{"Start":"04:39.020 ","End":"04:42.080","Text":"natural log of absolute value of x which is x."},{"Start":"04:42.080 ","End":"04:46.940","Text":"We\u0027ve basically solved the differential equation,"},{"Start":"04:46.940 ","End":"04:48.980","Text":"but there\u0027s still two things remaining."},{"Start":"04:48.980 ","End":"04:54.410","Text":"First of all, I have to substitute back from v to x and y. Secondly,"},{"Start":"04:54.410 ","End":"04:56.150","Text":"there was an initial condition."},{"Start":"04:56.150 ","End":"04:59.090","Text":"I\u0027ll just rearrange things slightly to make it"},{"Start":"04:59.090 ","End":"05:02.870","Text":"easier if I multiply everything by 2 and bring this over here,"},{"Start":"05:02.870 ","End":"05:04.855","Text":"this is what I get."},{"Start":"05:04.855 ","End":"05:09.890","Text":"I can also put the 2 from the natural log inside here based on"},{"Start":"05:09.890 ","End":"05:17.375","Text":"the logarithm and the other thing that\u0027s customary to do is when I set up a constant,"},{"Start":"05:17.375 ","End":"05:20.810","Text":"we say that this constant is the natural log of another"},{"Start":"05:20.810 ","End":"05:24.695","Text":"constant because every number can be the logarithm of something else."},{"Start":"05:24.695 ","End":"05:27.680","Text":"If we combine this with the logarithm,"},{"Start":"05:27.680 ","End":"05:32.480","Text":"the logarithm of a product is the sum of the logarithms we get this."},{"Start":"05:32.480 ","End":"05:39.050","Text":"Next, we can take the logarithm away from both sides because the logarithms are equal,"},{"Start":"05:39.050 ","End":"05:41.210","Text":"the quantities are equal."},{"Start":"05:41.210 ","End":"05:45.445","Text":"Then I replace v by y over x."},{"Start":"05:45.445 ","End":"05:50.775","Text":"2y over x times x squared is 2_yx and 1 times x squared is x squared."},{"Start":"05:50.775 ","End":"05:58.260","Text":"The initial condition said that the curve goes through the 0.1 comma 3."},{"Start":"05:58.260 ","End":"06:00.495","Text":"This means that when x is 1,"},{"Start":"06:00.495 ","End":"06:08.180","Text":"y is 3 and we can put that here and get 2 times 3 times 1"},{"Start":"06:08.180 ","End":"06:18.225","Text":"plus 1 squared equals c and this gives us that c equals 7."},{"Start":"06:18.225 ","End":"06:23.810","Text":"We finally get the solution to the differential equation as follows,"},{"Start":"06:23.810 ","End":"06:25.910","Text":"just by putting c equals 7 here,"},{"Start":"06:25.910 ","End":"06:28.620","Text":"and we are done."}],"ID":7832},{"Watched":false,"Name":"Exercise 3","Duration":"1m 40s","ChapterTopicVideoID":7759,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.060","Text":"In this exercise, we have to find the equation of"},{"Start":"00:03.060 ","End":"00:07.545","Text":"the curve such that the normal at each point passes through the origin."},{"Start":"00:07.545 ","End":"00:12.270","Text":"For example, at this point here we take the normal that goes through the origin."},{"Start":"00:12.270 ","End":"00:17.535","Text":"Now there\u0027s a formula for the equation of a normal passing through a given point."},{"Start":"00:17.535 ","End":"00:19.730","Text":"This is the equation,"},{"Start":"00:19.730 ","End":"00:24.770","Text":"is given by y minus y_1 equals something times x minus x_1,"},{"Start":"00:24.770 ","End":"00:29.230","Text":"and that\u0027s something is exactly the negative reciprocal of the slope."},{"Start":"00:29.230 ","End":"00:31.980","Text":"If it was the slope it would be dy by dx,"},{"Start":"00:31.980 ","End":"00:33.960","Text":"but since the slope is not the tangent,"},{"Start":"00:33.960 ","End":"00:35.640","Text":"but the normal negative reciprocal."},{"Start":"00:35.640 ","End":"00:38.775","Text":"Anyway, this is the formula and in our case, x_1,"},{"Start":"00:38.775 ","End":"00:42.030","Text":"y_1 is the origin so it means that x_1,"},{"Start":"00:42.030 ","End":"00:43.110","Text":"y_1 is 0, 0."},{"Start":"00:43.110 ","End":"00:46.500","Text":"When I plug it in here and I get this."},{"Start":"00:46.500 ","End":"00:50.105","Text":"After that I multiply both sides by dy,"},{"Start":"00:50.105 ","End":"00:52.690","Text":"I will get this."},{"Start":"00:52.690 ","End":"00:55.790","Text":"Then you can see the variables are separated I"},{"Start":"00:55.790 ","End":"00:58.505","Text":"just take an integral sign in front of each."},{"Start":"00:58.505 ","End":"01:01.205","Text":"Then let\u0027s actually do the integral."},{"Start":"01:01.205 ","End":"01:03.140","Text":"Here is y^2 over 2,"},{"Start":"01:03.140 ","End":"01:06.605","Text":"here is minus x^2 over 2 and there\u0027s the constant."},{"Start":"01:06.605 ","End":"01:13.375","Text":"Multiplied by 2 and then bring the x^2 to the other side."},{"Start":"01:13.375 ","End":"01:16.290","Text":"We get x^2 plus y^2 equals,"},{"Start":"01:16.290 ","End":"01:17.990","Text":"well, you say it should be 2c,"},{"Start":"01:17.990 ","End":"01:22.850","Text":"but twice a constant is just the constant which is k. In fact,"},{"Start":"01:22.850 ","End":"01:29.000","Text":"this is the answer, but k we can see is bigger than 0 or bigger or equal to anyway."},{"Start":"01:29.000 ","End":"01:33.920","Text":"This is a circle whose radius is the square root of k because"},{"Start":"01:33.920 ","End":"01:40.140","Text":"x^2 plus y^2 = r^2 would be a circle. We are done."}],"ID":7833},{"Watched":false,"Name":"Exercise 4","Duration":"2m 48s","ChapterTopicVideoID":7760,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.565","Text":"Here we have another word problem."},{"Start":"00:02.565 ","End":"00:04.800","Text":"Find the equation of the curve,"},{"Start":"00:04.800 ","End":"00:07.530","Text":"the slope of whose tangent at each point is equal to"},{"Start":"00:07.530 ","End":"00:11.025","Text":"half the slope of the segment from the origin to the point."},{"Start":"00:11.025 ","End":"00:13.604","Text":"Let\u0027s see what this means in terms of the picture."},{"Start":"00:13.604 ","End":"00:17.970","Text":"We have a curve which is the one in black."},{"Start":"00:17.970 ","End":"00:20.730","Text":"Then we have the tangent in red,"},{"Start":"00:20.730 ","End":"00:24.780","Text":"and we have the segment from the origin to the point in dotted blue."},{"Start":"00:24.780 ","End":"00:29.070","Text":"What we\u0027re saying is that the slope of this tangent in"},{"Start":"00:29.070 ","End":"00:34.840","Text":"red is half the slope of this dotted blue line and we have to find the curve."},{"Start":"00:35.360 ","End":"00:38.975","Text":"The derivative is the slope,"},{"Start":"00:38.975 ","End":"00:40.985","Text":"and that\u0027s dy over dx."},{"Start":"00:40.985 ","End":"00:42.559","Text":"That\u0027s the slope of the tangent."},{"Start":"00:42.559 ","End":"00:45.200","Text":"The slope of the dotted blue line,"},{"Start":"00:45.200 ","End":"00:48.770","Text":"the segment, is just y over x,"},{"Start":"00:48.770 ","End":"00:55.160","Text":"because this height of a triangle is y and the base is x,"},{"Start":"00:55.160 ","End":"00:56.990","Text":"it\u0027s y over x."},{"Start":"00:56.990 ","End":"01:00.320","Text":"The condition says that this one is half this one,"},{"Start":"01:00.320 ","End":"01:01.984","Text":"then this is the equation."},{"Start":"01:01.984 ","End":"01:03.950","Text":"Of course since x is in the denominator."},{"Start":"01:03.950 ","End":"01:07.174","Text":"We have to say that x is not equal to 0."},{"Start":"01:07.174 ","End":"01:10.150","Text":"We\u0027ll either have positive x or negative x,"},{"Start":"01:10.150 ","End":"01:12.875","Text":"in this case in the picture the x is positive."},{"Start":"01:12.875 ","End":"01:16.640","Text":"Actually, we have to remove this point strictly speaking,"},{"Start":"01:16.640 ","End":"01:18.170","Text":"because x is not 0,"},{"Start":"01:18.170 ","End":"01:19.685","Text":"but we\u0027ll let it be."},{"Start":"01:19.685 ","End":"01:23.870","Text":"Next thing we get from this is we can separate the variables."},{"Start":"01:23.870 ","End":"01:27.685","Text":"We bring the 2 over here and the dx over here."},{"Start":"01:27.685 ","End":"01:31.400","Text":"I guess we also have to say that y is not equal to 0."},{"Start":"01:31.400 ","End":"01:33.560","Text":"If we let y not equal to 0,"},{"Start":"01:33.560 ","End":"01:36.290","Text":"now we can take the integral of both sides,"},{"Start":"01:36.290 ","End":"01:40.400","Text":"and this is an easy one to solve because we know the integral"},{"Start":"01:40.400 ","End":"01:44.540","Text":"of 1 over x is natural log of x and similarly natural log of y."},{"Start":"01:44.540 ","End":"01:47.135","Text":"The 2 stays and we have a constant."},{"Start":"01:47.135 ","End":"01:50.460","Text":"Now we can bring the 2 inside the natural logarithm."},{"Start":"01:50.460 ","End":"01:53.530","Text":"It\u0027s the natural logarithm of y^2."},{"Start":"01:53.530 ","End":"01:56.225","Text":"Then we don\u0027t need the absolute value anymore."},{"Start":"01:56.225 ","End":"01:57.920","Text":"Here we still do,"},{"Start":"01:57.920 ","End":"02:00.110","Text":"and as is customary with logarithms,"},{"Start":"02:00.110 ","End":"02:05.255","Text":"we put c as the natural logarithm of some positive number."},{"Start":"02:05.255 ","End":"02:09.290","Text":"Then we can say that removing the log,"},{"Start":"02:09.290 ","End":"02:11.779","Text":"and then making the sum into a product,"},{"Start":"02:11.779 ","End":"02:16.220","Text":"we get the y^2 is absolute value of x times k,"},{"Start":"02:16.220 ","End":"02:18.305","Text":"or k times the absolute value of x."},{"Start":"02:18.305 ","End":"02:20.630","Text":"Now we don\u0027t want x to be 0."},{"Start":"02:20.630 ","End":"02:23.930","Text":"That would mean that x is either 0 or positive or all negative."},{"Start":"02:23.930 ","End":"02:25.460","Text":"If x is positive,"},{"Start":"02:25.460 ","End":"02:27.845","Text":"then we can drop the absolute value."},{"Start":"02:27.845 ","End":"02:29.390","Text":"If k is negative,"},{"Start":"02:29.390 ","End":"02:34.530","Text":"we just get a minus k instead of a k. If we let this constant be a,"},{"Start":"02:34.530 ","End":"02:36.350","Text":"which is k or minus k,"},{"Start":"02:36.350 ","End":"02:37.520","Text":"we can say in general,"},{"Start":"02:37.520 ","End":"02:40.940","Text":"without the absolute value that y^2 is ax."},{"Start":"02:40.940 ","End":"02:43.580","Text":"If a is positive, we get something like in the picture."},{"Start":"02:43.580 ","End":"02:45.770","Text":"If a is negative, we get a mirror image,"},{"Start":"02:45.770 ","End":"02:48.690","Text":"something like that. We are done."}],"ID":7834},{"Watched":false,"Name":"Exercise 5","Duration":"6m 36s","ChapterTopicVideoID":7761,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.174","Text":"Here we have a word problem."},{"Start":"00:02.174 ","End":"00:05.340","Text":"Find the equation of the curve which passes through the point."},{"Start":"00:05.340 ","End":"00:07.020","Text":"For each point on it,"},{"Start":"00:07.020 ","End":"00:09.345","Text":"the slope of the normal is."},{"Start":"00:09.345 ","End":"00:12.810","Text":"You have to remember what the slope of the normal is,"},{"Start":"00:12.810 ","End":"00:15.330","Text":"it\u0027s minus dx/dy,"},{"Start":"00:15.330 ","End":"00:21.135","Text":"and gives us our differential equation immediately because that\u0027s equal to this."},{"Start":"00:21.135 ","End":"00:25.815","Text":"Now we can cross-multiply because of the minus,"},{"Start":"00:25.815 ","End":"00:29.730","Text":"we just switched the order of the subtraction,"},{"Start":"00:29.730 ","End":"00:31.380","Text":"make it x^2 minus y^2,"},{"Start":"00:31.380 ","End":"00:32.820","Text":"and that will be okay."},{"Start":"00:32.820 ","End":"00:37.095","Text":"Then we also bring the 2xy dy over to the left,"},{"Start":"00:37.095 ","End":"00:42.090","Text":"and then we get it in the form of M dx plus N dy equals 0."},{"Start":"00:42.090 ","End":"00:45.750","Text":"I\u0027m aiming at is to show that this is a homogeneous differential equation,"},{"Start":"00:45.750 ","End":"00:47.270","Text":"and we do this by showing that each of"},{"Start":"00:47.270 ","End":"00:50.570","Text":"these 2 functions is homogeneous of the same order."},{"Start":"00:50.570 ","End":"00:53.765","Text":"What we get is that M is this."},{"Start":"00:53.765 ","End":"00:55.970","Text":"And I\u0027m not going to go into all the details,"},{"Start":"00:55.970 ","End":"00:57.425","Text":"you can follow it later."},{"Start":"00:57.425 ","End":"01:00.680","Text":"Turns out to be homogeneous of order 2."},{"Start":"01:00.680 ","End":"01:04.970","Text":"Similarly and also turns out to be homogeneous of order 2."},{"Start":"01:04.970 ","End":"01:09.380","Text":"This is homogeneous equation and we use our usual substitution,"},{"Start":"01:09.380 ","End":"01:13.250","Text":"which is the y equals vx and dy is given by this."},{"Start":"01:13.250 ","End":"01:20.870","Text":"I usually like to add that later on when we substitute back v is equal to y/x,"},{"Start":"01:20.870 ","End":"01:24.745","Text":"which means that we want the x naught to be 0."},{"Start":"01:24.745 ","End":"01:29.600","Text":"Continuing, we do the substitution and we get this."},{"Start":"01:29.600 ","End":"01:31.550","Text":"Now I\u0027ll show you how we got this."},{"Start":"01:31.550 ","End":"01:35.030","Text":"We substitute wherever we see y in this equation,"},{"Start":"01:35.030 ","End":"01:37.060","Text":"which would be here,"},{"Start":"01:37.060 ","End":"01:42.690","Text":"and here we substitute vx from y here."},{"Start":"01:42.690 ","End":"01:45.415","Text":"When we see dy,"},{"Start":"01:45.415 ","End":"01:50.120","Text":"which is here, we substitute what dy was here."},{"Start":"01:50.120 ","End":"01:54.150","Text":"For example, this y gives me this vx, this y gives me vx,"},{"Start":"01:54.650 ","End":"01:59.990","Text":"but the vx gets combined with the x and it gives"},{"Start":"01:59.990 ","End":"02:05.510","Text":"x^2 v. The dy is just what is written here,"},{"Start":"02:05.510 ","End":"02:06.680","Text":"and that\u0027s over here."},{"Start":"02:06.680 ","End":"02:09.020","Text":"That\u0027s how we get this equation."},{"Start":"02:09.020 ","End":"02:15.080","Text":"Then what we want to do is to expand after expansion and also"},{"Start":"02:15.080 ","End":"02:21.445","Text":"separating the dx here and the dx came from here,"},{"Start":"02:21.445 ","End":"02:28.070","Text":"and from here, which gave us x^2 minus v^2 x^2,"},{"Start":"02:28.070 ","End":"02:34.245","Text":"and this v with this 2x^2 v gives us the 2x^2 v^2."},{"Start":"02:34.245 ","End":"02:38.520","Text":"Then we also have the dv collected together."},{"Start":"02:38.520 ","End":"02:41.160","Text":"This dv came from,"},{"Start":"02:41.160 ","End":"02:43.230","Text":"I have a dv here,"},{"Start":"02:43.230 ","End":"02:49.880","Text":"and the coefficient is x times x^2 is x^3 and we still have a v,"},{"Start":"02:49.880 ","End":"02:51.335","Text":"and its minus 2."},{"Start":"02:51.335 ","End":"02:53.525","Text":"This gives us this equation."},{"Start":"02:53.525 ","End":"02:56.480","Text":"What we\u0027re aiming for is to separate the variables,"},{"Start":"02:56.480 ","End":"02:59.450","Text":"this is what we do with the homogeneous equations."},{"Start":"02:59.450 ","End":"03:03.530","Text":"That\u0027s what this substitution is supposed to do for us separate the variables."},{"Start":"03:03.530 ","End":"03:07.220","Text":"Continuing, I can take here x^2 outside"},{"Start":"03:07.220 ","End":"03:11.120","Text":"the brackets and I\u0027m left with 1 minus v^2 minus 2v^2,"},{"Start":"03:11.120 ","End":"03:13.010","Text":"which is just 3v^2."},{"Start":"03:13.010 ","End":"03:15.304","Text":"Bringing this to the other side,"},{"Start":"03:15.304 ","End":"03:16.775","Text":"this is what I get."},{"Start":"03:16.775 ","End":"03:25.680","Text":"Now we can really divide both sides by x^3 and also by 1 minus 3v^2."},{"Start":"03:25.680 ","End":"03:29.150","Text":"This gives us this equation, as we see,"},{"Start":"03:29.150 ","End":"03:31.130","Text":"the x\u0027s are all on the left with dx,"},{"Start":"03:31.130 ","End":"03:32.945","Text":"the v\u0027s are all on the right with dv."},{"Start":"03:32.945 ","End":"03:36.125","Text":"We just take an integral in front of each side,"},{"Start":"03:36.125 ","End":"03:40.835","Text":"and now we have a relatively simple integration problem."},{"Start":"03:40.835 ","End":"03:44.840","Text":"The integral of dx/x is natural log of x."},{"Start":"03:44.840 ","End":"03:48.200","Text":"But here, what we would like is to have"},{"Start":"03:48.200 ","End":"03:52.985","Text":"the derivative of the denominator in the numerator,"},{"Start":"03:52.985 ","End":"03:55.040","Text":"and we use our usual tricks."},{"Start":"03:55.040 ","End":"03:57.740","Text":"I\u0027d like to have minus 6v here,"},{"Start":"03:57.740 ","End":"04:00.860","Text":"so write minus 6 here then are going to fix it."},{"Start":"04:00.860 ","End":"04:05.360","Text":"This minus 3 with this minus 6 together give us 2,"},{"Start":"04:05.360 ","End":"04:06.710","Text":"so we haven\u0027t changed anything,"},{"Start":"04:06.710 ","End":"04:09.320","Text":"but now we have a derivative of this here."},{"Start":"04:09.320 ","End":"04:11.840","Text":"We use logarithms, because,"},{"Start":"04:11.840 ","End":"04:17.060","Text":"we have a formula that if we have the derivative of the denominator in the numerator,"},{"Start":"04:17.060 ","End":"04:20.240","Text":"the answer is just natural log of the denominator,"},{"Start":"04:20.240 ","End":"04:21.485","Text":"and that\u0027s what I\u0027ve done here."},{"Start":"04:21.485 ","End":"04:23.000","Text":"The minus a 1/3 stays,"},{"Start":"04:23.000 ","End":"04:24.140","Text":"this is natural log,"},{"Start":"04:24.140 ","End":"04:28.750","Text":"we add a constant and we\u0027re going to continue on the next page."},{"Start":"04:28.750 ","End":"04:31.310","Text":"This is what I had before,"},{"Start":"04:31.310 ","End":"04:33.905","Text":"except that I had v here,"},{"Start":"04:33.905 ","End":"04:38.765","Text":"and the place where I had v is where I put y/x."},{"Start":"04:38.765 ","End":"04:41.155","Text":"Remember we substitute back."},{"Start":"04:41.155 ","End":"04:44.370","Text":"Then we could just simplify this a bit."},{"Start":"04:44.370 ","End":"04:46.865","Text":"I have no idea why I wrote this twice."},{"Start":"04:46.865 ","End":"04:48.920","Text":"Okay, sorry, Let\u0027s continue."},{"Start":"04:48.920 ","End":"04:55.695","Text":"We have an initial condition that when x equals 1, y equals 2."},{"Start":"04:55.695 ","End":"04:58.190","Text":"If I plug that in,"},{"Start":"04:58.190 ","End":"05:01.220","Text":"what l\u0027ll get is that x is 1,"},{"Start":"05:01.220 ","End":"05:04.505","Text":"the absolute value of 1 is 1 and we get the minus a 1/3."},{"Start":"05:04.505 ","End":"05:07.570","Text":"Now here, y/x is 2,"},{"Start":"05:07.570 ","End":"05:12.734","Text":"2^2 is 4 times 3 is 12,"},{"Start":"05:12.734 ","End":"05:17.945","Text":"but 1 minus 12 is minus 11 in absolute values is still 11."},{"Start":"05:17.945 ","End":"05:20.005","Text":"That\u0027s how we get this."},{"Start":"05:20.005 ","End":"05:25.250","Text":"From here, c is equal to 1/3 natural log of 11,"},{"Start":"05:25.250 ","End":"05:27.260","Text":"because natural log of 1 is 0,"},{"Start":"05:27.260 ","End":"05:31.100","Text":"so I just bring this to the other side that gives me my constant."},{"Start":"05:31.100 ","End":"05:34.130","Text":"Now I can substitute the constant here,"},{"Start":"05:34.130 ","End":"05:36.890","Text":"and that will give me just the same as this,"},{"Start":"05:36.890 ","End":"05:40.070","Text":"but with the constant specifically stated."},{"Start":"05:40.070 ","End":"05:46.550","Text":"Let\u0027s bring this term to this side on the left and multiply everything by 3."},{"Start":"05:46.550 ","End":"05:49.670","Text":"Now, I can use laws of logarithm."},{"Start":"05:49.670 ","End":"05:52.040","Text":"Here we have the absolute value of x^3."},{"Start":"05:52.040 ","End":"05:54.785","Text":"The plus becomes a multiplication,"},{"Start":"05:54.785 ","End":"05:56.720","Text":"so we multiply by this,"},{"Start":"05:56.720 ","End":"05:58.510","Text":"and that\u0027s natural log of 11."},{"Start":"05:58.510 ","End":"06:01.730","Text":"If the logarithms are equal then the numbers are equal,"},{"Start":"06:01.730 ","End":"06:03.350","Text":"so we get this."},{"Start":"06:03.350 ","End":"06:06.245","Text":"There appear to be 2 possible solutions."},{"Start":"06:06.245 ","End":"06:09.635","Text":"Either x^3 minus 3y^2 x is 11,"},{"Start":"06:09.635 ","End":"06:11.885","Text":"or it\u0027s equal to minus 11."},{"Start":"06:11.885 ","End":"06:19.240","Text":"But since it passes through the point where x = 1 and y = 2,"},{"Start":"06:19.240 ","End":"06:23.990","Text":"then we see that this should be actually minus 11."},{"Start":"06:23.990 ","End":"06:26.060","Text":"If I drop the absolute value,"},{"Start":"06:26.060 ","End":"06:28.910","Text":"oops, I didn\u0027t write the minus here. Let me just fix that."},{"Start":"06:28.910 ","End":"06:30.395","Text":"I\u0027ll put a minus in here,"},{"Start":"06:30.395 ","End":"06:36.300","Text":"x^3 minus 3y^2x is minus 11 and this is our answer."}],"ID":7835},{"Watched":false,"Name":"Exercise 6","Duration":"3m 50s","ChapterTopicVideoID":7762,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.414","Text":"Here we have another word problem."},{"Start":"00:02.414 ","End":"00:04.470","Text":"We\u0027re given a curve in the first quadrant,"},{"Start":"00:04.470 ","End":"00:06.705","Text":"it passes through this point."},{"Start":"00:06.705 ","End":"00:11.850","Text":"We also know that for each point A on the curve,"},{"Start":"00:11.850 ","End":"00:16.980","Text":"the difference between the slope of the tangent to the curve at A and the slope of"},{"Start":"00:16.980 ","End":"00:23.010","Text":"the line connecting A with the origin is equal to the y-coordinate of A."},{"Start":"00:23.010 ","End":"00:25.875","Text":"We have to find the equation of the curve."},{"Start":"00:25.875 ","End":"00:28.905","Text":"Let\u0027s see, we have several quantities."},{"Start":"00:28.905 ","End":"00:32.695","Text":"We have the slope of the tangent to the curve at A."},{"Start":"00:32.695 ","End":"00:37.490","Text":"That would be given by just y\u0027 or dy over dx."},{"Start":"00:37.490 ","End":"00:43.922","Text":"Then we have the slope of the line connecting A with the origin,"},{"Start":"00:43.922 ","End":"00:46.455","Text":"and that would be y over x."},{"Start":"00:46.455 ","End":"00:50.010","Text":"From the point x, y to the origin, this is the slope."},{"Start":"00:50.010 ","End":"00:57.320","Text":"Then we also have the y-coordinate of the point P. Since the point is just x,"},{"Start":"00:57.320 ","End":"00:59.915","Text":"y, the y-coordinate is y."},{"Start":"00:59.915 ","End":"01:02.315","Text":"Now we have the 3 quantities."},{"Start":"01:02.315 ","End":"01:09.075","Text":"What we want to say is the difference between this and this is this."},{"Start":"01:09.075 ","End":"01:15.270","Text":"We get that this minus this is this."},{"Start":"01:15.270 ","End":"01:17.180","Text":"We also have an initial condition."},{"Start":"01:17.180 ","End":"01:18.815","Text":"The fact that it passes through 2,"},{"Start":"01:18.815 ","End":"01:21.830","Text":"4 gives us that y(2) is 4."},{"Start":"01:21.830 ","End":"01:28.220","Text":"Now we have a differential equation with an initial condition and let\u0027s start solving it."},{"Start":"01:28.220 ","End":"01:34.295","Text":"This is going to be a linear equation after we move this to the other side,"},{"Start":"01:34.295 ","End":"01:37.705","Text":"and we take y outside the brackets."},{"Start":"01:37.705 ","End":"01:40.575","Text":"There we are, I\u0027ve got some more room here."},{"Start":"01:40.575 ","End":"01:46.360","Text":"If we take this function of x to be our a(x) and this is b(x),"},{"Start":"01:46.360 ","End":"01:51.830","Text":"then we have the standard equation for y\u0027 plus a(x, y) is b(x)."},{"Start":"01:51.830 ","End":"01:55.445","Text":"We can bring the standard formula that solves this,"},{"Start":"01:55.445 ","End":"01:57.485","Text":"that gives us this."},{"Start":"01:57.485 ","End":"02:01.715","Text":"A(x) is just the integral of a(x)."},{"Start":"02:01.715 ","End":"02:04.010","Text":"The first bit of the formula is standard,"},{"Start":"02:04.010 ","End":"02:08.630","Text":"the last bit is true whenever b is 0,"},{"Start":"02:08.630 ","End":"02:10.580","Text":"which it is in our case."},{"Start":"02:10.580 ","End":"02:15.725","Text":"When b is 0, this bit just disappears and this is what we get."},{"Start":"02:15.725 ","End":"02:18.950","Text":"Now, we compute A(x),"},{"Start":"02:18.950 ","End":"02:23.840","Text":"which is just the integral of a(x)."},{"Start":"02:23.840 ","End":"02:26.135","Text":"I forgot to put the minus."},{"Start":"02:26.135 ","End":"02:27.740","Text":"I just put the minus here."},{"Start":"02:27.740 ","End":"02:29.410","Text":"It\u0027s more convenient."},{"Start":"02:29.410 ","End":"02:31.295","Text":"If a(x) is this,"},{"Start":"02:31.295 ","End":"02:34.170","Text":"then minus a(x) is without the minus,"},{"Start":"02:34.170 ","End":"02:36.390","Text":"and minus A(x) is this."},{"Start":"02:36.390 ","End":"02:40.490","Text":"Because when I put it into the formula in any event I need minus A."},{"Start":"02:40.490 ","End":"02:48.580","Text":"What I ended up getting is that y is equal to c e_^-A(x),"},{"Start":"02:48.580 ","End":"02:51.965","Text":"which is natural log of x plus x."},{"Start":"02:51.965 ","End":"02:54.020","Text":"Now if I expand this,"},{"Start":"02:54.020 ","End":"02:56.750","Text":"what I get is e^x,"},{"Start":"02:56.750 ","End":"02:58.910","Text":"e^ log of x."},{"Start":"02:58.910 ","End":"03:04.715","Text":"But e^ log of x is just x because e and natural log are inverse."},{"Start":"03:04.715 ","End":"03:10.320","Text":"I get y equals cxe^x."},{"Start":"03:10.900 ","End":"03:16.140","Text":"Because I have an initial condition that it passes through 2,"},{"Start":"03:16.140 ","End":"03:17.970","Text":"4, that when x is 2,"},{"Start":"03:17.970 ","End":"03:19.140","Text":"y is 4,"},{"Start":"03:19.140 ","End":"03:22.320","Text":"then if I put x=2 here and here,"},{"Start":"03:22.320 ","End":"03:23.880","Text":"and y is 4,"},{"Start":"03:23.880 ","End":"03:30.090","Text":"I end up getting that c is equal to 4e_^-2."},{"Start":"03:30.090 ","End":"03:33.635","Text":"If I put that into the equation here,"},{"Start":"03:33.635 ","End":"03:38.975","Text":"then I get 2e_ ^-(2x)e^x."},{"Start":"03:38.975 ","End":"03:45.800","Text":"I can just combine e_^-2 with e^x and get e^x minus 2."},{"Start":"03:45.800 ","End":"03:48.155","Text":"This is our answer."},{"Start":"03:48.155 ","End":"03:50.400","Text":"This is the solution."}],"ID":7836},{"Watched":false,"Name":"Exercise 7","Duration":"2m 54s","ChapterTopicVideoID":7763,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.815","Text":"Here we have another word problem in which turns into a differential equation."},{"Start":"00:04.815 ","End":"00:09.090","Text":"Find the equation of the curve that passes through the origin and which is"},{"Start":"00:09.090 ","End":"00:15.870","Text":"perpendicular to each line connecting a point on the curve to the point 3, 4."},{"Start":"00:15.870 ","End":"00:19.919","Text":"First thing I want to say is that lines are perpendicular,"},{"Start":"00:19.919 ","End":"00:22.710","Text":"2 curves are mutually perpendicular."},{"Start":"00:22.710 ","End":"00:25.245","Text":"The product of the slopes is minus 1."},{"Start":"00:25.245 ","End":"00:26.930","Text":"Second thing I want to say is,"},{"Start":"00:26.930 ","End":"00:34.670","Text":"what is the slope of the lines connecting a point to 3, 4?"},{"Start":"00:34.670 ","End":"00:39.200","Text":"Well, that\u0027s just the difference of the y\u0027s over the difference in the x\u0027s,"},{"Start":"00:39.200 ","End":"00:41.690","Text":"rise over run, as we say,"},{"Start":"00:41.690 ","End":"00:45.215","Text":"and that\u0027s y minus 4 over x minus 3."},{"Start":"00:45.215 ","End":"00:47.450","Text":"Now, at the point x, y,"},{"Start":"00:47.450 ","End":"00:50.330","Text":"the curve is perpendicular to the line."},{"Start":"00:50.330 ","End":"00:54.425","Text":"It must be that the product of the slopes is minus 1."},{"Start":"00:54.425 ","End":"00:57.365","Text":"Instead of dy over dx,"},{"Start":"00:57.365 ","End":"01:00.725","Text":"basically what we get is dx over dy."},{"Start":"01:00.725 ","End":"01:04.760","Text":"We get that this slope is minus dx"},{"Start":"01:04.760 ","End":"01:08.945","Text":"over dy for the curve and then it\u0027s going to be perpendicular."},{"Start":"01:08.945 ","End":"01:11.060","Text":"Besides this differential equation,"},{"Start":"01:11.060 ","End":"01:14.285","Text":"we also have an initial condition because it says here,"},{"Start":"01:14.285 ","End":"01:16.025","Text":"passes through the origin,"},{"Start":"01:16.025 ","End":"01:17.940","Text":"which means that y(0) is 0."},{"Start":"01:17.940 ","End":"01:21.620","Text":"Here\u0027s a differential equation with an initial condition and"},{"Start":"01:21.620 ","End":"01:26.170","Text":"it looks like we\u0027re going to do it by separation of values."},{"Start":"01:26.170 ","End":"01:30.055","Text":"Continuing, just give myself some more room here,"},{"Start":"01:30.055 ","End":"01:38.700","Text":"we get that y minus 4 times dy is equal to x minus 3 dx."},{"Start":"01:38.700 ","End":"01:44.375","Text":"Then we can put an integral sign in front of each of them."},{"Start":"01:44.375 ","End":"01:46.290","Text":"Here\u0027s the integral."},{"Start":"01:46.290 ","End":"01:48.725","Text":"These are both easy integrals."},{"Start":"01:48.725 ","End":"01:52.435","Text":"This one is y minus 4^2 over 2,"},{"Start":"01:52.435 ","End":"01:55.630","Text":"and this one is x minus 3^2 over 2,"},{"Start":"01:55.630 ","End":"01:59.015","Text":"but with the minus in front and the constant of integration."},{"Start":"01:59.015 ","End":"02:03.620","Text":"Multiply both sides by 2 and this is what we get, 2c,"},{"Start":"02:03.620 ","End":"02:07.160","Text":"we rename as k. Now we got to remember"},{"Start":"02:07.160 ","End":"02:11.300","Text":"that we have an initial condition that when x is 0, y is 0."},{"Start":"02:11.300 ","End":"02:13.340","Text":"Plug in in 0,"},{"Start":"02:13.340 ","End":"02:15.050","Text":"0 here and we get,"},{"Start":"02:15.050 ","End":"02:17.500","Text":"minus 4^2 is 16,"},{"Start":"02:17.500 ","End":"02:20.690","Text":"minus 9 plus k equals,"},{"Start":"02:20.690 ","End":"02:23.495","Text":"so that gives us k is 25."},{"Start":"02:23.495 ","End":"02:26.795","Text":"Put k equals 25 back here,"},{"Start":"02:26.795 ","End":"02:34.250","Text":"and we get y minus 4^2 equals minus x minus 3^2 plus k is 25."},{"Start":"02:34.250 ","End":"02:37.915","Text":"In this case, we could isolate y"},{"Start":"02:37.915 ","End":"02:42.590","Text":"by taking the square roots of both sides and then adding 4."},{"Start":"02:42.590 ","End":"02:46.880","Text":"What we get is that y is 4 plus or minus the square root,"},{"Start":"02:46.880 ","End":"02:49.370","Text":"which basically gives us 2 solutions,"},{"Start":"02:49.370 ","End":"02:54.840","Text":"one with the plus and one with the minus. We are done."}],"ID":7837},{"Watched":false,"Name":"Exercise 8","Duration":"4m 30s","ChapterTopicVideoID":7764,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.175","Text":"Here we have another word problem,"},{"Start":"00:02.175 ","End":"00:07.185","Text":"and it\u0027s best to look at the diagram while reading the description."},{"Start":"00:07.185 ","End":"00:10.455","Text":"The area S is bounded by the curve,"},{"Start":"00:10.455 ","End":"00:13.980","Text":"so S is obviously the area here,"},{"Start":"00:13.980 ","End":"00:15.525","Text":"though it isn\u0027t labeled."},{"Start":"00:15.525 ","End":"00:17.010","Text":"Maybe I should label it."},{"Start":"00:17.010 ","End":"00:23.910","Text":"That\u0027s S here by the curve y equals y(x) and the lines x=a."},{"Start":"00:23.910 ","End":"00:27.080","Text":"Now notice x equals x. I mean just in general,"},{"Start":"00:27.080 ","End":"00:32.210","Text":"x between a and x and the x-axis and the curve."},{"Start":"00:32.210 ","End":"00:35.885","Text":"We have this bound S. What we know is that S"},{"Start":"00:35.885 ","End":"00:39.805","Text":"is proportional to the arc length between the point,"},{"Start":"00:39.805 ","End":"00:42.740","Text":"this is a y of a and this is x, y(x)."},{"Start":"00:42.740 ","End":"00:44.930","Text":"In other words, as x progresses,"},{"Start":"00:44.930 ","End":"00:48.725","Text":"the area S is proportional to the arc length,"},{"Start":"00:48.725 ","End":"00:55.130","Text":"call it l and this is going to give us our equation because we know what the area is."},{"Start":"00:55.130 ","End":"00:57.365","Text":"The area is the integral."},{"Start":"00:57.365 ","End":"01:03.950","Text":"The area is given by the integral from a to x of the function and the arc length,"},{"Start":"01:03.950 ","End":"01:06.230","Text":"l, is given by a different formula."},{"Start":"01:06.230 ","End":"01:09.350","Text":"It\u0027s the square root of 1 plus the derivative squared,"},{"Start":"01:09.350 ","End":"01:10.945","Text":"the integral of that."},{"Start":"01:10.945 ","End":"01:14.230","Text":"If S is proportional to l,"},{"Start":"01:14.230 ","End":"01:17.569","Text":"that means there\u0027s some constant of proportionality k,"},{"Start":"01:17.569 ","End":"01:23.645","Text":"such that S is k times l. I don\u0027t need the diagram anymore."},{"Start":"01:23.645 ","End":"01:28.345","Text":"Now, let\u0027s write that in terms of the integrals S is this,"},{"Start":"01:28.345 ","End":"01:30.150","Text":"and l is this,"},{"Start":"01:30.150 ","End":"01:33.090","Text":"so this equals k times this."},{"Start":"01:33.090 ","End":"01:37.040","Text":"Now the delicate part, the integral,"},{"Start":"01:37.040 ","End":"01:39.215","Text":"although it\u0027s a definite integral,"},{"Start":"01:39.215 ","End":"01:41.795","Text":"the upper limit is a variable,"},{"Start":"01:41.795 ","End":"01:45.260","Text":"so ends up getting a function of x and actually I get a"},{"Start":"01:45.260 ","End":"01:48.680","Text":"primitive of the function y(x) in this case,"},{"Start":"01:48.680 ","End":"01:51.515","Text":"on the primitive of this function in the other case."},{"Start":"01:51.515 ","End":"01:55.640","Text":"If I differentiate, I should get back to the function itself."},{"Start":"01:55.640 ","End":"02:00.455","Text":"Again, if I take the integral with an upper limit variable x,"},{"Start":"02:00.455 ","End":"02:03.830","Text":"then I get a primitive of y(x) here,"},{"Start":"02:03.830 ","End":"02:09.215","Text":"and I get a primitive of the square root of 1 plus y\u0027^2 over here."},{"Start":"02:09.215 ","End":"02:11.240","Text":"If I differentiate the primitive,"},{"Start":"02:11.240 ","End":"02:12.950","Text":"I get back to the function itself."},{"Start":"02:12.950 ","End":"02:16.430","Text":"In other words, what I\u0027m saying is that we get, well,"},{"Start":"02:16.430 ","End":"02:20.600","Text":"first of all, I\u0027ll just write that this is, I\u0027m differentiating."},{"Start":"02:20.600 ","End":"02:24.320","Text":"I should be a bit clearer that this is derivative should be at an angle,"},{"Start":"02:24.320 ","End":"02:25.535","Text":"let\u0027s say like this."},{"Start":"02:25.535 ","End":"02:28.250","Text":"This is the derivative with respect to x."},{"Start":"02:28.250 ","End":"02:32.970","Text":"That brings me back to the original y(x) here,"},{"Start":"02:32.970 ","End":"02:35.189","Text":"which I\u0027ll just call y."},{"Start":"02:35.189 ","End":"02:41.100","Text":"Here I get the square root of 1 plus y\u0027^2 and just drop the integral,"},{"Start":"02:41.100 ","End":"02:43.775","Text":"and this is what we get basically."},{"Start":"02:43.775 ","End":"02:46.445","Text":"If I square both sides,"},{"Start":"02:46.445 ","End":"02:56.070","Text":"I will get that y^2 is k squared times 1 plus y\u0027^2 and I\u0027d like to isolate y\u0027."},{"Start":"02:56.070 ","End":"03:00.290","Text":"If I bring the k^2 y\u0027^2 to"},{"Start":"03:00.290 ","End":"03:04.940","Text":"the other side and divide by k^2 and take the square root in other words,"},{"Start":"03:04.940 ","End":"03:06.290","Text":"if you do the algebra,"},{"Start":"03:06.290 ","End":"03:12.860","Text":"you end up with plus or minus 1 over k square root of y^2 minus k^2."},{"Start":"03:12.860 ","End":"03:14.540","Text":"But we\u0027re not done yet."},{"Start":"03:14.540 ","End":"03:18.325","Text":"I\u0027m going to replace y\u0027 by dy over dx."},{"Start":"03:18.325 ","End":"03:20.644","Text":"Here we are in a fresh page."},{"Start":"03:20.644 ","End":"03:23.615","Text":"I want to separate the variables here."},{"Start":"03:23.615 ","End":"03:29.315","Text":"What I\u0027m going to do is take the stuff with y and dy on the left,"},{"Start":"03:29.315 ","End":"03:32.770","Text":"and we end up with this."},{"Start":"03:32.770 ","End":"03:37.910","Text":"Now we take the integral of each side, the integral of this,"},{"Start":"03:37.910 ","End":"03:39.110","Text":"the integral of this,"},{"Start":"03:39.110 ","End":"03:43.625","Text":"this we look up in the integral table."},{"Start":"03:43.625 ","End":"03:48.580","Text":"It\u0027s one of those which is the arc cosine hyperbolic though."},{"Start":"03:48.580 ","End":"03:51.755","Text":"We have the arc cosine hyperbolic,"},{"Start":"03:51.755 ","End":"03:57.470","Text":"or the inverse hyperbolic cosine of y over k is equal to plus or"},{"Start":"03:57.470 ","End":"04:03.935","Text":"minus 1 over k stays integral of d x is just x plus the constant of integration."},{"Start":"04:03.935 ","End":"04:08.270","Text":"Since arc cosine hyperbolic is the inverse of cosine hyperbolic,"},{"Start":"04:08.270 ","End":"04:10.750","Text":"I can bring this to the other side."},{"Start":"04:10.750 ","End":"04:13.550","Text":"It was I just take the cosine hyperbolic of both sides."},{"Start":"04:13.550 ","End":"04:16.865","Text":"I\u0027m left with y over k is cosine hyperbolic of this,"},{"Start":"04:16.865 ","End":"04:18.575","Text":"and then I multiply it by k,"},{"Start":"04:18.575 ","End":"04:23.870","Text":"ensure to believe algebra trigonometry brings us to this solution and"},{"Start":"04:23.870 ","End":"04:30.410","Text":"this is the answer. We\u0027re done."}],"ID":7838},{"Watched":false,"Name":"Exercise 9","Duration":"1m 27s","ChapterTopicVideoID":7765,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.150","Text":"In this exercise, we have to find the family of curves which is"},{"Start":"00:03.150 ","End":"00:06.810","Text":"orthogonal to this family, x plus 2y=c."},{"Start":"00:06.810 ","End":"00:10.440","Text":"It\u0027s a family because c is a parameter, so it can change."},{"Start":"00:10.440 ","End":"00:14.940","Text":"What we want to do is find the differential equations satisfied by members of the family"},{"Start":"00:14.940 ","End":"00:21.070","Text":"and get rid of c. The easiest thing to do is just differentiate implicitly."},{"Start":"00:21.770 ","End":"00:24.465","Text":"In this is Part 1 of the exercise,"},{"Start":"00:24.465 ","End":"00:33.435","Text":"we get 1 plus 2y\u0027=0 from the c. Let\u0027s change y\u0027 to d over dx."},{"Start":"00:33.435 ","End":"00:35.070","Text":"This is what we get."},{"Start":"00:35.070 ","End":"00:37.500","Text":"This is the differential equation for this family."},{"Start":"00:37.500 ","End":"00:39.344","Text":"Now for the orthogonal family,"},{"Start":"00:39.344 ","End":"00:44.300","Text":"we replace dy over dx by its negative reciprocal,"},{"Start":"00:44.300 ","End":"00:47.020","Text":"which is minus dx over dy."},{"Start":"00:47.020 ","End":"00:53.330","Text":"If we do that, then this equation after the replacement becomes this."},{"Start":"00:53.330 ","End":"00:56.765","Text":"This is a differential equation of the orthogonal family,"},{"Start":"00:56.765 ","End":"00:59.585","Text":"and now we have to try and solve this."},{"Start":"00:59.585 ","End":"01:02.630","Text":"This is the next stage, Stage 3."},{"Start":"01:02.630 ","End":"01:05.855","Text":"We just bring this to the other side and"},{"Start":"01:05.855 ","End":"01:10.605","Text":"then we can say that if we multiply by dy over dx,"},{"Start":"01:10.605 ","End":"01:12.535","Text":"we get this =2,"},{"Start":"01:12.535 ","End":"01:16.295","Text":"separate the variables and dy is 2dx."},{"Start":"01:16.295 ","End":"01:25.310","Text":"Take the integral of both sides and we get that y=2x plus k and that\u0027s the answer."},{"Start":"01:25.310 ","End":"01:27.630","Text":"That\u0027s all there is to it."}],"ID":7839},{"Watched":false,"Name":"Exercise 10","Duration":"1m 30s","ChapterTopicVideoID":7766,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.430","Text":"Here we have to find the family of curves orthogonal to the family xy=c."},{"Start":"00:05.430 ","End":"00:07.560","Text":"It\u0027s a family because c varies."},{"Start":"00:07.560 ","End":"00:12.260","Text":"What we want to do first is find the differential equations satisfied by the families,"},{"Start":"00:12.260 ","End":"00:14.625","Text":"so we do an implicit differentiation."},{"Start":"00:14.625 ","End":"00:19.395","Text":"We have here a product rule and the derivative of"},{"Start":"00:19.395 ","End":"00:26.265","Text":"x times y plus the derivative of the y times the x equals 0."},{"Start":"00:26.265 ","End":"00:28.065","Text":"This just gives us,"},{"Start":"00:28.065 ","End":"00:30.660","Text":"if we replace y\u0027 by dy over dx,"},{"Start":"00:30.660 ","End":"00:33.785","Text":"this is the equation we get. That was step 1."},{"Start":"00:33.785 ","End":"00:37.550","Text":"Now in step 2, we get the differential equation of the orthogonal family."},{"Start":"00:37.550 ","End":"00:43.205","Text":"We do this by replacing dy over dx by its negative reciprocal,"},{"Start":"00:43.205 ","End":"00:45.620","Text":"which is minus dx over dy."},{"Start":"00:45.620 ","End":"00:48.980","Text":"If we substitute this in here,"},{"Start":"00:48.980 ","End":"00:51.380","Text":"then this is what we get."},{"Start":"00:51.380 ","End":"00:53.770","Text":"Just simply y,"},{"Start":"00:53.770 ","End":"00:59.150","Text":"here we have minus dx over dy instead of dy over dx times x equals 0."},{"Start":"00:59.150 ","End":"01:01.540","Text":"Now we have to start solving it."},{"Start":"01:01.540 ","End":"01:04.580","Text":"Bring this to the other side and multiply by dy."},{"Start":"01:04.580 ","End":"01:07.760","Text":"You can see we\u0027re heading for separation of variables."},{"Start":"01:07.760 ","End":"01:09.140","Text":"When we get to this point,"},{"Start":"01:09.140 ","End":"01:11.480","Text":"we just take an integral in front of each side."},{"Start":"01:11.480 ","End":"01:12.920","Text":"They\u0027re easy integrals."},{"Start":"01:12.920 ","End":"01:14.660","Text":"Here it\u0027s y^2 over 2,"},{"Start":"01:14.660 ","End":"01:16.250","Text":"here it\u0027s x^2 over 2,"},{"Start":"01:16.250 ","End":"01:17.890","Text":"but we have the constant."},{"Start":"01:17.890 ","End":"01:22.515","Text":"Multiply both sides by 2 and rename the constant to k,"},{"Start":"01:22.515 ","End":"01:24.360","Text":"and then bring the x^2 over,"},{"Start":"01:24.360 ","End":"01:26.400","Text":"and this is what we get."},{"Start":"01:26.400 ","End":"01:30.390","Text":"This is the family of orthogonal curves and we\u0027re done."}],"ID":7840},{"Watched":false,"Name":"Exercise 11","Duration":"3m 23s","ChapterTopicVideoID":7767,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.950","Text":"Here we have a 2 part exercise."},{"Start":"00:01.950 ","End":"00:04.695","Text":"In the first part, we have to find the family of curves"},{"Start":"00:04.695 ","End":"00:08.760","Text":"orthogonal to the family x^2 plus 2y^2=C."},{"Start":"00:08.760 ","End":"00:12.240","Text":"It\u0027s a family because C varies. When we\u0027ve done this."},{"Start":"00:12.240 ","End":"00:15.345","Text":"Second part will be to find a specific curve,"},{"Start":"00:15.345 ","End":"00:21.705","Text":"in the family that\u0027s orthogonal to this curve in the original family at this point."},{"Start":"00:21.705 ","End":"00:23.775","Text":"Well anyway, let\u0027s start with a."},{"Start":"00:23.775 ","End":"00:26.400","Text":"In Part A, we start off by trying to find"},{"Start":"00:26.400 ","End":"00:29.220","Text":"the differential equations satisfied by this family."},{"Start":"00:29.220 ","End":"00:31.710","Text":"We do this by implicit differentiation."},{"Start":"00:31.710 ","End":"00:36.030","Text":"We get 2x plus 4y but times y\u0027,"},{"Start":"00:36.030 ","End":"00:38.425","Text":"and the C gives 0."},{"Start":"00:38.425 ","End":"00:42.485","Text":"Then we write the y\u0027 as dy over dx."},{"Start":"00:42.485 ","End":"00:44.210","Text":"This is what we get."},{"Start":"00:44.210 ","End":"00:48.695","Text":"In Part 2, we need to find the differential equation of the orthogonal family."},{"Start":"00:48.695 ","End":"00:54.240","Text":"This we do by replacing dy over dx here by minus dx over dy."},{"Start":"00:54.240 ","End":"00:56.255","Text":"If we do this replacement,"},{"Start":"00:56.255 ","End":"00:58.130","Text":"what we get is similar to this,"},{"Start":"00:58.130 ","End":"01:01.480","Text":"but with the minus and with this inverted."},{"Start":"01:01.480 ","End":"01:03.350","Text":"In Part 3,"},{"Start":"01:03.350 ","End":"01:06.005","Text":"we have to start solving this differential equation."},{"Start":"01:06.005 ","End":"01:07.520","Text":"So bring this to the other side."},{"Start":"01:07.520 ","End":"01:10.120","Text":"We\u0027re aiming for separation of variables."},{"Start":"01:10.120 ","End":"01:13.400","Text":"Now bring the x\u0027s to the left,"},{"Start":"01:13.400 ","End":"01:15.730","Text":"I think, and y\u0027s to the right."},{"Start":"01:15.730 ","End":"01:17.660","Text":"This is what we get, but of course,"},{"Start":"01:17.660 ","End":"01:21.590","Text":"we must remember the restriction y not 0, x not 0."},{"Start":"01:21.590 ","End":"01:26.615","Text":"At this point I can just stick the integral sign in front of each side."},{"Start":"01:26.615 ","End":"01:28.735","Text":"Integral equals integral."},{"Start":"01:28.735 ","End":"01:30.889","Text":"These are easy integrals."},{"Start":"01:30.889 ","End":"01:36.490","Text":"What we get here is natural log of y and here natural log of x absolute value actually."},{"Start":"01:36.490 ","End":"01:38.479","Text":"When we deal with logarithms,"},{"Start":"01:38.479 ","End":"01:41.720","Text":"we usually pre-talk constant as the logarithm of"},{"Start":"01:41.720 ","End":"01:45.440","Text":"some positive constant because any number can be the logarithm of the constant."},{"Start":"01:45.440 ","End":"01:49.775","Text":"In any event, what we get is that, just simplify,"},{"Start":"01:49.775 ","End":"01:51.920","Text":"I can put the 2 inside and then I can get rid of"},{"Start":"01:51.920 ","End":"01:55.070","Text":"the absolute value because I\u0027ll get x^2 here."},{"Start":"01:55.070 ","End":"01:57.740","Text":"Now we get rid of the logarithm,"},{"Start":"01:57.740 ","End":"02:02.020","Text":"but make the plus into a product using the rules of logarithms."},{"Start":"02:02.020 ","End":"02:03.710","Text":"Sorry, yeah, that was the next step."},{"Start":"02:03.710 ","End":"02:06.920","Text":"The first step is to combine these using the laws of logarithms."},{"Start":"02:06.920 ","End":"02:08.435","Text":"So plus becomes a product,"},{"Start":"02:08.435 ","End":"02:15.655","Text":"and then we can drop the logarithm and get that absolute value of y is Cx^2."},{"Start":"02:15.655 ","End":"02:18.950","Text":"Well, this possibly gives 2 solutions."},{"Start":"02:18.950 ","End":"02:21.220","Text":"This could be y or minus y."},{"Start":"02:21.220 ","End":"02:25.190","Text":"We either have y is Cx^2 or y is minus Cx^2."},{"Start":"02:25.190 ","End":"02:27.440","Text":"But depending on y positive or negative."},{"Start":"02:27.440 ","End":"02:31.220","Text":"But if we just reverse the sign of the C in the other case,"},{"Start":"02:31.220 ","End":"02:32.870","Text":"we get that y is ax^2,"},{"Start":"02:32.870 ","End":"02:36.455","Text":"while a could be positive or negative and that covers both of these."},{"Start":"02:36.455 ","End":"02:38.600","Text":"Maybe think about it, you\u0027ll see this so."},{"Start":"02:38.600 ","End":"02:41.390","Text":"That\u0027s only answered Part A of the question."},{"Start":"02:41.390 ","End":"02:42.785","Text":"Let\u0027s get onto Part B,"},{"Start":"02:42.785 ","End":"02:44.155","Text":"it\u0027s on the next page."},{"Start":"02:44.155 ","End":"02:50.870","Text":"If you remember in Part A we found the family to be y equals ax^2."},{"Start":"02:50.870 ","End":"02:56.980","Text":"Now we wanted to find one of these curves that passes through 1, 2."},{"Start":"02:56.980 ","End":"03:00.840","Text":"If we put in here y equals 2 and x equals 1,"},{"Start":"03:00.840 ","End":"03:04.635","Text":"you\u0027ll see that 2 is a times 1^2 so a equals 2."},{"Start":"03:04.635 ","End":"03:06.380","Text":"So if we plug that back in here,"},{"Start":"03:06.380 ","End":"03:08.900","Text":"we get y equals 2x^2."},{"Start":"03:08.900 ","End":"03:11.615","Text":"That\u0027s the curve in this family that will be"},{"Start":"03:11.615 ","End":"03:15.005","Text":"orthogonal to the original family at this point."},{"Start":"03:15.005 ","End":"03:24.400","Text":"We are done. Except possibly to make a nice little box for the final answer, done."}],"ID":7841},{"Watched":false,"Name":"Exercise 12","Duration":"2m 10s","ChapterTopicVideoID":7768,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.415","Text":"In this exercise, we have a family of curves, x plus 2_y=c."},{"Start":"00:05.415 ","End":"00:08.445","Text":"As c varies, we get different curves."},{"Start":"00:08.445 ","End":"00:14.205","Text":"We want to find another family which is orthogonal to this family."},{"Start":"00:14.205 ","End":"00:20.339","Text":"Every curve of the new family will intersect every curve of the old family at 90 degrees."},{"Start":"00:20.339 ","End":"00:22.515","Text":"We do it in 3 steps."},{"Start":"00:22.515 ","End":"00:27.180","Text":"The first step is to differentiate the family,"},{"Start":"00:27.180 ","End":"00:30.360","Text":"but we use implicit differentiation."},{"Start":"00:30.360 ","End":"00:32.700","Text":"The derivative of x is 1,"},{"Start":"00:32.700 ","End":"00:37.120","Text":"the derivative of 2y because the function of y is not just 2,"},{"Start":"00:37.120 ","End":"00:42.500","Text":"but we have to multiply it by y\u0027 and the constant goes to 0."},{"Start":"00:42.500 ","End":"00:43.850","Text":"For this exercise,"},{"Start":"00:43.850 ","End":"00:45.635","Text":"you want the other notation."},{"Start":"00:45.635 ","End":"00:50.470","Text":"We want to put it in dy by dx notation instead of y\u0027."},{"Start":"00:50.470 ","End":"00:54.245","Text":"Now here\u0027s the key thing for moving to the orthogonal."},{"Start":"00:54.245 ","End":"01:00.065","Text":"We replace dy by dx by minus dx by dy."},{"Start":"01:00.065 ","End":"01:04.250","Text":"That\u0027s the negative reciprocal and that\u0027s because lines are"},{"Start":"01:04.250 ","End":"01:10.235","Text":"orthogonal if and only if their slopes are negative reciprocals of each other."},{"Start":"01:10.235 ","End":"01:16.730","Text":"Replace dy by dx here by minus dx by dy and this is what we get,"},{"Start":"01:16.730 ","End":"01:19.565","Text":"this inverts as a minus in front."},{"Start":"01:19.565 ","End":"01:26.850","Text":"Finally, we want to solve this differential equation and get a new family of solutions,"},{"Start":"01:26.850 ","End":"01:30.645","Text":"so we\u0027ll use separation of variables."},{"Start":"01:30.645 ","End":"01:35.150","Text":"Bringing the whole dx by dy to the other side and invert it."},{"Start":"01:35.150 ","End":"01:37.670","Text":"Actually, that\u0027s a wasted step because I could have"},{"Start":"01:37.670 ","End":"01:41.540","Text":"straightaway just multiplied by dy, Oh, never mind."},{"Start":"01:41.540 ","End":"01:43.735","Text":"Anyway, now we have this."},{"Start":"01:43.735 ","End":"01:45.410","Text":"When we have this thing,"},{"Start":"01:45.410 ","End":"01:48.350","Text":"we just stick an integral sign in front of each side of"},{"Start":"01:48.350 ","End":"01:51.680","Text":"the equation and perform an integral."},{"Start":"01:51.680 ","End":"01:56.190","Text":"The integral of 1dy is just y."},{"Start":"01:56.190 ","End":"01:59.560","Text":"The integral of 2dx is 2x,"},{"Start":"01:59.560 ","End":"02:02.000","Text":"and there\u0027s always the constant of integration."},{"Start":"02:02.000 ","End":"02:05.195","Text":"I didn\u0027t use the same letter c called this one k."},{"Start":"02:05.195 ","End":"02:10.120","Text":"This is a family of orthogonal curves, and we\u0027re done."}],"ID":7842},{"Watched":false,"Name":"Exercise 13","Duration":"5m 45s","ChapterTopicVideoID":7769,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.380","Text":"In this exercise we have to find a family of curves which forms"},{"Start":"00:04.380 ","End":"00:09.870","Text":"a 45-degree angle with a given family which is x^2 plus y^2=c."},{"Start":"00:09.870 ","End":"00:14.085","Text":"Let\u0027s first of all find the differential equation of the given family."},{"Start":"00:14.085 ","End":"00:16.058","Text":"It\u0027s the first part of the question,"},{"Start":"00:16.058 ","End":"00:18.570","Text":"and what we do is implicit differentiation."},{"Start":"00:18.570 ","End":"00:21.480","Text":"Here we get 2x, here 2y times y\u0027,"},{"Start":"00:21.480 ","End":"00:23.490","Text":"and the c gives us nothing."},{"Start":"00:23.490 ","End":"00:25.560","Text":"Just divide by 2,"},{"Start":"00:25.560 ","End":"00:32.325","Text":"and this is the differential equation for this family where y\u0027 is the slope."},{"Start":"00:32.325 ","End":"00:38.848","Text":"Now, if we want to change the slope and move it so that It\u0027s 45 degrees;"},{"Start":"00:38.848 ","End":"00:40.400","Text":"it doesn\u0027t say to the left or right,"},{"Start":"00:40.400 ","End":"00:43.400","Text":"but have a formula from trigonometry that says that"},{"Start":"00:43.400 ","End":"00:48.529","Text":"the tangent of the difference of angles is such and such a formula."},{"Start":"00:48.529 ","End":"00:52.010","Text":"Now in our case what I\u0027m going to be doing is taking"},{"Start":"00:52.010 ","End":"00:55.762","Text":"the tangent Alpha to be our derivative,"},{"Start":"00:55.762 ","End":"00:56.875","Text":"so this is going to be y\u0027."},{"Start":"00:56.875 ","End":"01:03.920","Text":"Lets say Alpha is the angle of the slope then we want to shift it by 45 degrees,"},{"Start":"01:03.920 ","End":"01:12.889","Text":"so Theta is 45 degrees and so this will give us the slope of the new family of curves."},{"Start":"01:12.889 ","End":"01:19.640","Text":"What we get is that we have to replace the original y\u0027 which is"},{"Start":"01:19.640 ","End":"01:23.180","Text":"the slope of the first family with the slope from"},{"Start":"01:23.180 ","End":"01:27.020","Text":"the new family which is this thing moved by 45 degrees."},{"Start":"01:27.020 ","End":"01:28.250","Text":"According to the formula,"},{"Start":"01:28.250 ","End":"01:33.695","Text":"this is what we get except that in our case Theta is 45 degrees."},{"Start":"01:33.695 ","End":"01:36.110","Text":"Well, tangent of 45 degrees is 1."},{"Start":"01:36.110 ","End":"01:41.460","Text":"First of all, we replace this by tangent 45 and then replace tangent 45 by 1,"},{"Start":"01:41.460 ","End":"01:47.790","Text":"so what we\u0027re left with is that we have to replace y\u0027 by this."},{"Start":"01:47.790 ","End":"01:52.847","Text":"Then what I\u0027m going to be doing is replacing the y\u0027 that\u0027s"},{"Start":"01:52.847 ","End":"01:59.045","Text":"hear by this expression here that will give us the new family."},{"Start":"01:59.045 ","End":"02:04.070","Text":"We get x as is plus y as is instead of y\u0027,"},{"Start":"02:04.070 ","End":"02:10.950","Text":"we replace it by this expression which is the slope that rotated 45 degrees."},{"Start":"02:11.090 ","End":"02:15.300","Text":"Now multiply both sides by 1 plus y\u0027,"},{"Start":"02:15.300 ","End":"02:18.930","Text":"and this is what we get."},{"Start":"02:18.930 ","End":"02:23.060","Text":"Then we write y\u0027 as dy over dx,"},{"Start":"02:23.060 ","End":"02:24.545","Text":"so this is what we get."},{"Start":"02:24.545 ","End":"02:29.375","Text":"Next, multiply both sides by dx and open brackets,"},{"Start":"02:29.375 ","End":"02:31.670","Text":"and this is what you\u0027ll get."},{"Start":"02:31.670 ","End":"02:37.815","Text":"If you separate the dx and the dy we get this expression."},{"Start":"02:37.815 ","End":"02:42.560","Text":"As usual we let this bit be M, this bit be N,"},{"Start":"02:42.560 ","End":"02:45.995","Text":"and we go to try and show this is homogeneous"},{"Start":"02:45.995 ","End":"02:49.940","Text":"for m. Here\u0027s the computation which I\u0027m not going to get into,"},{"Start":"02:49.940 ","End":"02:53.525","Text":"but you see Lambda to the power of 1 so homogeneous of Degree 1."},{"Start":"02:53.525 ","End":"02:56.370","Text":"The N also turns out to be homogeneous of Degree 1"},{"Start":"02:56.370 ","End":"02:59.743","Text":"which means this is a homogeneous differential equation,"},{"Start":"02:59.743 ","End":"03:04.790","Text":"and that means that our usual substitution will work."},{"Start":"03:04.790 ","End":"03:09.364","Text":"If we do this substitution we\u0027re supposed to get a separable equation."},{"Start":"03:09.364 ","End":"03:12.200","Text":"Let\u0027s just remember that when we substitute back,"},{"Start":"03:12.200 ","End":"03:17.630","Text":"the v is equal to y over x. I\u0027m going to do the substitution,"},{"Start":"03:17.630 ","End":"03:19.700","Text":"I\u0027m going to continue on the next page."},{"Start":"03:19.700 ","End":"03:23.215","Text":"After the substitution we get this."},{"Start":"03:23.215 ","End":"03:27.790","Text":"There up above we had a y here and here which are replaced by vx,"},{"Start":"03:27.790 ","End":"03:30.850","Text":"and the dy was here which are replaced by this,"},{"Start":"03:30.850 ","End":"03:33.250","Text":"and then we just do some algebra."},{"Start":"03:33.250 ","End":"03:39.085","Text":"Let\u0027s open up brackets and collect dx and dv separately."},{"Start":"03:39.085 ","End":"03:45.310","Text":"Next thing to do is to simplify by taking stuff out of the brackets where we can."},{"Start":"03:45.310 ","End":"03:47.350","Text":"This thing cancels with this thing."},{"Start":"03:47.350 ","End":"03:48.670","Text":"If we take x out,"},{"Start":"03:48.670 ","End":"03:51.040","Text":"then we get 1 plus v^2 dx."},{"Start":"03:51.040 ","End":"03:52.990","Text":"Here we take x squared out,"},{"Start":"03:52.990 ","End":"03:57.705","Text":"so I then get 1 plus v. Then we divide by x,"},{"Start":"03:57.705 ","End":"04:00.225","Text":"both sides that simplifies it."},{"Start":"04:00.225 ","End":"04:09.230","Text":"I\u0027m going to leave the xs on the left,"},{"Start":"04:09.230 ","End":"04:11.070","Text":"so let\u0027s move the x over here,"},{"Start":"04:11.070 ","End":"04:12.523","Text":"the 1 plus v^2 over here."},{"Start":"04:12.523 ","End":"04:14.375","Text":"Anyway, we end up with this."},{"Start":"04:14.375 ","End":"04:21.955","Text":"This is now completely separated so that we can put an integral sign in front of each."},{"Start":"04:21.955 ","End":"04:24.110","Text":"Before we do the integral,"},{"Start":"04:24.110 ","End":"04:31.805","Text":"we should simplify the bit with the v. If we separate it,"},{"Start":"04:31.805 ","End":"04:37.720","Text":"we get 1 over 1 plus v^2."},{"Start":"04:37.720 ","End":"04:42.260","Text":"The other bit is v over 1 plus v^2."},{"Start":"04:42.260 ","End":"04:46.340","Text":"If you ignore the stuff in bold it\u0027s v over 1 plus v^2,"},{"Start":"04:46.340 ","End":"04:50.720","Text":"but because our plan is here to have"},{"Start":"04:50.720 ","End":"04:53.720","Text":"a function on the denominator with its"},{"Start":"04:53.720 ","End":"04:57.110","Text":"derivative in the numerator instead of v we prefer 2v."},{"Start":"04:57.110 ","End":"04:59.930","Text":"The 2 is artificially put in"},{"Start":"04:59.930 ","End":"05:02.960","Text":"here but of course we compensate by multiplying also by 1/2,"},{"Start":"05:02.960 ","End":"05:05.450","Text":"so we haven\u0027t changed anything."},{"Start":"05:05.450 ","End":"05:07.580","Text":"Now as for the integral,"},{"Start":"05:07.580 ","End":"05:12.260","Text":"here we have the natural log of x."},{"Start":"05:12.260 ","End":"05:16.400","Text":"Here this is an immediate thing which is arc tangent."},{"Start":"05:16.400 ","End":"05:18.812","Text":"The minus I\u0027m just putting it in front of each,"},{"Start":"05:18.812 ","End":"05:23.405","Text":"here we have the 1/2 and here\u0027s the natural log of the denominator."},{"Start":"05:23.405 ","End":"05:25.652","Text":"I don\u0027t need absolute value because it\u0027s positive."},{"Start":"05:25.652 ","End":"05:27.710","Text":"Here\u0027s where we add the constant,"},{"Start":"05:27.710 ","End":"05:30.440","Text":"then we have to substitute back."},{"Start":"05:30.440 ","End":"05:32.810","Text":"Instead of v, we put y over x,"},{"Start":"05:32.810 ","End":"05:36.110","Text":"and this is the expression we get."},{"Start":"05:36.110 ","End":"05:37.670","Text":"I think we\u0027ll just leave it at that,"},{"Start":"05:37.670 ","End":"05:40.910","Text":"we won\u0027t try and isolate 1 variable in terms of another."},{"Start":"05:40.910 ","End":"05:45.720","Text":"We\u0027ll just say that this is the answer and we are done."}],"ID":7843},{"Watched":false,"Name":"Exercise 14","Duration":"3m 23s","ChapterTopicVideoID":7770,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.090","Text":"Here we have another world problem and it\u0027s a good job."},{"Start":"00:03.090 ","End":"00:05.090","Text":"We have a diagram too."},{"Start":"00:05.090 ","End":"00:06.560","Text":"I\u0027ll explain it."},{"Start":"00:06.560 ","End":"00:08.835","Text":"At each point on a curve,"},{"Start":"00:08.835 ","End":"00:10.500","Text":"this is the curve,"},{"Start":"00:10.500 ","End":"00:12.195","Text":"looks like an ellipse."},{"Start":"00:12.195 ","End":"00:16.710","Text":"This is the typical point A, which is x,y."},{"Start":"00:16.710 ","End":"00:18.090","Text":"At each point on a curve,"},{"Start":"00:18.090 ","End":"00:19.245","Text":"that will be A,"},{"Start":"00:19.245 ","End":"00:23.475","Text":"the segment of the normal between the point on the x-axis."},{"Start":"00:23.475 ","End":"00:24.840","Text":"First, the normal,"},{"Start":"00:24.840 ","End":"00:30.530","Text":"which is perpendicular to the tangent and the segment between the point and the x-axis,"},{"Start":"00:30.530 ","End":"00:35.315","Text":"that would be the segment AC is bisected by the y-axis."},{"Start":"00:35.315 ","End":"00:40.190","Text":"Here it cuts the y-axis at the point B and bisected means the distance from A to"},{"Start":"00:40.190 ","End":"00:46.355","Text":"B is the same as the distance from B to C. We have to find the equation of this curve."},{"Start":"00:46.355 ","End":"00:49.430","Text":"Let\u0027s just scroll down a bit."},{"Start":"00:49.430 ","End":"00:55.350","Text":"First thing is that the equation of the normal at the point A"},{"Start":"00:55.350 ","End":"01:01.550","Text":"is given by y minus y_1 equals the slope times x minus x_1."},{"Start":"01:01.550 ","End":"01:05.600","Text":"Now the slope is not of the curve because that would be dy over dx."},{"Start":"01:05.600 ","End":"01:07.025","Text":"It\u0027s a slope with a normal."},{"Start":"01:07.025 ","End":"01:09.275","Text":"That\u0027s why it\u0027s a negative reciprocal."},{"Start":"01:09.275 ","End":"01:13.610","Text":"The next thing is that the intersection with the y-axis,"},{"Start":"01:13.610 ","End":"01:16.249","Text":"at the y-axis, x is 0."},{"Start":"01:16.249 ","End":"01:19.925","Text":"We\u0027re talking about x_1, y_1 here."},{"Start":"01:19.925 ","End":"01:22.055","Text":"If we put x_1=0,"},{"Start":"01:22.055 ","End":"01:25.140","Text":"and you isolate y_1, then you\u0027ll get this."},{"Start":"01:25.140 ","End":"01:28.620","Text":"Here\u0027s x_1, here\u0027s y_1 at this point,"},{"Start":"01:28.620 ","End":"01:30.570","Text":"again, in terms of a general x,"},{"Start":"01:30.570 ","End":"01:34.790","Text":"y, and the intersection with the x-axis,"},{"Start":"01:34.790 ","End":"01:38.465","Text":"that\u0027s to find C that we get by putting y=0."},{"Start":"01:38.465 ","End":"01:41.300","Text":"Here\u0027s y_1, that is, in our case."},{"Start":"01:41.300 ","End":"01:43.655","Text":"If y_1 is 0, this is 0."},{"Start":"01:43.655 ","End":"01:49.440","Text":"What you get if you isolate x_1 is this here if you can do the computation."},{"Start":"01:49.440 ","End":"01:52.340","Text":"Now we have everything pretty much."},{"Start":"01:52.340 ","End":"01:56.510","Text":"We just haven\u0027t produced the condition that this is the bisector,"},{"Start":"01:56.510 ","End":"01:58.490","Text":"that this B is midway."},{"Start":"01:58.490 ","End":"02:03.710","Text":"All that we need to do is say that the x coordinate of B is"},{"Start":"02:03.710 ","End":"02:06.740","Text":"halfway between the x coordinates"},{"Start":"02:06.740 ","End":"02:10.100","Text":"of A and C. That\u0027s what we\u0027re going to do for our equation,"},{"Start":"02:10.100 ","End":"02:14.390","Text":"is to say that the x(B) is the average,"},{"Start":"02:14.390 ","End":"02:17.665","Text":"is the halfway point between the x(A) and the x(C)."},{"Start":"02:17.665 ","End":"02:19.505","Text":"Again, let\u0027s look at the picture."},{"Start":"02:19.505 ","End":"02:22.130","Text":"To say that B is halfway,"},{"Start":"02:22.130 ","End":"02:27.920","Text":"it\u0027s enough to say that the x of this is the average of the x of this and the x of this."},{"Start":"02:27.920 ","End":"02:29.870","Text":"That\u0027s what we\u0027re saying here."},{"Start":"02:29.870 ","End":"02:31.535","Text":"Now let\u0027s spell it out."},{"Start":"02:31.535 ","End":"02:38.660","Text":"So x_B is 0 because the B is on the y-axis,"},{"Start":"02:38.660 ","End":"02:40.600","Text":"so it\u0027s x is 0 and this is half."},{"Start":"02:40.600 ","End":"02:44.770","Text":"This is what we computed for x_A and this is what we computed for x_C."},{"Start":"02:44.770 ","End":"02:48.200","Text":"Opening it up, multiplying by 2 and everything,"},{"Start":"02:48.200 ","End":"02:50.570","Text":"we eventually get to the separated,"},{"Start":"02:50.570 ","End":"02:54.395","Text":"which is this, put the integral sign in front,"},{"Start":"02:54.395 ","End":"02:56.420","Text":"do the integration for 2x,"},{"Start":"02:56.420 ","End":"03:02.960","Text":"it\u0027s x^2 and for minus y it\u0027s minus y^2 over 2 or we can put it as a decimal,"},{"Start":"03:02.960 ","End":"03:04.915","Text":"a 1/2 plus C,"},{"Start":"03:04.915 ","End":"03:08.240","Text":"and bring this over to the other side,"},{"Start":"03:08.240 ","End":"03:09.605","Text":"multiply by 2,"},{"Start":"03:09.605 ","End":"03:16.550","Text":"change 2C to the letter k. This is the solution which is actually a family of curves."},{"Start":"03:16.550 ","End":"03:19.550","Text":"I can tell by looking at it that it really is an ellipse,"},{"Start":"03:19.550 ","End":"03:23.880","Text":"just like it looked to us. Done."}],"ID":7844},{"Watched":false,"Name":"Exercise 15","Duration":"5m 53s","ChapterTopicVideoID":7771,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.015","Text":"Here\u0027s an exercise with a word problem."},{"Start":"00:03.015 ","End":"00:08.430","Text":"Notice here it says illustrate the problem with a sketch in the first quadrant"},{"Start":"00:08.430 ","End":"00:10.290","Text":"and I\u0027ve actually started with"},{"Start":"00:10.290 ","End":"00:14.010","Text":"the sketch because otherwise it\u0027s hard to make sense of it."},{"Start":"00:14.010 ","End":"00:16.920","Text":"But later on your own you should practice taking"},{"Start":"00:16.920 ","End":"00:20.535","Text":"written description and converting it into a sketch,"},{"Start":"00:20.535 ","End":"00:22.410","Text":"a figure on your own."},{"Start":"00:22.410 ","End":"00:27.270","Text":"Anyway, we have to find the equation of the curve and the curve here"},{"Start":"00:27.270 ","End":"00:33.090","Text":"is marked in this black thin line passing through 0, 1."},{"Start":"00:33.090 ","End":"00:34.530","Text":"That\u0027s our point a,"},{"Start":"00:34.530 ","End":"00:36.087","Text":"which is 0, 1,"},{"Start":"00:36.087 ","End":"00:39.200","Text":"and such at the triangle bounded by the y-axis,"},{"Start":"00:39.200 ","End":"00:40.459","Text":"this is the y-axis,"},{"Start":"00:40.459 ","End":"00:42.230","Text":"the tangent to the curve at"},{"Start":"00:42.230 ","End":"00:47.555","Text":"any point M. Here\u0027s point M on the curve and here\u0027s the tangent,"},{"Start":"00:47.555 ","End":"00:49.940","Text":"which is red, then blue, then red,"},{"Start":"00:49.940 ","End":"00:56.960","Text":"and the segment OM from the origin to M. That gives us 3 sides of a triangle,"},{"Start":"00:56.960 ","End":"00:58.610","Text":"the one marked in blue,"},{"Start":"00:58.610 ","End":"01:00.140","Text":"this is the bit on the y-axis."},{"Start":"01:00.140 ","End":"01:03.859","Text":"This is the segment from O to M and this is part of the tangent."},{"Start":"01:03.859 ","End":"01:07.010","Text":"Now we\u0027re given that this is an isosceles triangle,"},{"Start":"01:07.010 ","End":"01:09.875","Text":"where the base is a segment MN,"},{"Start":"01:09.875 ","End":"01:12.365","Text":"which means that if this is the base,"},{"Start":"01:12.365 ","End":"01:14.569","Text":"then these are the 2 equal sides."},{"Start":"01:14.569 ","End":"01:16.730","Text":"Let me just indicate that."},{"Start":"01:16.730 ","End":"01:22.990","Text":"This side here and this side here are equal because MM is the base or MN."},{"Start":"01:22.990 ","End":"01:29.270","Text":"What we have to do is find the length of this and the length of this and compare them."},{"Start":"01:29.270 ","End":"01:31.160","Text":"I added some information here."},{"Start":"01:31.160 ","End":"01:32.525","Text":"Let me explain."},{"Start":"01:32.525 ","End":"01:36.725","Text":"The length of this is just square root of x^ plus y^2."},{"Start":"01:36.725 ","End":"01:38.945","Text":"That\u0027s Pythagoras\u0027 theorem."},{"Start":"01:38.945 ","End":"01:43.145","Text":"Little bit harder is to figure out the length of this side."},{"Start":"01:43.145 ","End":"01:47.185","Text":"But if we draw horizontal dotted line from here to here,"},{"Start":"01:47.185 ","End":"01:54.800","Text":"it\u0027s y and from here to here we just use the rise over run equals slope formula."},{"Start":"01:54.800 ","End":"01:58.490","Text":"This bit here over this bit here is the slope."},{"Start":"01:58.490 ","End":"02:02.630","Text":"This bit here is the slope times the run."},{"Start":"02:02.630 ","End":"02:07.460","Text":"The run is minus x because we\u0027re going from here to here and the slope,"},{"Start":"02:07.460 ","End":"02:09.440","Text":"which in this case is negative,"},{"Start":"02:09.440 ","End":"02:11.330","Text":"is just dy over dx."},{"Start":"02:11.330 ","End":"02:16.009","Text":"We get the run times slope equals rise."},{"Start":"02:16.009 ","End":"02:17.540","Text":"That\u0027s just from here to here."},{"Start":"02:17.540 ","End":"02:20.030","Text":"If we add together this thing,"},{"Start":"02:20.030 ","End":"02:22.590","Text":"including the minus with y, we get this."},{"Start":"02:22.590 ","End":"02:29.195","Text":"These are the 2 things that we need to compare when we say that these 2 sides are equal."},{"Start":"02:29.195 ","End":"02:32.120","Text":"That gives us our differential equation and"},{"Start":"02:32.120 ","End":"02:36.215","Text":"the initial condition is given by the fact that the point 0,"},{"Start":"02:36.215 ","End":"02:38.195","Text":"1 is on the curve."},{"Start":"02:38.195 ","End":"02:41.005","Text":"Scrolling down a bit."},{"Start":"02:41.005 ","End":"02:48.666","Text":"Here\u0027s the equation that this equals this with condition that it goes through 0,"},{"Start":"02:48.666 ","End":"02:50.390","Text":"1, which is this."},{"Start":"02:50.390 ","End":"02:53.900","Text":"Now we can ignore the diagram because we\u0027ve got"},{"Start":"02:53.900 ","End":"02:57.410","Text":"all that we need here to solve differential equation."},{"Start":"02:57.410 ","End":"03:05.495","Text":"I can multiply both sides by dx and bring everything to the left-hand side."},{"Start":"03:05.495 ","End":"03:10.220","Text":"Now I call this part M and this part N and it\u0027s"},{"Start":"03:10.220 ","End":"03:14.810","Text":"easy to check that each of these 2 functions is homogeneous of degree 1."},{"Start":"03:14.810 ","End":"03:18.020","Text":"That makes this a homogeneous differential equation."},{"Start":"03:18.020 ","End":"03:22.485","Text":"What we do with that is we have our usual substitution that y equals"},{"Start":"03:22.485 ","End":"03:28.610","Text":"vx and dy from the product rule is this."},{"Start":"03:28.610 ","End":"03:35.330","Text":"Let\u0027s not forget that at the end we need to substitute back that v is y over x."},{"Start":"03:35.330 ","End":"03:37.880","Text":"Now once we make the substitution and I\u0027m"},{"Start":"03:37.880 ","End":"03:40.670","Text":"doing this quickly because you\u0027ve seen these things before,"},{"Start":"03:40.670 ","End":"03:49.695","Text":"just replacing y by vx in this equation and so on and dy by this, we get this."},{"Start":"03:49.695 ","End":"03:54.050","Text":"The next step is to separate the variables."},{"Start":"03:54.050 ","End":"03:57.170","Text":"Just get vs on the right with dv,"},{"Start":"03:57.170 ","End":"03:58.550","Text":"xs on the left with dx,"},{"Start":"03:58.550 ","End":"03:59.690","Text":"a bit of algebra here."},{"Start":"03:59.690 ","End":"04:01.265","Text":"I\u0027m not getting into that."},{"Start":"04:01.265 ","End":"04:05.875","Text":"Then we can put the integral sign in front of each."},{"Start":"04:05.875 ","End":"04:11.050","Text":"We can do the integration that minus natural log of x."},{"Start":"04:11.050 ","End":"04:13.220","Text":"This is also an immediate exactly."},{"Start":"04:13.220 ","End":"04:16.880","Text":"It\u0027s a table of integrals lookup, it\u0027s a Standard 1."},{"Start":"04:16.880 ","End":"04:22.040","Text":"The integral happens to be this and c is our constant of integration."},{"Start":"04:22.040 ","End":"04:26.545","Text":"But we usually write the c as natural log of something positive."},{"Start":"04:26.545 ","End":"04:28.185","Text":"Done it on the next page."},{"Start":"04:28.185 ","End":"04:32.060","Text":"Let\u0027s see what\u0027s natural log of k. We\u0027re up to here now."},{"Start":"04:32.060 ","End":"04:34.820","Text":"Now let\u0027s use the rules of the logarithm."},{"Start":"04:34.820 ","End":"04:37.461","Text":"This is the logarithm of this times this."},{"Start":"04:37.461 ","End":"04:39.985","Text":"The plus, we can multiply these 2."},{"Start":"04:39.985 ","End":"04:44.360","Text":"The natural log of x when it\u0027s minus is natural log of 1 over x."},{"Start":"04:44.360 ","End":"04:48.470","Text":"If we do all that and throw out the logarithm, we\u0027ll get this."},{"Start":"04:48.470 ","End":"04:51.255","Text":"We had natural log of this equals natural log of this and"},{"Start":"04:51.255 ","End":"04:55.100","Text":"then 2 numbers have equal logs and the numbers are equal."},{"Start":"04:55.100 ","End":"04:59.885","Text":"Just condensing a bit here because we\u0027ve done this thing before."},{"Start":"04:59.885 ","End":"05:04.730","Text":"Now we remember to replace v here and here by y over x."},{"Start":"05:04.730 ","End":"05:06.740","Text":"This is what we get."},{"Start":"05:06.740 ","End":"05:14.660","Text":"Then a bit of simplification to take 1 over x outside the square root and it comes out."},{"Start":"05:14.660 ","End":"05:16.445","Text":"Anyway, this is what we get,"},{"Start":"05:16.445 ","End":"05:19.110","Text":"multiply by x and I get rid of the fractions."},{"Start":"05:19.110 ","End":"05:26.210","Text":"Now use the initial condition that y(0) is 1 and so we get,"},{"Start":"05:26.210 ","End":"05:27.845","Text":"if we plug all that in,"},{"Start":"05:27.845 ","End":"05:30.605","Text":"that 1 is k times 2,"},{"Start":"05:30.605 ","End":"05:35.569","Text":"which makes k to be 1/2 or 0.5 in decimal."},{"Start":"05:35.569 ","End":"05:40.085","Text":"Now if we put this k back here is 0.5,"},{"Start":"05:40.085 ","End":"05:42.680","Text":"just as equally, put it on the other side is 2,"},{"Start":"05:42.680 ","End":"05:45.005","Text":"1 over 0.5 is 2."},{"Start":"05:45.005 ","End":"05:53.340","Text":"We can get our final answer in this form and we are done."}],"ID":7845},{"Watched":false,"Name":"Exercise 16","Duration":"4m 40s","ChapterTopicVideoID":7772,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.800","Text":"Here we have a word problem and accompanying diagram, let see."},{"Start":"00:04.800 ","End":"00:13.380","Text":"The area S must be this area here is bounded by the curve y=y(x),"},{"Start":"00:13.380 ","End":"00:18.510","Text":"on the x-axis and the lines x=1."},{"Start":"00:18.510 ","End":"00:20.895","Text":"This I wrote just as x=x,"},{"Start":"00:20.895 ","End":"00:22.590","Text":"meaning some vertical lines through"},{"Start":"00:22.590 ","End":"00:26.175","Text":"an arbitrary x and x is going to be a variable in fact."},{"Start":"00:26.175 ","End":"00:28.590","Text":"We also know that y(1) is 2,"},{"Start":"00:28.590 ","End":"00:31.035","Text":"which means that this must be 2 here."},{"Start":"00:31.035 ","End":"00:33.675","Text":"Just put some dotted line in."},{"Start":"00:33.675 ","End":"00:38.570","Text":"We\u0027re asking if a curve exists such that the area of S=2y(x),"},{"Start":"00:38.570 ","End":"00:41.185","Text":"y(x) is the height here."},{"Start":"00:41.185 ","End":"00:44.015","Text":"This would be y(x)."},{"Start":"00:44.015 ","End":"00:49.110","Text":"What I want is S=2y."},{"Start":"00:49.110 ","End":"00:51.150","Text":"What we need is a differential equation."},{"Start":"00:51.150 ","End":"00:52.760","Text":"Let\u0027s lead up to it."},{"Start":"00:52.760 ","End":"00:56.555","Text":"First, we have to check whether the equation S,"},{"Start":"00:56.555 ","End":"01:04.680","Text":"which is the integral from 1 to x(y)=2y the area is twice the height."},{"Start":"01:04.680 ","End":"01:06.505","Text":"The initial condition y(1)=2,"},{"Start":"01:06.505 ","End":"01:09.500","Text":"we have to check whether this has a solution that\u0027s just"},{"Start":"01:09.500 ","End":"01:13.510","Text":"rephrasing it and close to mathematical terms."},{"Start":"01:13.510 ","End":"01:16.170","Text":"We get simply that the integral,"},{"Start":"01:16.170 ","End":"01:18.030","Text":"just writing this down again here,"},{"Start":"01:18.030 ","End":"01:21.305","Text":"then we\u0027ll come later to the initial condition."},{"Start":"01:21.305 ","End":"01:23.780","Text":"What we do in these cases is differentiate"},{"Start":"01:23.780 ","End":"01:26.390","Text":"because when I have an integral up to a variable x,"},{"Start":"01:26.390 ","End":"01:31.650","Text":"and differentiate it, I get back to the original function."},{"Start":"01:31.650 ","End":"01:34.880","Text":"What this gives me, if I differentiate it with respect to x,"},{"Start":"01:34.880 ","End":"01:36.470","Text":"that\u0027s the derivative sign."},{"Start":"01:36.470 ","End":"01:40.530","Text":"Maybe I should write it diagonal. There it is."},{"Start":"01:40.530 ","End":"01:43.700","Text":"The derivative of this thing is just y itself,"},{"Start":"01:43.700 ","End":"01:47.135","Text":"the integral, and then the derivative with respect to x is y."},{"Start":"01:47.135 ","End":"01:50.690","Text":"Here we just differentiate y, which is y\u0027."},{"Start":"01:50.690 ","End":"01:54.380","Text":"What we get is a linear differential equation."},{"Start":"01:54.380 ","End":"01:57.605","Text":"If I divide by 2 and bring this to the other side,"},{"Start":"01:57.605 ","End":"02:02.150","Text":"I get y\u0027 plus minus 1/2y=0,"},{"Start":"02:02.150 ","End":"02:05.780","Text":"when this is our a(x) and b(x) in terms of"},{"Start":"02:05.780 ","End":"02:09.200","Text":"the formula for differential equations which are linear."},{"Start":"02:09.200 ","End":"02:12.560","Text":"When b is 0, there\u0027s a special formula."},{"Start":"02:12.560 ","End":"02:14.750","Text":"I move to another page,"},{"Start":"02:14.750 ","End":"02:16.610","Text":"just copied this thing."},{"Start":"02:16.610 ","End":"02:25.070","Text":"Now, we use the formula that when you have b_0 y\u0027 plus a(x) y is 0,"},{"Start":"02:25.070 ","End":"02:27.995","Text":"then the solution is of this form."},{"Start":"02:27.995 ","End":"02:37.675","Text":"In our case, what it gives us is that y=ce and the integral of a(x) dx,"},{"Start":"02:37.675 ","End":"02:44.930","Text":"is plus 1/2x because this minus goes with this minus and the 1/2, write a 0.5."},{"Start":"02:44.930 ","End":"02:48.026","Text":"Now, we get to the initial condition."},{"Start":"02:48.026 ","End":"02:51.645","Text":"Y(1) is 2 so if we put y is 2,"},{"Start":"02:51.645 ","End":"02:52.830","Text":"x is 1,"},{"Start":"02:52.830 ","End":"02:55.020","Text":"we get the 2 here,"},{"Start":"02:55.020 ","End":"03:04.110","Text":"and x is 1 here so 2=ce^0.5 so c=2e^ minus 0.5."},{"Start":"03:04.110 ","End":"03:07.350","Text":"Then we replace c here,"},{"Start":"03:07.350 ","End":"03:09.015","Text":"so this is what we get."},{"Start":"03:09.015 ","End":"03:11.000","Text":"Now, we have to ask the question,"},{"Start":"03:11.000 ","End":"03:13.220","Text":"can this equal to y?"},{"Start":"03:13.220 ","End":"03:15.260","Text":"This is our original question."},{"Start":"03:15.260 ","End":"03:17.300","Text":"If this is the case,"},{"Start":"03:17.300 ","End":"03:26.030","Text":"then we have that the integral and y is here of this dx is equal to twice this thing."},{"Start":"03:26.030 ","End":"03:27.440","Text":"This would give us,"},{"Start":"03:27.440 ","End":"03:32.135","Text":"if we integrated the integral of this"},{"Start":"03:32.135 ","End":"03:37.700","Text":"would be 2e^minus 0.5 because these parts are constant,"},{"Start":"03:37.700 ","End":"03:41.150","Text":"e^ 0.5x divided by 0.5,"},{"Start":"03:41.150 ","End":"03:43.550","Text":"which means multiplying by 2."},{"Start":"03:43.550 ","End":"03:45.275","Text":"I\u0027ll just write that at the side,"},{"Start":"03:45.275 ","End":"03:50.000","Text":"the integral of e^0.5x is"},{"Start":"03:50.000 ","End":"04:00.510","Text":"just 1 over 0.5 e^0.5x, which is 2e^0.5x."},{"Start":"04:00.510 ","End":"04:05.385","Text":"This 2 from this integral goes with this 2 to give 4."},{"Start":"04:05.385 ","End":"04:11.680","Text":"We have 4 times this and taken between x and 1."},{"Start":"04:11.680 ","End":"04:14.330","Text":"If we substitute x,"},{"Start":"04:14.330 ","End":"04:16.595","Text":"we exactly get this."},{"Start":"04:16.595 ","End":"04:21.640","Text":"But this thing is equal to this minus what happens when we substitute 1."},{"Start":"04:21.640 ","End":"04:23.915","Text":"When we substitute 1,"},{"Start":"04:23.915 ","End":"04:25.940","Text":"we don\u0027t get 0 you can check,"},{"Start":"04:25.940 ","End":"04:30.245","Text":"you actually get that x is 1 this cancels with this and the answer is 4."},{"Start":"04:30.245 ","End":"04:33.620","Text":"This thing is equal to this minus 4,"},{"Start":"04:33.620 ","End":"04:34.760","Text":"so they\u0027re not equal,"},{"Start":"04:34.760 ","End":"04:40.360","Text":"and so the answer is no, and we\u0027re done."}],"ID":7846},{"Watched":false,"Name":"Exercise 17","Duration":"3m 48s","ChapterTopicVideoID":7773,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:06.955","Text":"Here we have another word problem very similar to a previous one where we have that S,"},{"Start":"00:06.955 ","End":"00:11.903","Text":"which is this is bounded by the curve y(x),"},{"Start":"00:11.903 ","End":"00:13.320","Text":"x equals 1,"},{"Start":"00:13.320 ","End":"00:15.700","Text":"and let\u0027s call this x equals x,"},{"Start":"00:15.700 ","End":"00:20.290","Text":"it means a variable x, and the y-axis."},{"Start":"00:20.290 ","End":"00:22.885","Text":"We\u0027re given that y(1) is 2,"},{"Start":"00:22.885 ","End":"00:25.355","Text":"it means that it\u0027s 2 here."},{"Start":"00:25.355 ","End":"00:28.105","Text":"We asked the question,"},{"Start":"00:28.105 ","End":"00:34.240","Text":"if such a curve exists where the area of S is simply y minus 2,"},{"Start":"00:34.240 ","End":"00:36.430","Text":"y(x) is the height here, y,"},{"Start":"00:36.430 ","End":"00:37.990","Text":"which is y(x),"},{"Start":"00:37.990 ","End":"00:41.490","Text":"and the area is going to equal y minus 2."},{"Start":"00:41.490 ","End":"00:44.930","Text":"We put that in terms of an equation,"},{"Start":"00:44.930 ","End":"00:48.485","Text":"what we get is that S,"},{"Start":"00:48.485 ","End":"00:52.535","Text":"which is the integral from 1 to x of y,"},{"Start":"00:52.535 ","End":"00:55.800","Text":"is going to equal y minus 2."},{"Start":"00:56.110 ","End":"01:01.970","Text":"What we\u0027re going to do now is differentiate both sides of this equation."},{"Start":"01:01.970 ","End":"01:03.555","Text":"Well, first I\u0027ll write it again,"},{"Start":"01:03.555 ","End":"01:06.965","Text":"and we also should remember there\u0027s an initial condition."},{"Start":"01:06.965 ","End":"01:11.045","Text":"Just copying this and I\u0027m going to differentiate this."},{"Start":"01:11.045 ","End":"01:14.960","Text":"If you take a variable upper limit x and differentiate with"},{"Start":"01:14.960 ","End":"01:19.505","Text":"respect to get back to the original function, which is y(x)."},{"Start":"01:19.505 ","End":"01:22.880","Text":"Just going to indicate that I\u0027m differentiating both sides."},{"Start":"01:22.880 ","End":"01:24.470","Text":"This is a derivative sine,"},{"Start":"01:24.470 ","End":"01:26.165","Text":"I should write it clearer,"},{"Start":"01:26.165 ","End":"01:27.905","Text":"we are differentiate it."},{"Start":"01:27.905 ","End":"01:29.795","Text":"Here we just get y,"},{"Start":"01:29.795 ","End":"01:32.300","Text":"and here if we differentiate this,"},{"Start":"01:32.300 ","End":"01:35.495","Text":"the derivative of y minus 2 is y\u0027."},{"Start":"01:35.495 ","End":"01:38.615","Text":"This is very simply y equals y\u0027."},{"Start":"01:38.615 ","End":"01:40.880","Text":"If we write that as a linear equation,"},{"Start":"01:40.880 ","End":"01:45.500","Text":"then it\u0027s y\u0027 plus minus 1, y equals 0."},{"Start":"01:45.500 ","End":"01:47.210","Text":"Here our A(x),"},{"Start":"01:47.210 ","End":"01:49.160","Text":"B(x) from the linear."},{"Start":"01:49.160 ","End":"01:53.409","Text":"Let me just move to another page."},{"Start":"01:53.409 ","End":"01:56.630","Text":"I was saying if B(x) is 0,"},{"Start":"01:56.630 ","End":"01:59.060","Text":"then there\u0027s a simplified formula,"},{"Start":"01:59.060 ","End":"02:01.265","Text":"and that is this."},{"Start":"02:01.265 ","End":"02:06.290","Text":"If we apply it to our case where A(x) is minus 1,"},{"Start":"02:06.290 ","End":"02:11.975","Text":"and notice that because of the minus integral of A(x),"},{"Start":"02:11.975 ","End":"02:14.360","Text":"we get the integral of this is minus x,"},{"Start":"02:14.360 ","End":"02:16.070","Text":"minus minus x is x."},{"Start":"02:16.070 ","End":"02:19.610","Text":"In short, we just get y equals ce^x."},{"Start":"02:19.610 ","End":"02:21.815","Text":"Now let\u0027s bring in the initial condition."},{"Start":"02:21.815 ","End":"02:25.175","Text":"Initial condition when x is 1, y is 2."},{"Start":"02:25.175 ","End":"02:26.645","Text":"If we plug that in,"},{"Start":"02:26.645 ","End":"02:31.030","Text":"we just get c is equal to e to the minus 1,"},{"Start":"02:31.030 ","End":"02:33.815","Text":"which we then put back in here,"},{"Start":"02:33.815 ","End":"02:37.420","Text":"and we get y equals 2e to the minus 1 is 3x."},{"Start":"02:37.420 ","End":"02:41.730","Text":"Even just combine the exponents and get 2e to the x minus 1."},{"Start":"02:41.730 ","End":"02:43.550","Text":"Now we come to our question."},{"Start":"02:43.550 ","End":"02:47.060","Text":"Now that we\u0027ve solved for y, we have to ask the original question."},{"Start":"02:47.060 ","End":"02:49.430","Text":"Here we\u0027ve replaced y,"},{"Start":"02:49.430 ","End":"02:52.610","Text":"here and here by what y equals,"},{"Start":"02:52.610 ","End":"02:55.250","Text":"which is 2e to the x minus 1."},{"Start":"02:55.250 ","End":"02:57.440","Text":"This equation becomes this equation."},{"Start":"02:57.440 ","End":"02:59.120","Text":"Can this equal this?"},{"Start":"02:59.120 ","End":"03:01.115","Text":"Now if we do this integral,"},{"Start":"03:01.115 ","End":"03:04.160","Text":"we get this integral is the thing itself."},{"Start":"03:04.160 ","End":"03:07.225","Text":"It\u0027s just 2e to the x minus 1."},{"Start":"03:07.225 ","End":"03:10.910","Text":"We have to substitute it between the limits of x and 1."},{"Start":"03:10.910 ","End":"03:14.510","Text":"Here we get 2e to the x minus 1 minus 2 as before."},{"Start":"03:14.510 ","End":"03:16.880","Text":"If we put in x, we get the thing itself."},{"Start":"03:16.880 ","End":"03:18.155","Text":"If we put in 1,"},{"Start":"03:18.155 ","End":"03:20.290","Text":"we get 2e^0,"},{"Start":"03:20.290 ","End":"03:22.260","Text":"which is just 2."},{"Start":"03:22.260 ","End":"03:25.565","Text":"The question is, can this equal this?"},{"Start":"03:25.565 ","End":"03:28.385","Text":"And the answer is yes."},{"Start":"03:28.385 ","End":"03:30.005","Text":"There is such a curve."},{"Start":"03:30.005 ","End":"03:32.225","Text":"The curve is our curve,"},{"Start":"03:32.225 ","End":"03:36.960","Text":"y equals 2e to the x minus 1."},{"Start":"03:36.960 ","End":"03:40.845","Text":"Then this curve satisfies the condition."},{"Start":"03:40.845 ","End":"03:48.570","Text":"Yes, indeed there exists a curve which satisfies the given conditions and we\u0027re done."}],"ID":7847},{"Watched":false,"Name":"Exercise 18","Duration":"3m 24s","ChapterTopicVideoID":7774,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.790","Text":"Here we have another word problem which will turn out to be a differential equation."},{"Start":"00:04.790 ","End":"00:06.285","Text":"It comes with a diagram."},{"Start":"00:06.285 ","End":"00:08.910","Text":"Given a curve passing through the point B,"},{"Start":"00:08.910 ","End":"00:11.128","Text":"so B is a fixed point, 0,"},{"Start":"00:11.128 ","End":"00:15.000","Text":"1, and A is a point on the curve,"},{"Start":"00:15.000 ","End":"00:17.310","Text":"but it\u0027s a variable point on the curve."},{"Start":"00:17.310 ","End":"00:20.310","Text":"At each such point A, the slope,"},{"Start":"00:20.310 ","End":"00:23.844","Text":"let me just say that the slope in the pictures,"},{"Start":"00:23.844 ","End":"00:25.440","Text":"haven\u0027t drawn it well."},{"Start":"00:25.440 ","End":"00:28.665","Text":"Anyway, the slope here is y\u0027,"},{"Start":"00:28.665 ","End":"00:34.440","Text":"and we\u0027re given that y\u0027 is equal to the area of the trapezoid."},{"Start":"00:34.440 ","End":"00:37.530","Text":"I\u0027ll give it a letter T for trapezoids."},{"Start":"00:37.530 ","End":"00:43.190","Text":"So the area of the trapezoid is equal to the derivative here."},{"Start":"00:43.190 ","End":"00:45.800","Text":"Let\u0027s recall some geometry."},{"Start":"00:45.800 ","End":"00:48.575","Text":"In order to compute the area of a trapezoid,"},{"Start":"00:48.575 ","End":"00:50.690","Text":"we look at it as if it\u0027s on its side."},{"Start":"00:50.690 ","End":"00:53.155","Text":"This in fact is the height,"},{"Start":"00:53.155 ","End":"00:56.345","Text":"which is equal to the distance from zero to x."},{"Start":"00:56.345 ","End":"00:59.930","Text":"So this is x from here to here,"},{"Start":"00:59.930 ","End":"01:05.855","Text":"from C to D. This bit is just y and this is always equal to 1."},{"Start":"01:05.855 ","End":"01:09.380","Text":"So these are the two bases and this is the height."},{"Start":"01:09.380 ","End":"01:11.180","Text":"If you recall,"},{"Start":"01:11.180 ","End":"01:13.745","Text":"or I\u0027ll remind you from geometry,"},{"Start":"01:13.745 ","End":"01:18.185","Text":"that the area of a trapezoid is,"},{"Start":"01:18.185 ","End":"01:22.145","Text":"one way of looking at it is the average of the two bases times the height."},{"Start":"01:22.145 ","End":"01:26.690","Text":"So 1 plus y over 2x,"},{"Start":"01:26.690 ","End":"01:29.285","Text":"which is what I\u0027m going to write in the next line."},{"Start":"01:29.285 ","End":"01:33.950","Text":"Y plus 1 over 2 average base times height x,"},{"Start":"01:33.950 ","End":"01:41.540","Text":"and the given condition is that it\u0027s equal to y\u0027 or dy over dx."},{"Start":"01:41.540 ","End":"01:43.280","Text":"This is our differential equation."},{"Start":"01:43.280 ","End":"01:48.305","Text":"Here\u0027s the initial condition that it goes through the point B, which is 0,1."},{"Start":"01:48.305 ","End":"01:54.320","Text":"So what we can do is separate the variables here and put"},{"Start":"01:54.320 ","End":"02:00.635","Text":"the y plus 1 in the denominator and the x over here with the 1.5."},{"Start":"02:00.635 ","End":"02:02.735","Text":"This is what we get."},{"Start":"02:02.735 ","End":"02:04.069","Text":"Now that it\u0027s separated,"},{"Start":"02:04.069 ","End":"02:11.330","Text":"I can put the integral sign in front of each of them and then do the integration."},{"Start":"02:11.330 ","End":"02:15.130","Text":"On this side I get the natural logarithm of 1 plus y."},{"Start":"02:15.130 ","End":"02:16.760","Text":"Here it\u0027s just a polynomial,"},{"Start":"02:16.760 ","End":"02:20.480","Text":"so it\u0027s x squared over 2 over 2, x squared over 4."},{"Start":"02:20.480 ","End":"02:21.980","Text":"When we have logarithm problems,"},{"Start":"02:21.980 ","End":"02:25.610","Text":"we usually write the constant as natural log of a positive"},{"Start":"02:25.610 ","End":"02:30.140","Text":"constant k. If I bring this to the other side,"},{"Start":"02:30.140 ","End":"02:32.750","Text":"the difference in logs is the log of the quotient,"},{"Start":"02:32.750 ","End":"02:36.830","Text":"so we get 1 plus y over k is x squared over 4."},{"Start":"02:36.830 ","End":"02:40.580","Text":"Now take the exponent of each side and that gets rid of the logarithm."},{"Start":"02:40.580 ","End":"02:45.835","Text":"So we get 1 plus y over k is e to the power of this."},{"Start":"02:45.835 ","End":"02:50.234","Text":"Now we can isolate y multiplied by k and subtract 1,"},{"Start":"02:50.234 ","End":"02:52.505","Text":"and we\u0027ve isolated this."},{"Start":"02:52.505 ","End":"02:56.580","Text":"Now it\u0027s time for the initial condition that y of zero is 1."},{"Start":"02:56.580 ","End":"02:59.190","Text":"When x is 0, y is 1,"},{"Start":"02:59.190 ","End":"03:05.810","Text":"so we get that 1 is equal to k times e to the 0 minus 1."},{"Start":"03:05.810 ","End":"03:07.355","Text":"In other words, k is 2."},{"Start":"03:07.355 ","End":"03:14.180","Text":"When k is 2, we put that back here and we get this as the answer."},{"Start":"03:14.180 ","End":"03:18.710","Text":"This is the equation of the curve which satisfies that condition that"},{"Start":"03:18.710 ","End":"03:24.480","Text":"the derivative is equal to the area of the trapezoid and so on. We are done."}],"ID":7848},{"Watched":false,"Name":"Exercise 19","Duration":"2m 41s","ChapterTopicVideoID":7775,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.355","Text":"Here we have a word problem with exponential growth or decay."},{"Start":"00:05.355 ","End":"00:10.770","Text":"We just redefining what it means to grow or decay exponentially is when"},{"Start":"00:10.770 ","End":"00:15.765","Text":"the rate of growth or decay is proportional to the value of the thing itself."},{"Start":"00:15.765 ","End":"00:21.890","Text":"Now, we\u0027re given abstractly that at start time, that\u0027s t=0,"},{"Start":"00:21.890 ","End":"00:26.940","Text":"we have a quantity y_0 and it\u0027s constant"},{"Start":"00:26.940 ","End":"00:31.970","Text":"to proportionality is k. We have to find the formula for the quantity"},{"Start":"00:31.970 ","End":"00:39.320","Text":"as a function of t. What we have is that the condition that y"},{"Start":"00:39.320 ","End":"00:46.720","Text":"is exponential with constant of proportionality k means that y\u0027 is equal to k times y."},{"Start":"00:46.720 ","End":"00:49.160","Text":"It\u0027s proportional with constant of proportionality k,"},{"Start":"00:49.160 ","End":"00:51.380","Text":"which means that it\u0027s equal to k times this."},{"Start":"00:51.380 ","End":"00:56.425","Text":"We also have times 0 that is equal to y_0."},{"Start":"00:56.425 ","End":"00:58.760","Text":"That\u0027s the differential equation."},{"Start":"00:58.760 ","End":"01:02.435","Text":"We can make it a linear differential equation."},{"Start":"01:02.435 ","End":"01:07.165","Text":"That\u0027s y\u0027 and I bring the minus k(y) =0."},{"Start":"01:07.165 ","End":"01:12.859","Text":"It\u0027s one of those special ones where this is our a(x) and b is 0."},{"Start":"01:12.859 ","End":"01:20.990","Text":"It has a formula that y is equal to a constant times e^kt."},{"Start":"01:20.990 ","End":"01:25.040","Text":"It\u0027s really the integral of minus k and here it\u0027s minus."},{"Start":"01:25.040 ","End":"01:27.500","Text":"The formula was given earlier."},{"Start":"01:27.500 ","End":"01:30.380","Text":"I suppose I should really remind you what it was,"},{"Start":"01:30.380 ","End":"01:38.660","Text":"is that y is equal to a constant times e to the power of the minus"},{"Start":"01:38.660 ","End":"01:47.380","Text":"the integral of a(t)dt and this is the a(t),"},{"Start":"01:47.380 ","End":"01:49.890","Text":"which doesn\u0027t really depend on t at all."},{"Start":"01:49.890 ","End":"01:54.860","Text":"This is usually a b(t) which equals 0 and that\u0027s when we have the shortcut formula."},{"Start":"01:54.860 ","End":"01:58.230","Text":"If a(t) is minus k,"},{"Start":"01:58.230 ","End":"02:01.940","Text":"then we have the integral of minus minus k. The integral of k,"},{"Start":"02:01.940 ","End":"02:06.140","Text":"integral of constant is just k times the variable kt."},{"Start":"02:06.140 ","End":"02:10.340","Text":"We have t instead of x here because independent variable is time."},{"Start":"02:10.340 ","End":"02:13.000","Text":"That explains that."},{"Start":"02:13.000 ","End":"02:19.280","Text":"Then we put in the initial condition that when x is 0, y is y_0."},{"Start":"02:19.280 ","End":"02:20.900","Text":"So that if we do that,"},{"Start":"02:20.900 ","End":"02:25.130","Text":"then we get that the constant is equal to c,"},{"Start":"02:25.130 ","End":"02:27.890","Text":"put t=0 and y=y_0,"},{"Start":"02:27.890 ","End":"02:31.685","Text":"and that\u0027s equal to C. Then finally put the C back in here."},{"Start":"02:31.685 ","End":"02:38.540","Text":"We get a function of the amount as a function of time and the initial quantity."},{"Start":"02:38.540 ","End":"02:41.910","Text":"This is our answer. We are done."}],"ID":7849},{"Watched":false,"Name":"Exercise 20","Duration":"3m 33s","ChapterTopicVideoID":7776,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.490","Text":"Here we have a word problem."},{"Start":"00:02.490 ","End":"00:04.410","Text":"One of these growth decay,"},{"Start":"00:04.410 ","End":"00:07.140","Text":"and actually its growth and its population."},{"Start":"00:07.140 ","End":"00:11.535","Text":"The population of the earth is increasing at a rate of 2 percent per year,"},{"Start":"00:11.535 ","End":"00:14.835","Text":"and it was found to be 4 billion in 1980."},{"Start":"00:14.835 ","End":"00:16.785","Text":"Then we have three questions."},{"Start":"00:16.785 ","End":"00:19.875","Text":"What will it be in 2010,"},{"Start":"00:19.875 ","End":"00:22.635","Text":"what was it in 1974,"},{"Start":"00:22.635 ","End":"00:26.313","Text":"and when will we have a population of 50 billion,"},{"Start":"00:26.313 ","End":"00:28.575","Text":"assuming this rate stays the same?"},{"Start":"00:28.575 ","End":"00:32.400","Text":"Now, we\u0027re told also that population grows exponentially,"},{"Start":"00:32.400 ","End":"00:38.168","Text":"which means that at any given moment the rate of growth is proportional to its value,"},{"Start":"00:38.168 ","End":"00:42.260","Text":"and that constant of proportionality is what later we\u0027ll be given as"},{"Start":"00:42.260 ","End":"00:47.720","Text":"k. Here we\u0027re reminded that if the quantity grows exponentially,"},{"Start":"00:47.720 ","End":"00:52.070","Text":"and at time 0 we have an amount y_0, y naught,"},{"Start":"00:52.070 ","End":"00:55.535","Text":"then at any given time you have a quantity,"},{"Start":"00:55.535 ","End":"00:59.285","Text":"y given by this formula y naught e^kt."},{"Start":"00:59.285 ","End":"01:06.320","Text":"What I think we\u0027d better do is measure time in terms of 1980."},{"Start":"01:06.320 ","End":"01:13.485","Text":"We\u0027ll let 1980 be our time t = 0 and measure it in years,"},{"Start":"01:13.485 ","End":"01:17.595","Text":"and that way this 2010 will be time 30,"},{"Start":"01:17.595 ","End":"01:20.925","Text":"and 1974 will be time minus 6,"},{"Start":"01:20.925 ","End":"01:23.640","Text":"and here we don\u0027t know, it\u0027s a question mark."},{"Start":"01:23.640 ","End":"01:25.305","Text":"Continuing."},{"Start":"01:25.305 ","End":"01:29.129","Text":"In our case we\u0027re given that y naught is 4 million,"},{"Start":"01:29.129 ","End":"01:31.670","Text":"now this should be 4 billion,"},{"Start":"01:31.670 ","End":"01:33.185","Text":"anyway, leave it like that,"},{"Start":"01:33.185 ","End":"01:36.380","Text":"and k is 0.02,"},{"Start":"01:36.380 ","End":"01:38.045","Text":"which is 2 percent."},{"Start":"01:38.045 ","End":"01:40.160","Text":"In our given particular case,"},{"Start":"01:40.160 ","End":"01:45.095","Text":"we have the y is 4e^0.02t,"},{"Start":"01:45.095 ","End":"01:48.245","Text":"where t is number of years since 1980,"},{"Start":"01:48.245 ","End":"01:51.700","Text":"and the answer is going to be in billion."},{"Start":"01:51.700 ","End":"01:57.050","Text":"The three questions, let me just say that when we write mill,"},{"Start":"01:57.050 ","End":"01:59.200","Text":"it really means billion,"},{"Start":"01:59.200 ","End":"02:00.830","Text":"and just ignore that."},{"Start":"02:00.830 ","End":"02:03.650","Text":"It\u0027s not essential to anything."},{"Start":"02:03.650 ","End":"02:08.944","Text":"At time 30, we just substitute 30 into this formula."},{"Start":"02:08.944 ","End":"02:12.680","Text":"T is 30, so it\u0027s 0.02 times 30."},{"Start":"02:12.680 ","End":"02:14.060","Text":"Do it on the calculator,"},{"Start":"02:14.060 ","End":"02:18.010","Text":"comes out to about 7.28 billion."},{"Start":"02:18.010 ","End":"02:21.150","Text":"In part 2, we let the time be minus 6,"},{"Start":"02:21.150 ","End":"02:22.710","Text":"which was 1974,"},{"Start":"02:22.710 ","End":"02:24.390","Text":"substitute minus 6,"},{"Start":"02:24.390 ","End":"02:28.270","Text":"and we get from the calculator about 4.51."},{"Start":"02:28.270 ","End":"02:31.635","Text":"In part c we don\u0027t know the time,"},{"Start":"02:31.635 ","End":"02:33.555","Text":"we do know the amount,"},{"Start":"02:33.555 ","End":"02:37.340","Text":"y is equal to 50 billion."},{"Start":"02:37.340 ","End":"02:39.290","Text":"I don\u0027t know why the inconsistency,"},{"Start":"02:39.290 ","End":"02:43.055","Text":"I should have just written the same mill, meaning billion."},{"Start":"02:43.055 ","End":"02:47.015","Text":"Anyway, it\u0027s 50 because we\u0027re counting in billions."},{"Start":"02:47.015 ","End":"02:50.050","Text":"For part c, we don\u0027t have an immediate answer,"},{"Start":"02:50.050 ","End":"02:57.330","Text":"we just write the equation that 4e^0.02t is equal to 50."},{"Start":"02:57.330 ","End":"03:01.460","Text":"What we do is we just divide by 4,"},{"Start":"03:01.460 ","End":"03:07.790","Text":"and then we take the logarithm of that and we get this is equal to this,"},{"Start":"03:07.790 ","End":"03:16.805","Text":"and then we get that 0.02t is natural log of 12.5 divided by the 0.02."},{"Start":"03:16.805 ","End":"03:20.720","Text":"On the calculator it comes out to roughly 126,"},{"Start":"03:20.720 ","End":"03:23.585","Text":"and remember this is years since 1980,"},{"Start":"03:23.585 ","End":"03:26.360","Text":"so we have to add 1980 to this,"},{"Start":"03:26.360 ","End":"03:30.840","Text":"and we get approximately the year 2106."},{"Start":"03:30.840 ","End":"03:33.400","Text":"We\u0027re done here."}],"ID":7850},{"Watched":false,"Name":"Exercise 21","Duration":"2m 56s","ChapterTopicVideoID":7777,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.590","Text":"Here we have another word problem which we convert into a differential equation."},{"Start":"00:04.590 ","End":"00:08.430","Text":"The population in a certain city grows exponentially."},{"Start":"00:08.430 ","End":"00:11.265","Text":"In a certain year there were 400,000 residents,"},{"Start":"00:11.265 ","End":"00:14.685","Text":"and 4 years later, there were 440,000."},{"Start":"00:14.685 ","End":"00:18.885","Text":"In Part A, we have to find the annual growth rate as a percent,"},{"Start":"00:18.885 ","End":"00:20.130","Text":"and in Part B,"},{"Start":"00:20.130 ","End":"00:23.610","Text":"we have to find that after how many years from the initial year,"},{"Start":"00:23.610 ","End":"00:27.045","Text":"there were 550,000 residents."},{"Start":"00:27.045 ","End":"00:32.025","Text":"As usual, we\u0027re reminded that if the quantity t grows exponentially,"},{"Start":"00:32.025 ","End":"00:34.320","Text":"and I\u0027ll remind you that exponentially means the rate"},{"Start":"00:34.320 ","End":"00:36.555","Text":"of change is proportional to the amount,"},{"Start":"00:36.555 ","End":"00:39.765","Text":"and at the start time t=0,"},{"Start":"00:39.765 ","End":"00:41.255","Text":"the amount is y_0,"},{"Start":"00:41.255 ","End":"00:45.800","Text":"then the quantity at any given time t is given by this formula."},{"Start":"00:45.800 ","End":"00:51.155","Text":"The k here is the constant of proportionality from the exponential growth."},{"Start":"00:51.155 ","End":"00:53.930","Text":"Let\u0027s start solving it."},{"Start":"00:53.930 ","End":"01:02.360","Text":"First of all, we have the formula that y(t) is y_0, which is 400e^kt."},{"Start":"01:02.360 ","End":"01:07.650","Text":"We still don\u0027t know k. We know that after 4 years,"},{"Start":"01:07.650 ","End":"01:10.185","Text":"the y is 440."},{"Start":"01:10.185 ","End":"01:13.700","Text":"I forgot to say that we\u0027re working in thousands,"},{"Start":"01:13.700 ","End":"01:15.610","Text":"so when I write thou,"},{"Start":"01:15.610 ","End":"01:19.090","Text":"it\u0027s just short for 1,000,"},{"Start":"01:19.090 ","End":"01:22.625","Text":"just so we don\u0027t have to deal in large numbers all the time."},{"Start":"01:22.625 ","End":"01:25.870","Text":"If we substitute 4 in here,"},{"Start":"01:25.870 ","End":"01:30.195","Text":"then we get 400e^4k is 440."},{"Start":"01:30.195 ","End":"01:33.650","Text":"Divide both sides by 400,"},{"Start":"01:33.650 ","End":"01:35.210","Text":"and we get this."},{"Start":"01:35.210 ","End":"01:37.715","Text":"Then take the logarithm."},{"Start":"01:37.715 ","End":"01:41.560","Text":"Logarithm of e^4k is just 4k,"},{"Start":"01:41.560 ","End":"01:45.590","Text":"because logarithm and exponent are inverse."},{"Start":"01:45.590 ","End":"01:51.990","Text":"We get that 4k is natural log of this and if we divide by 4, we get this,"},{"Start":"01:51.990 ","End":"01:54.150","Text":"time to bring out the calculator,"},{"Start":"01:54.150 ","End":"01:58.575","Text":"comes out to be 0.02 roughly."},{"Start":"01:58.575 ","End":"02:00.450","Text":"We were asked in percent,"},{"Start":"02:00.450 ","End":"02:04.565","Text":"so that gives us 2 percent growth rate."},{"Start":"02:04.565 ","End":"02:07.820","Text":"It also enables us to fill in the formula."},{"Start":"02:07.820 ","End":"02:09.560","Text":"Now that we have the k,"},{"Start":"02:09.560 ","End":"02:15.650","Text":"we can then write it completely as 400e^0.02t."},{"Start":"02:16.190 ","End":"02:20.165","Text":"Again, approximately because this k was approximate."},{"Start":"02:20.165 ","End":"02:25.730","Text":"Now in part b, we have to find out when the population is 550,000."},{"Start":"02:25.730 ","End":"02:30.065","Text":"We write the equation that y(t) is 550,000."},{"Start":"02:30.065 ","End":"02:32.090","Text":"Filling it in from here,"},{"Start":"02:32.090 ","End":"02:33.800","Text":"we get this equation."},{"Start":"02:33.800 ","End":"02:35.810","Text":"We divide by 400,"},{"Start":"02:35.810 ","End":"02:39.080","Text":"and 550/400 is 1.375."},{"Start":"02:39.080 ","End":"02:43.100","Text":"Take the logarithm and then finally do the computation."},{"Start":"02:43.100 ","End":"02:47.220","Text":"Bring this to the other side and bring out the calculator."},{"Start":"02:47.220 ","End":"02:53.280","Text":"We see it\u0027s just short of 16 years we\u0027re going to have 550,000 population."},{"Start":"02:53.280 ","End":"02:55.780","Text":"We are done."}],"ID":7851},{"Watched":false,"Name":"Exercise 22","Duration":"3m 37s","ChapterTopicVideoID":7778,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.980","Text":"Here we have another word problem involving money in a bank."},{"Start":"00:04.980 ","End":"00:11.430","Text":"A man deposited money in the bank at an interest rate of 4 percent compounded annually."},{"Start":"00:11.430 ","End":"00:15.036","Text":"This actually means that the money is growing exponentially."},{"Start":"00:15.036 ","End":"00:18.539","Text":"When we have an interest rate that\u0027s being compounded,"},{"Start":"00:18.539 ","End":"00:20.145","Text":"then we have exponential growth."},{"Start":"00:20.145 ","End":"00:24.330","Text":"We\u0027re told that after 5 years he had accumulated $5,000."},{"Start":"00:24.330 ","End":"00:25.950","Text":"Then we have 2 questions, a,"},{"Start":"00:25.950 ","End":"00:28.255","Text":"how much did he initially deposit, and b,"},{"Start":"00:28.255 ","End":"00:32.710","Text":"after how many years will he have accumulated $7,000?"},{"Start":"00:32.710 ","End":"00:37.760","Text":"Here we\u0027re reminded that if the quantity y(t) grows exponentially,"},{"Start":"00:37.760 ","End":"00:38.990","Text":"t is time here,"},{"Start":"00:38.990 ","End":"00:43.100","Text":"and at the start time t=0 the amount is y_0,"},{"Start":"00:43.100 ","End":"00:47.365","Text":"then the quantity at any given time is given by the following formula,"},{"Start":"00:47.365 ","End":"00:53.250","Text":"and k is that constant of proportionality which in our case is 4 percent."},{"Start":"00:53.250 ","End":"00:56.930","Text":"Let\u0027s start solving. First of all,"},{"Start":"00:56.930 ","End":"01:04.331","Text":"because k is 4 percent and 4 percent is 0.04 the same thing,"},{"Start":"01:04.331 ","End":"01:07.920","Text":"then we have that y(t) equals according to this formula"},{"Start":"01:07.920 ","End":"01:12.125","Text":"we substitute k as 0.04 and this is what we get."},{"Start":"01:12.125 ","End":"01:14.300","Text":"Now in part a,"},{"Start":"01:14.300 ","End":"01:20.235","Text":"we know that after 5 years we have $5,000,"},{"Start":"01:20.235 ","End":"01:23.448","Text":"so if we put t=5,"},{"Start":"01:23.448 ","End":"01:29.535","Text":"then we will get that y(5) is 5,000 which means that"},{"Start":"01:29.535 ","End":"01:35.970","Text":"putting 5 in here that 0.04 times 5 is 0.2."},{"Start":"01:35.970 ","End":"01:41.190","Text":"Anyway, and that\u0027s going to equal to 5,000 so we divide"},{"Start":"01:41.190 ","End":"01:46.860","Text":"by e^0.2 and do this on the calculator."},{"Start":"01:46.860 ","End":"01:53.885","Text":"It comes out to be approximately $4,093.65 and"},{"Start":"01:53.885 ","End":"01:57.770","Text":"that gives us the formula also for"},{"Start":"01:57.770 ","End":"02:03.215","Text":"the general equation of y in terms of t because now we have the y_0."},{"Start":"02:03.215 ","End":"02:05.000","Text":"But just to answer the question,"},{"Start":"02:05.000 ","End":"02:06.320","Text":"this was the amount."},{"Start":"02:06.320 ","End":"02:08.025","Text":"What did they ask initially?"},{"Start":"02:08.025 ","End":"02:12.725","Text":"Deposit. Yes, that was $4,093 and so on."},{"Start":"02:12.725 ","End":"02:16.370","Text":"But now that we have the full formula we can now tackle"},{"Start":"02:16.370 ","End":"02:22.810","Text":"part b which wants to know when we\u0027re going to have $7,000?"},{"Start":"02:22.810 ","End":"02:28.275","Text":"We just substitute for y(t) as 7,000 here,"},{"Start":"02:28.275 ","End":"02:30.825","Text":"so we get that this is 7,000."},{"Start":"02:30.825 ","End":"02:36.960","Text":"Then we divide by the initial amount and so we get this over this,"},{"Start":"02:36.960 ","End":"02:42.830","Text":"with the calculator 1.7; 1 years approximately."},{"Start":"02:42.830 ","End":"02:44.626","Text":"Sorry, that\u0027s just part of the answer."},{"Start":"02:44.626 ","End":"02:46.040","Text":"That\u0027s not the year, sorry."},{"Start":"02:46.040 ","End":"02:50.360","Text":"We have to take the logarithm of both sides and get"},{"Start":"02:50.360 ","End":"02:57.965","Text":"that 0.04t is the natural logarithm of this then divide by 0.04."},{"Start":"02:57.965 ","End":"03:00.085","Text":"After the division we get,"},{"Start":"03:00.085 ","End":"03:06.705","Text":"that sounds more reasonable 13.41 years to get the $7,000."},{"Start":"03:06.705 ","End":"03:09.920","Text":"That\u0027s an approximation, but we are done."},{"Start":"03:09.920 ","End":"03:13.355","Text":"I just like to add there is a certain ambiguity in the question."},{"Start":"03:13.355 ","End":"03:17.405","Text":"This is how many years from the time he initially deposited,"},{"Start":"03:17.405 ","End":"03:22.310","Text":"but if we wanted to count from the time he had 5,000,"},{"Start":"03:22.310 ","End":"03:30.230","Text":"then we can subtract 5 years and get just 8.41 years from the current time,"},{"Start":"03:30.230 ","End":"03:31.790","Text":"not from the previous 5 years."},{"Start":"03:31.790 ","End":"03:37.890","Text":"Well, it\u0027s a matter of semantics whether we guessed our counting from. Okay, that\u0027s it."}],"ID":7852},{"Watched":false,"Name":"Exercise 23","Duration":"2m 48s","ChapterTopicVideoID":7779,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.625","Text":"Another word problem exercise."},{"Start":"00:02.625 ","End":"00:06.840","Text":"The number of wild animals at a nature reserve grows exponentially."},{"Start":"00:06.840 ","End":"00:11.700","Text":"I\u0027ll remind you that this means that the rate of change is proportional to the amount."},{"Start":"00:11.700 ","End":"00:15.240","Text":"That proportionality, later we\u0027ll see it\u0027s called"},{"Start":"00:15.240 ","End":"00:18.630","Text":"k. There were 1,000 animals at the initial count."},{"Start":"00:18.630 ","End":"00:20.880","Text":"At a second count, 20 months later,"},{"Start":"00:20.880 ","End":"00:23.280","Text":"there were 1,400 wild animals."},{"Start":"00:23.280 ","End":"00:29.280","Text":"How many months after the initial count will reserve have 2,000 animals?"},{"Start":"00:29.280 ","End":"00:33.065","Text":"Just to remind you that if a quantity grows exponentially"},{"Start":"00:33.065 ","End":"00:36.890","Text":"and we have initial amount of y_0,"},{"Start":"00:36.890 ","End":"00:38.600","Text":"then after any given time t,"},{"Start":"00:38.600 ","End":"00:42.535","Text":"it\u0027s given by the formula y(t) is y_0 e^kt,"},{"Start":"00:42.535 ","End":"00:47.615","Text":"where k is that constant of proportionality that relates to exponential growth."},{"Start":"00:47.615 ","End":"00:49.535","Text":"At the very beginning,"},{"Start":"00:49.535 ","End":"00:52.895","Text":"we have 1,000, so we already know y_0."},{"Start":"00:52.895 ","End":"00:58.310","Text":"We can write the formula that y is 1,000e^kt because"},{"Start":"00:58.310 ","End":"01:03.440","Text":"we still don\u0027t know k. But we do know that after 20 months,"},{"Start":"01:03.440 ","End":"01:05.945","Text":"and we\u0027re measuring time in months here,"},{"Start":"01:05.945 ","End":"01:08.905","Text":"that y is 1,400,"},{"Start":"01:08.905 ","End":"01:12.285","Text":"so if I plug 20 into here,"},{"Start":"01:12.285 ","End":"01:18.435","Text":"I can get that 1,000 times e^20k is 1,400."},{"Start":"01:18.435 ","End":"01:20.595","Text":"We want to solve this and get k,"},{"Start":"01:20.595 ","End":"01:23.235","Text":"so divide by 1,000,"},{"Start":"01:23.235 ","End":"01:25.490","Text":"you get e^20k is 1.4."},{"Start":"01:25.490 ","End":"01:29.720","Text":"Take logarithms and then we got 20k is the natural log of this,"},{"Start":"01:29.720 ","End":"01:33.420","Text":"so k is natural log of 1.4 over 20,"},{"Start":"01:33.420 ","End":"01:38.720","Text":"and with the calculator gives us 0.017,"},{"Start":"01:38.720 ","End":"01:47.055","Text":"which actually means that there\u0027s a growth rate by 1.7% growth rate."},{"Start":"01:47.055 ","End":"01:54.305","Text":"That means that we can now write the full formula for y in terms of t,"},{"Start":"01:54.305 ","End":"02:01.400","Text":"which is 1,000 times e to the power of this constant times t. Now we"},{"Start":"02:01.400 ","End":"02:08.720","Text":"want to continue because we want to find out when the population is 2,000,"},{"Start":"02:08.720 ","End":"02:10.430","Text":"so we just write an equation,"},{"Start":"02:10.430 ","End":"02:12.020","Text":"y(t) is 2,000,"},{"Start":"02:12.020 ","End":"02:18.950","Text":"and then we try to solve for t. What we get is we plug it into the equation."},{"Start":"02:18.950 ","End":"02:22.670","Text":"We know what y of t is and also here it\u0027s 2,000,"},{"Start":"02:22.670 ","End":"02:24.140","Text":"so we just work on this,"},{"Start":"02:24.140 ","End":"02:25.640","Text":"divide it by 1,000."},{"Start":"02:25.640 ","End":"02:28.110","Text":"We get this is 2, take the logarithm,"},{"Start":"02:28.110 ","End":"02:33.410","Text":"we get this, this is already becoming familiar, divided by 0.017."},{"Start":"02:33.410 ","End":"02:39.360","Text":"Use the calculator to approximate and we get 40.77 months, 3 years,"},{"Start":"02:39.360 ","End":"02:43.640","Text":"and 4.77 months, then we\u0027re going to have 2,000,"},{"Start":"02:43.640 ","End":"02:45.770","Text":"which is double the original population."},{"Start":"02:45.770 ","End":"02:48.480","Text":"Okay, we\u0027re done."}],"ID":7853},{"Watched":false,"Name":"Exercise 24","Duration":"3m 51s","ChapterTopicVideoID":7780,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:05.580","Text":"Here we have another word problem this time involving radioactive decay."},{"Start":"00:05.580 ","End":"00:12.300","Text":"Radioactive isotope carbon 14 has a half-life of 5,750 years."},{"Start":"00:12.300 ","End":"00:14.085","Text":"At any given moment,"},{"Start":"00:14.085 ","End":"00:17.415","Text":"its rate of decay is proportional to the present amount."},{"Start":"00:17.415 ","End":"00:20.070","Text":"That\u0027s what we mean by exponential decay,"},{"Start":"00:20.070 ","End":"00:25.455","Text":"is proportional referred to later when we get to here with constant or proportionality."},{"Start":"00:25.455 ","End":"00:26.970","Text":"Now, 2 questions."},{"Start":"00:26.970 ","End":"00:29.640","Text":"How many grams of this isotope will survive after"},{"Start":"00:29.640 ","End":"00:33.390","Text":"1000 years if there were 100 grams initially?"},{"Start":"00:33.390 ","End":"00:35.000","Text":"The second part of the question,"},{"Start":"00:35.000 ","End":"00:39.950","Text":"after how many years will they remain just 10 grams of the initial 100?"},{"Start":"00:39.950 ","End":"00:43.010","Text":"Now the half-life is going to help us to find the decay,"},{"Start":"00:43.010 ","End":"00:44.480","Text":"which is the proportionality."},{"Start":"00:44.480 ","End":"00:47.660","Text":"I want to remind you of a general formula that if we have"},{"Start":"00:47.660 ","End":"00:51.470","Text":"a quantity which decay is exponentially and at the start time,"},{"Start":"00:51.470 ","End":"00:53.270","Text":"we know the amount y naught."},{"Start":"00:53.270 ","End":"00:54.650","Text":"Then at any given time t,"},{"Start":"00:54.650 ","End":"00:58.325","Text":"we know that y is given by y naught e to the kt,"},{"Start":"00:58.325 ","End":"01:01.250","Text":"where k is the constant of proportionality."},{"Start":"01:01.250 ","End":"01:02.510","Text":"But in the case of decay,"},{"Start":"01:02.510 ","End":"01:06.335","Text":"it\u0027s going to be negative if it\u0027s not negative and it\u0027ll be suspicious."},{"Start":"01:06.335 ","End":"01:09.500","Text":"In case you forgot, the half-life is the time it will"},{"Start":"01:09.500 ","End":"01:14.160","Text":"take till they only remain half of the initial amount."},{"Start":"01:14.350 ","End":"01:16.700","Text":"Let\u0027s begin."},{"Start":"01:16.700 ","End":"01:24.260","Text":"First thing is that we know that after 5,750 years,"},{"Start":"01:24.260 ","End":"01:26.570","Text":"we\u0027re only going to have half the initial amount,"},{"Start":"01:26.570 ","End":"01:28.775","Text":"which is y_0 over 2."},{"Start":"01:28.775 ","End":"01:33.515","Text":"In general, we also can plug in the 5,750 here."},{"Start":"01:33.515 ","End":"01:41.990","Text":"We get the equation that y naught e^k times 5,750 is equal to this."},{"Start":"01:41.990 ","End":"01:44.060","Text":"The y naught will cancel."},{"Start":"01:44.060 ","End":"01:45.170","Text":"When we solve the equation,"},{"Start":"01:45.170 ","End":"01:47.690","Text":"we\u0027ll get k. First of all,"},{"Start":"01:47.690 ","End":"01:53.720","Text":"we cancel the y naught and then we take the log of both sides."},{"Start":"01:53.720 ","End":"01:58.550","Text":"Then the logarithm of e to the power of something is"},{"Start":"01:58.550 ","End":"02:03.740","Text":"just that thing itself because natural log is the inverse of the exponential function."},{"Start":"02:03.740 ","End":"02:11.165","Text":"We\u0027ll just bring this over to the other side and use a calculator to do this computation."},{"Start":"02:11.165 ","End":"02:14.780","Text":"We get that the constant of proportionality is roughly this."},{"Start":"02:14.780 ","End":"02:15.950","Text":"Certainly, it\u0027s negative,"},{"Start":"02:15.950 ","End":"02:18.370","Text":"which is what we were expecting."},{"Start":"02:18.370 ","End":"02:23.750","Text":"Once we have k, we can write down the formula that we originally had except with"},{"Start":"02:23.750 ","End":"02:28.670","Text":"k plugged in so we fully know now at any time t what the amount is."},{"Start":"02:28.670 ","End":"02:31.250","Text":"Let\u0027s see. In part a,"},{"Start":"02:31.250 ","End":"02:38.165","Text":"we had to find a 1000 years later how much of this stuff would be left,"},{"Start":"02:38.165 ","End":"02:39.950","Text":"from the original 100s."},{"Start":"02:39.950 ","End":"02:45.638","Text":"We just do 100e to the power of this and this and this times 1000,"},{"Start":"02:45.638 ","End":"02:47.089","Text":"and on the calculator,"},{"Start":"02:47.089 ","End":"02:50.930","Text":"we get 88.69 grams."},{"Start":"02:50.930 ","End":"02:54.455","Text":"Hasn\u0027t dropped by very much in 1000 years."},{"Start":"02:54.455 ","End":"02:58.955","Text":"Let\u0027s get on to part b. I\u0027ll do it on another page. Here we are."},{"Start":"02:58.955 ","End":"03:01.160","Text":"I\u0027ll just remind you of the formula again,"},{"Start":"03:01.160 ","End":"03:03.530","Text":"that after any time t,"},{"Start":"03:03.530 ","End":"03:06.199","Text":"the amount is given by this formula."},{"Start":"03:06.199 ","End":"03:09.350","Text":"The 100 was the initial amount and this constant is the"},{"Start":"03:09.350 ","End":"03:13.505","Text":"k. But we want to know when it\u0027s going to be 10 grams here."},{"Start":"03:13.505 ","End":"03:16.850","Text":"We get the equation that this thing is equal to 10."},{"Start":"03:16.850 ","End":"03:19.010","Text":"Now divide by a 100."},{"Start":"03:19.010 ","End":"03:22.805","Text":"We get here on the right, 0.1."},{"Start":"03:22.805 ","End":"03:25.820","Text":"Then take the logarithm of both sides."},{"Start":"03:25.820 ","End":"03:27.410","Text":"This is what we get,"},{"Start":"03:27.410 ","End":"03:32.270","Text":"and then divide by the coefficient of t and we get this over this,"},{"Start":"03:32.270 ","End":"03:40.145","Text":"which on the calculator gives us that the answer is 19,188 years."},{"Start":"03:40.145 ","End":"03:43.700","Text":"In other words, it takes almost 20,000 years for it to go"},{"Start":"03:43.700 ","End":"03:47.285","Text":"down from 100 grams to 10 grams."},{"Start":"03:47.285 ","End":"03:49.430","Text":"Anyway, that\u0027s the answer to part b,"},{"Start":"03:49.430 ","End":"03:51.660","Text":"and we are done."}],"ID":7854},{"Watched":false,"Name":"Exercise 25","Duration":"4m 13s","ChapterTopicVideoID":7781,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.880","Text":"Here we have another word problem exercise."},{"Start":"00:02.880 ","End":"00:06.525","Text":"In a certain pool there are 240 tons of fish"},{"Start":"00:06.525 ","End":"00:10.635","Text":"and the quantity of fish in it increases by 4 percent each week."},{"Start":"00:10.635 ","End":"00:12.825","Text":"Because it\u0027s 4 percent of the amount,"},{"Start":"00:12.825 ","End":"00:18.900","Text":"and that means that the growth is exponential and the constant is 4 percent."},{"Start":"00:18.900 ","End":"00:22.320","Text":"In a second pool, there are 200 tons of fish and"},{"Start":"00:22.320 ","End":"00:25.965","Text":"the quantity of fish in it increases by 10 percent each week."},{"Start":"00:25.965 ","End":"00:27.765","Text":"That\u0027s also exponential."},{"Start":"00:27.765 ","End":"00:33.165","Text":"The question is, after how many weeks will both pools have the same quantity of fish?"},{"Start":"00:33.165 ","End":"00:38.585","Text":"This is reasonable because the first pool has more fish but it grows more slowly,"},{"Start":"00:38.585 ","End":"00:40.535","Text":"so eventually it will catch up."},{"Start":"00:40.535 ","End":"00:42.530","Text":"Second question."},{"Start":"00:42.530 ","End":"00:44.930","Text":"After how many weeks will the second pool have"},{"Start":"00:44.930 ","End":"00:47.945","Text":"twice the quantity of fish as the first pool?"},{"Start":"00:47.945 ","End":"00:49.290","Text":"Now we have 2 pools,"},{"Start":"00:49.290 ","End":"00:55.145","Text":"so we\u0027re going to call the amount in one pool x and the other one y."},{"Start":"00:55.145 ","End":"00:56.840","Text":"Let\u0027s start with the first pool."},{"Start":"00:56.840 ","End":"00:58.310","Text":"The amount will be x,"},{"Start":"00:58.310 ","End":"01:02.138","Text":"we\u0027ll call it and k will be the constant of proportionality,"},{"Start":"01:02.138 ","End":"01:06.635","Text":"and in our case x_0 the starting amount is 240,"},{"Start":"01:06.635 ","End":"01:10.400","Text":"which will be in tons and the constant of"},{"Start":"01:10.400 ","End":"01:16.850","Text":"proportionality is 0.04 because that\u0027s exactly 4 percent."},{"Start":"01:16.850 ","End":"01:20.389","Text":"Of course t we\u0027re measuring in weeks,"},{"Start":"01:20.389 ","End":"01:23.015","Text":"I should say the unit of time for t is weeks."},{"Start":"01:23.015 ","End":"01:24.995","Text":"Makes sense everything is in weeks."},{"Start":"01:24.995 ","End":"01:26.720","Text":"As for the second pool,"},{"Start":"01:26.720 ","End":"01:32.990","Text":"we\u0027ll call that one y and we\u0027ll call the constant of proportionality m. In this case,"},{"Start":"01:32.990 ","End":"01:37.580","Text":"m being 10 percent is 0.1 and the initial amount"},{"Start":"01:37.580 ","End":"01:43.100","Text":"is 200 and we\u0027re measuring it in ton, that\u0027s what this subscript ton means."},{"Start":"01:43.100 ","End":"01:47.865","Text":"Now the equation that we\u0027re looking for in Part A is,"},{"Start":"01:47.865 ","End":"01:50.040","Text":"we want to know at what time t,"},{"Start":"01:50.040 ","End":"01:51.720","Text":"t is in question mark,"},{"Start":"01:51.720 ","End":"01:53.515","Text":"does x(t) equal y(t),"},{"Start":"01:53.515 ","End":"01:56.420","Text":"the quantity in both pools, the same."},{"Start":"01:56.420 ","End":"01:58.310","Text":"Scroll down a bit."},{"Start":"01:58.310 ","End":"02:02.750","Text":"So what we get is we just plug in into the formula."},{"Start":"02:02.750 ","End":"02:05.645","Text":"Here, x(t) is given by this formula,"},{"Start":"02:05.645 ","End":"02:14.640","Text":"but I\u0027m going to copy it and also replace x_0 and K. It\u0027s 240e^0.04t,"},{"Start":"02:14.640 ","End":"02:18.880","Text":"and here it\u0027s 200 times e^0.1t."},{"Start":"02:18.880 ","End":"02:21.065","Text":"So now we have an equation."},{"Start":"02:21.065 ","End":"02:25.560","Text":"We\u0027ll take the numbers to the left and the exponents to the right."},{"Start":"02:25.560 ","End":"02:27.140","Text":"We get this."},{"Start":"02:27.140 ","End":"02:30.620","Text":"Now this over this is 1.2,"},{"Start":"02:30.620 ","End":"02:32.840","Text":"and remember when we divide,"},{"Start":"02:32.840 ","End":"02:39.500","Text":"we can subtract the exponent 0.1 minus 0.04 is 0.06."},{"Start":"02:39.500 ","End":"02:43.106","Text":"Then we can take the logarithm of both sides,"},{"Start":"02:43.106 ","End":"02:48.170","Text":"and finally, we can isolate t by dividing this by this,"},{"Start":"02:48.170 ","End":"02:52.020","Text":"which gives us that t is approximately,"},{"Start":"02:52.020 ","End":"02:58.115","Text":"just slightly over 3 weeks in order for it to equal the quantity in the first pool."},{"Start":"02:58.115 ","End":"03:01.235","Text":"Now let\u0027s get to the second problem."},{"Start":"03:01.235 ","End":"03:03.140","Text":"Second problem B,"},{"Start":"03:03.140 ","End":"03:05.390","Text":"is what we want to know is the fish in"},{"Start":"03:05.390 ","End":"03:09.050","Text":"the second pool will be twice the amount of the first pool."},{"Start":"03:09.050 ","End":"03:15.145","Text":"So y(t) is twice x(t), and that gives us the equation,"},{"Start":"03:15.145 ","End":"03:16.970","Text":"pretty much like we had in the beginning,"},{"Start":"03:16.970 ","End":"03:19.610","Text":"except there\u0027s an extra factor of 2 here."},{"Start":"03:19.610 ","End":"03:21.935","Text":"Amount in second pool,"},{"Start":"03:21.935 ","End":"03:25.265","Text":"amount in first pool at the time t, and this is twice."},{"Start":"03:25.265 ","End":"03:34.835","Text":"This time we just have to divide this times this is 480 divide it here and bring the 0.1."},{"Start":"03:34.835 ","End":"03:36.260","Text":"We\u0027ll do it the other way round."},{"Start":"03:36.260 ","End":"03:38.930","Text":"Sorry. This one goes over to this side,"},{"Start":"03:38.930 ","End":"03:40.010","Text":"this goes to this side."},{"Start":"03:40.010 ","End":"03:41.885","Text":"Here we get 480/200,"},{"Start":"03:41.885 ","End":"03:46.400","Text":"and here we get e^0.1 minus 0.04,"},{"Start":"03:46.400 ","End":"03:50.120","Text":"again 0.06, and this thing is 2.4."},{"Start":"03:50.120 ","End":"03:54.485","Text":"We take the log of both sides and we get this."},{"Start":"03:54.485 ","End":"03:57.340","Text":"Then we isolate t by just taking this,"},{"Start":"03:57.340 ","End":"04:00.170","Text":"log(2.4), dividing by 0.6,"},{"Start":"04:00.170 ","End":"04:05.540","Text":"pulling out a calculator and we see that it occurs after about 14.5"},{"Start":"04:05.540 ","End":"04:13.260","Text":"weeks that the second pool get double the fish that the first pool did and we\u0027re done."}],"ID":7855},{"Watched":false,"Name":"Exercise 26","Duration":"7m 51s","ChapterTopicVideoID":7782,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.675","Text":"Here we have quite a lengthy word problem. Let\u0027s read it."},{"Start":"00:03.675 ","End":"00:05.955","Text":"At time t=0,"},{"Start":"00:05.955 ","End":"00:11.655","Text":"a tank contains 4 kilograms of salt dissolved in 200 liters of water."},{"Start":"00:11.655 ","End":"00:17.040","Text":"Saltwater at 0.2 kilograms per liter of water is flowing"},{"Start":"00:17.040 ","End":"00:22.890","Text":"into the tank at a rate of 25 liters per minute and simultaneously,"},{"Start":"00:22.890 ","End":"00:27.480","Text":"the mixed solution is draining out of the tank at the same rate."},{"Start":"00:27.480 ","End":"00:29.655","Text":"Then there are questions."},{"Start":"00:29.655 ","End":"00:32.909","Text":"Compute the amount of salt in the tank after 8 minutes."},{"Start":"00:32.909 ","End":"00:37.905","Text":"After how long will the amount of salt in the tank be twice the initial amount?"},{"Start":"00:37.905 ","End":"00:40.250","Text":"Let\u0027s decide what our units are going to be."},{"Start":"00:40.250 ","End":"00:43.805","Text":"Well, it looks like time is going to be measured in minutes."},{"Start":"00:43.805 ","End":"00:46.700","Text":"We\u0027re going to have minutes for time."},{"Start":"00:46.700 ","End":"00:50.750","Text":"Volume is going to be in liters everywhere it talks about"},{"Start":"00:50.750 ","End":"00:55.170","Text":"liters and mass will be in kilograms."},{"Start":"00:55.170 ","End":"00:57.379","Text":"These are the units we\u0027re going to be using,"},{"Start":"00:57.379 ","End":"01:00.965","Text":"minutes, liters, and kilograms."},{"Start":"01:00.965 ","End":"01:07.490","Text":"The general idea is that a tank contains saltwater at a certain concentration."},{"Start":"01:07.490 ","End":"01:11.600","Text":"Water\u0027s flowing in and water\u0027s flowing out at the same rate."},{"Start":"01:11.600 ","End":"01:16.820","Text":"The concentration of the salt in the water is going to change over time and we\u0027re"},{"Start":"01:16.820 ","End":"01:19.190","Text":"going to get a differential equation to express"},{"Start":"01:19.190 ","End":"01:22.055","Text":"how the amount of salt in the water is changing."},{"Start":"01:22.055 ","End":"01:29.870","Text":"Let y be the amount of salt in kilograms at time t. That\u0027s our dependent variable."},{"Start":"01:29.870 ","End":"01:31.520","Text":"The independent variable is time,"},{"Start":"01:31.520 ","End":"01:35.615","Text":"the dependent variable is the number of kilograms of salt in the water."},{"Start":"01:35.615 ","End":"01:40.845","Text":"What we know already is that we have at time 0,"},{"Start":"01:40.845 ","End":"01:42.740","Text":"there are 4 kilograms."},{"Start":"01:42.740 ","End":"01:48.110","Text":"Part a asks us to compute the amount of salt after 8 minutes,"},{"Start":"01:48.110 ","End":"01:54.390","Text":"which is y(8) and I guess part b is at what time?"},{"Start":"01:54.410 ","End":"01:58.370","Text":"Will the amount be twice the initial amount?"},{"Start":"01:58.370 ","End":"02:01.070","Text":"Well, let\u0027s see. Initial amount is 4, 4 times 2 is 8."},{"Start":"02:01.070 ","End":"02:04.695","Text":"This is part a,"},{"Start":"02:04.695 ","End":"02:07.520","Text":"this is part b and all we need is"},{"Start":"02:07.520 ","End":"02:11.510","Text":"a differential equation to apply this initial condition to and solve."},{"Start":"02:11.510 ","End":"02:14.600","Text":"What we can say is that the dy/dt"},{"Start":"02:14.600 ","End":"02:17.840","Text":"is the rate of change of the amount of salt in the tank."},{"Start":"02:17.840 ","End":"02:22.315","Text":"Now, this rate of change is made up of 2 bits."},{"Start":"02:22.315 ","End":"02:24.800","Text":"There\u0027s the rate of change that\u0027s coming"},{"Start":"02:24.800 ","End":"02:28.745","Text":"in and the rate of change of the salt that\u0027s leaving."},{"Start":"02:28.745 ","End":"02:33.455","Text":"What we\u0027re going to call it is the influx rate and the outflux rate."},{"Start":"02:33.455 ","End":"02:38.260","Text":"This influx rate is of the amount of salt and also the outflux rate."},{"Start":"02:38.260 ","End":"02:40.730","Text":"We\u0027re talking about amount of salt."},{"Start":"02:40.730 ","End":"02:43.145","Text":"Let\u0027s get each of these separately."},{"Start":"02:43.145 ","End":"02:48.200","Text":"The influx rate we can compute because the water is"},{"Start":"02:48.200 ","End":"02:53.425","Text":"flowing in at a concentration of 0.2 kilograms per liter."},{"Start":"02:53.425 ","End":"02:55.610","Text":"At a rate of 25 liters per minute,"},{"Start":"02:55.610 ","End":"02:58.985","Text":"we can cancel the liters and get it in kilograms per minute,"},{"Start":"02:58.985 ","End":"03:02.000","Text":"so the salt is coming in at 5 kilograms a minute."},{"Start":"03:02.000 ","End":"03:05.825","Text":"The outflux rate is dependent on y."},{"Start":"03:05.825 ","End":"03:08.975","Text":"Let me explain these quantities."},{"Start":"03:08.975 ","End":"03:11.690","Text":"Y(t)/200 needs explanation."},{"Start":"03:11.690 ","End":"03:14.150","Text":"At any given moment t,"},{"Start":"03:14.150 ","End":"03:16.220","Text":"the amount of salt is y(t)."},{"Start":"03:16.220 ","End":"03:19.820","Text":"Its concentration is the number of"},{"Start":"03:19.820 ","End":"03:24.410","Text":"kilograms over the volume number of liters, which is 200."},{"Start":"03:24.410 ","End":"03:28.715","Text":"Y(t)/200 is the concentration of salt at any given moment."},{"Start":"03:28.715 ","End":"03:33.350","Text":"Then we multiply this concentration by the number of liters per minute."},{"Start":"03:33.350 ","End":"03:36.450","Text":"The 25 is going to be the same as this because that was given."},{"Start":"03:36.450 ","End":"03:38.150","Text":"If you multiply this by this,"},{"Start":"03:38.150 ","End":"03:41.930","Text":"this is what we get y(t)/8 and this is in kilograms per minute."},{"Start":"03:41.930 ","End":"03:43.310","Text":"To get the rate of change,"},{"Start":"03:43.310 ","End":"03:45.440","Text":"we have to subtract these 2."},{"Start":"03:45.440 ","End":"03:47.060","Text":"I\u0027m going to continue on another page,"},{"Start":"03:47.060 ","End":"03:52.760","Text":"but let\u0027s just remember what we said that the rate of change is influx minus outflux."},{"Start":"03:52.760 ","End":"04:00.355","Text":"It\u0027s going to be 5 minus y/8 and our initial condition is going to be y(0)=4."},{"Start":"04:00.355 ","End":"04:02.325","Text":"Here we are, like I said,"},{"Start":"04:02.325 ","End":"04:06.480","Text":"the rate of change dy/dt is 5 minus y/8,"},{"Start":"04:06.480 ","End":"04:10.065","Text":"and our initial condition, y(0) equals 4."},{"Start":"04:10.065 ","End":"04:14.570","Text":"I\u0027m going to bring something to the other side, the y/8."},{"Start":"04:14.570 ","End":"04:18.170","Text":"What I get is a linear differential equation."},{"Start":"04:18.170 ","End":"04:21.710","Text":"We get this as I call it y prime plus 1/8(y),"},{"Start":"04:21.710 ","End":"04:24.575","Text":"which is this equals 5."},{"Start":"04:24.575 ","End":"04:29.670","Text":"This is the standard form of a linear differential equation of degree 1."},{"Start":"04:29.670 ","End":"04:32.395","Text":"This is our a(t) and this is b(t)."},{"Start":"04:32.395 ","End":"04:34.535","Text":"They both happen to be constants."},{"Start":"04:34.535 ","End":"04:36.500","Text":"Let\u0027s remember the formula,"},{"Start":"04:36.500 ","End":"04:40.560","Text":"which is that y(t) equals this expression,"},{"Start":"04:40.560 ","End":"04:44.865","Text":"where A(t) is just the integral of a(t)."},{"Start":"04:44.865 ","End":"04:47.925","Text":"First, let\u0027s compute a(t)."},{"Start":"04:47.925 ","End":"04:54.390","Text":"The words I\u0027m basically saying that this bit A is my asterisk and this expression here,"},{"Start":"04:54.390 ","End":"04:56.900","Text":"I\u0027ll call this 1 double asterisk,"},{"Start":"04:56.900 ","End":"04:59.060","Text":"and these will be side calculations."},{"Start":"04:59.060 ","End":"05:03.395","Text":"Now, the integral of this is just the integral"},{"Start":"05:03.395 ","End":"05:08.970","Text":"of A(t) is 1/8 and the integral of 1/8 is just 1/8t."},{"Start":"05:08.970 ","End":"05:12.410","Text":"This expression, the double asterisk is the integral."},{"Start":"05:12.410 ","End":"05:17.615","Text":"Now b(t) is 5 and e^a(t),"},{"Start":"05:17.615 ","End":"05:19.620","Text":"we\u0027ve already computed a(t) here,"},{"Start":"05:19.620 ","End":"05:24.320","Text":"so we substitute this in here to maybe highlight that to emphasize it."},{"Start":"05:24.320 ","End":"05:26.210","Text":"This is the integral."},{"Start":"05:26.210 ","End":"05:28.595","Text":"It\u0027s 5 divided by 1/8,"},{"Start":"05:28.595 ","End":"05:30.395","Text":"which is like 5 times 8,"},{"Start":"05:30.395 ","End":"05:33.610","Text":"which is 40 e^1/8t."},{"Start":"05:33.610 ","End":"05:35.925","Text":"Now just have to put it all together,"},{"Start":"05:35.925 ","End":"05:42.855","Text":"scroll up a bit and get that y(t) is from the formula,"},{"Start":"05:42.855 ","End":"05:45.810","Text":"e^ minus a(t),"},{"Start":"05:45.810 ","End":"05:50.565","Text":"which is e^minus 1/8t times this expression."},{"Start":"05:50.565 ","End":"05:54.565","Text":"We should add the class constants of this is plus C,"},{"Start":"05:54.565 ","End":"06:00.960","Text":"so when we multiply this with this just gives 40 because minus 8t and plus 8t,"},{"Start":"06:00.960 ","End":"06:05.685","Text":"they cancel and we just get Ce^ minus 1/8t."},{"Start":"06:05.685 ","End":"06:07.995","Text":"This is the expression for y(t)."},{"Start":"06:07.995 ","End":"06:17.260","Text":"Now we can plug in the initial condition that y(0)=4 and so we put y is 4 and t is 0,"},{"Start":"06:17.260 ","End":"06:19.230","Text":"so we get t is 0,"},{"Start":"06:19.230 ","End":"06:20.850","Text":"this is e^0 is 1,"},{"Start":"06:20.850 ","End":"06:23.310","Text":"so we get 40 plus C=4."},{"Start":"06:23.310 ","End":"06:25.610","Text":"C is minus 36."},{"Start":"06:25.610 ","End":"06:30.605","Text":"Then put that minus 36 back here and we get that"},{"Start":"06:30.605 ","End":"06:37.210","Text":"y(t) is 40 minus 36 e^minus 1/8t."},{"Start":"06:37.210 ","End":"06:42.065","Text":"This is the expression for y in a given time but we had 2 questions."},{"Start":"06:42.065 ","End":"06:44.060","Text":"We had question a,"},{"Start":"06:44.060 ","End":"06:52.520","Text":"which was to figure out what is y(8) and so will we have to do is put t=8 in here."},{"Start":"06:52.520 ","End":"06:55.505","Text":"T=8 e^ minus 1."},{"Start":"06:55.505 ","End":"06:58.250","Text":"This is what it is as an expression and with the calculator,"},{"Start":"06:58.250 ","End":"07:03.050","Text":"we get 26.75 and the answer is in kilograms."},{"Start":"07:03.050 ","End":"07:06.245","Text":"The second question we asked was,"},{"Start":"07:06.245 ","End":"07:10.320","Text":"when is y(t)=8? What is t?"},{"Start":"07:10.320 ","End":"07:15.740","Text":"We have an equation that this general term for t is equal to 8"},{"Start":"07:15.740 ","End":"07:21.425","Text":"and we have to find or solve for t. If we bring 8 to the other side,"},{"Start":"07:21.425 ","End":"07:24.170","Text":"it\u0027s 32 and then divide by 36."},{"Start":"07:24.170 ","End":"07:29.725","Text":"Basically, we get this and then we get the minus 1/8(t)."},{"Start":"07:29.725 ","End":"07:33.060","Text":"32 over 36 is 8/9 if we cancel by 4,"},{"Start":"07:33.060 ","End":"07:34.710","Text":"so we get this expression."},{"Start":"07:34.710 ","End":"07:39.360","Text":"Finally, the t is equal to minus 8,"},{"Start":"07:39.360 ","End":"07:40.910","Text":"you bring the minus 8 to the other side,"},{"Start":"07:40.910 ","End":"07:42.830","Text":"natural log of 8/9,"},{"Start":"07:42.830 ","End":"07:47.660","Text":"and the answer comes out to be 0.942 minutes."},{"Start":"07:47.660 ","End":"07:51.240","Text":"This was long, but we\u0027re finally done."}],"ID":7856},{"Watched":false,"Name":"Exercise 27","Duration":"8m 13s","ChapterTopicVideoID":7783,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.080","Text":"Here we have another word problem in physics."},{"Start":"00:04.080 ","End":"00:10.005","Text":"Let\u0027s read it. A rowboat is initially towed at a rate of 12 kilometers per hour."},{"Start":"00:10.005 ","End":"00:11.985","Text":"At time t equals 0,"},{"Start":"00:11.985 ","End":"00:15.800","Text":"the cable is released and the man in the boat starts rowing in"},{"Start":"00:15.800 ","End":"00:20.810","Text":"the direction of the motion and applies a force of 20 newtons to the boat."},{"Start":"00:20.810 ","End":"00:24.890","Text":"The mass of the boat and rower is 500 kilograms"},{"Start":"00:24.890 ","End":"00:29.580","Text":"and the resistance in newton is given by 2v,"},{"Start":"00:29.580 ","End":"00:33.725","Text":"where v is the velocity of the boat in meters per second."},{"Start":"00:33.725 ","End":"00:35.705","Text":"There are 3 questions."},{"Start":"00:35.705 ","End":"00:37.505","Text":"The first question is,"},{"Start":"00:37.505 ","End":"00:41.090","Text":"what is the velocity of the boat after 30 seconds?"},{"Start":"00:41.090 ","End":"00:43.070","Text":"The second question is,"},{"Start":"00:43.070 ","End":"00:46.985","Text":"when will the velocity of the boat be 5 meters per second?"},{"Start":"00:46.985 ","End":"00:48.800","Text":"The third question is,"},{"Start":"00:48.800 ","End":"00:51.680","Text":"to find the asymptotic velocity of the boat."},{"Start":"00:51.680 ","End":"00:57.590","Text":"That means what the velocity of the boat tends to as t goes to infinity,"},{"Start":"00:57.590 ","End":"01:00.275","Text":"what it eventually goes toward."},{"Start":"01:00.275 ","End":"01:01.670","Text":"One of the questions in physics,"},{"Start":"01:01.670 ","End":"01:04.670","Text":"I usually like to say what the units are that we\u0027re using"},{"Start":"01:04.670 ","End":"01:09.650","Text":"so we\u0027re going to measure time in seconds."},{"Start":"01:09.650 ","End":"01:11.330","Text":"Distance will be in meters,"},{"Start":"01:11.330 ","End":"01:15.155","Text":"meaning that velocity is going to be in meters per second."},{"Start":"01:15.155 ","End":"01:19.550","Text":"Which also means that when we see kilometers per hour,"},{"Start":"01:19.550 ","End":"01:23.255","Text":"we\u0027re going to have to convert that to meters per second."},{"Start":"01:23.255 ","End":"01:27.455","Text":"Force is going to be measured in newtons,"},{"Start":"01:27.455 ","End":"01:32.630","Text":"and mass is going to be measured in kilograms."},{"Start":"01:32.630 ","End":"01:34.370","Text":"Those are the units we\u0027re using;"},{"Start":"01:34.370 ","End":"01:35.900","Text":"velocity in meters per second,"},{"Start":"01:35.900 ","End":"01:37.909","Text":"force in newtons, weight in kilograms,"},{"Start":"01:37.909 ","End":"01:39.965","Text":"and there\u0027s one conversion to do."},{"Start":"01:39.965 ","End":"01:44.870","Text":"Let\u0027s start and we\u0027re going to be using Newton\u0027s second law,"},{"Start":"01:44.870 ","End":"01:48.155","Text":"basically says that force equals mass times acceleration."},{"Start":"01:48.155 ","End":"01:50.000","Text":"He talks always about a body."},{"Start":"01:50.000 ","End":"01:53.220","Text":"In our case, the body is the boat."},{"Start":"01:53.930 ","End":"01:58.040","Text":"V is the velocity as a function of time,"},{"Start":"01:58.040 ","End":"02:02.765","Text":"and let\u0027s say velocity of the body or the boat and we know that the mass"},{"Start":"02:02.765 ","End":"02:08.575","Text":"is 500 kilograms and acceleration is,"},{"Start":"02:08.575 ","End":"02:09.920","Text":"what we talked about here,"},{"Start":"02:09.920 ","End":"02:14.000","Text":"is actually the rate of change of velocity is acceleration is"},{"Start":"02:14.000 ","End":"02:19.705","Text":"dv/dt and it\u0027s actually measured in meters per second squared."},{"Start":"02:19.705 ","End":"02:23.089","Text":"There\u0027s two forces acting upon the body."},{"Start":"02:23.089 ","End":"02:27.425","Text":"There\u0027s one that moves it forwards because the man is rowing,"},{"Start":"02:27.425 ","End":"02:30.210","Text":"but there\u0027s also the resistance of the water."},{"Start":"02:30.210 ","End":"02:34.070","Text":"What I\u0027m saying is that force on the body is the resistance which"},{"Start":"02:34.070 ","End":"02:37.880","Text":"is pushing it one way and the rowing force moving it the other way."},{"Start":"02:37.880 ","End":"02:40.040","Text":"You have something that is reversed of course."},{"Start":"02:40.040 ","End":"02:45.590","Text":"I\u0027m sorry, it\u0027s actually the rowing force minus the resistance."},{"Start":"02:45.590 ","End":"02:47.870","Text":"Of course, this is moving it forward, this is moving it backward."},{"Start":"02:47.870 ","End":"02:49.745","Text":"The rowing force is 20."},{"Start":"02:49.745 ","End":"02:56.210","Text":"The resistance we are given is 2v so this is the expression for the force on the body."},{"Start":"02:56.210 ","End":"02:57.725","Text":"Sorry, I got it backwards."},{"Start":"02:57.725 ","End":"03:06.320","Text":"Next, we can say that we have the differential equation using the Newton\u0027s second law,"},{"Start":"03:06.320 ","End":"03:08.825","Text":"mass which is 500,"},{"Start":"03:08.825 ","End":"03:14.760","Text":"that\u0027s the mass and the acceleration is dv/dt is equal to the force."},{"Start":"03:14.760 ","End":"03:17.750","Text":"This already we found was the force, which is this."},{"Start":"03:17.750 ","End":"03:20.450","Text":"This is Newton\u0027s second law and it\u0027s"},{"Start":"03:20.450 ","End":"03:25.700","Text":"a differential equation in v. We know the initial condition that at time 0,"},{"Start":"03:25.700 ","End":"03:29.305","Text":"this is 12 kilometers per hour."},{"Start":"03:29.305 ","End":"03:34.970","Text":"Later on, we\u0027ll convert this to meters per second when we\u0027re ready for it too,"},{"Start":"03:34.970 ","End":"03:36.380","Text":"will do the conversion."},{"Start":"03:36.380 ","End":"03:44.105","Text":"If I divide both sides here by 500 and bring the terms with v over to the left."},{"Start":"03:44.105 ","End":"03:45.380","Text":"Then what I\u0027ll get is"},{"Start":"03:45.380 ","End":"03:49.590","Text":"a linear differential equation"},{"Start":"03:49.590 ","End":"03:54.500","Text":"and we\u0027re going to use the formula for linear differential equations."},{"Start":"03:54.500 ","End":"03:58.850","Text":"This is it. Big A is the integral of little a."},{"Start":"03:58.850 ","End":"04:04.905","Text":"What we should do is I\u0027m going to call the A(t) will be an asterisk."},{"Start":"04:04.905 ","End":"04:09.835","Text":"This expression is going to be double asterisk."},{"Start":"04:09.835 ","End":"04:12.740","Text":"I\u0027ll compute those as side exercises."},{"Start":"04:12.740 ","End":"04:17.450","Text":"In fact, I\u0027ll expose them both at the same time A(t), as I say,"},{"Start":"04:17.450 ","End":"04:21.130","Text":"it\u0027s the integral of little a(t) it\u0027s an integral of a constant so"},{"Start":"04:21.130 ","End":"04:25.115","Text":"it\u0027s that constant times t double asterisk expression,"},{"Start":"04:25.115 ","End":"04:30.305","Text":"which actually should extend to all the way up to here."},{"Start":"04:30.305 ","End":"04:34.850","Text":"This is equal to 1/25,"},{"Start":"04:34.850 ","End":"04:37.715","Text":"which is b(t), that\u0027s this."},{"Start":"04:37.715 ","End":"04:43.390","Text":"Then e^A(t) we found here 1/250t."},{"Start":"04:43.390 ","End":"04:45.710","Text":"I just copied it from here to here,"},{"Start":"04:45.710 ","End":"04:51.015","Text":"dt which is equal to the 1/25 stays,"},{"Start":"04:51.015 ","End":"04:52.130","Text":"and the integral of this,"},{"Start":"04:52.130 ","End":"04:57.165","Text":"we have to divide itself divided by 1/250,"},{"Start":"04:57.165 ","End":"04:59.720","Text":"which means multiplying by 250."},{"Start":"04:59.720 ","End":"05:06.065","Text":"Basically what I\u0027m saying is if I get 1/25 and divide it by 1/250,"},{"Start":"05:06.065 ","End":"05:10.550","Text":"it\u0027s like 1/5 times 250,"},{"Start":"05:10.550 ","End":"05:13.160","Text":"and that equals 10, and that\u0027s the 10 here."},{"Start":"05:13.160 ","End":"05:18.275","Text":"We\u0027ve got this bit and we\u0027ve got the A(t) so now we just have to plug everything in."},{"Start":"05:18.275 ","End":"05:21.855","Text":"What we get is the following."},{"Start":"05:21.855 ","End":"05:24.830","Text":"That v(t) equals e to the minus,"},{"Start":"05:24.830 ","End":"05:27.455","Text":"I\u0027m just substituting A(t) is this,"},{"Start":"05:27.455 ","End":"05:29.945","Text":"double asterisk is this,"},{"Start":"05:29.945 ","End":"05:31.775","Text":"and plus the constant."},{"Start":"05:31.775 ","End":"05:33.350","Text":"Sorry, I\u0027m running out of page."},{"Start":"05:33.350 ","End":"05:35.270","Text":"I\u0027m going to move to the next page."},{"Start":"05:35.270 ","End":"05:37.130","Text":"If you multiply out,"},{"Start":"05:37.130 ","End":"05:39.080","Text":"you\u0027ll see that this term and this term"},{"Start":"05:39.080 ","End":"05:42.320","Text":"cancel because of the minus and plus the exponent."},{"Start":"05:42.320 ","End":"05:50.630","Text":"What we get is just 10c e to the minus 1/250t."},{"Start":"05:50.630 ","End":"05:55.955","Text":"Now we know that v(0) and already in kilometers per hour."},{"Start":"05:55.955 ","End":"06:01.940","Text":"If we convert a kilometer is 1,000 meters and an hour is 3,600 seconds,"},{"Start":"06:01.940 ","End":"06:08.285","Text":"we get the velocity at time 0 in terms of meters per second, it\u0027s 3.33."},{"Start":"06:08.285 ","End":"06:10.670","Text":"This is going to help us to find the constant."},{"Start":"06:10.670 ","End":"06:13.895","Text":"If we put t equals 0 here,"},{"Start":"06:13.895 ","End":"06:16.280","Text":"then what we get is v(0),"},{"Start":"06:16.280 ","End":"06:19.984","Text":"which is 3.33, approximately."},{"Start":"06:19.984 ","End":"06:23.160","Text":"That is going to equal 10 plus, its speed is 1."},{"Start":"06:23.160 ","End":"06:25.360","Text":"If t is 0, e^0 is 1,"},{"Start":"06:25.360 ","End":"06:29.855","Text":"so it\u0027s plus c and somehow the constant has shrunk to a little c, never mind."},{"Start":"06:29.855 ","End":"06:34.840","Text":"That gives us c is approximately minus 6.67."},{"Start":"06:34.840 ","End":"06:38.515","Text":"We have now the expression that we can put c in here."},{"Start":"06:38.515 ","End":"06:46.100","Text":"We have v(t) is approximately equal to 10 minus 6.67 e to the minus t over 250."},{"Start":"06:46.100 ","End":"06:49.730","Text":"Now we can finally get to the 3 questions A,"},{"Start":"06:49.730 ","End":"06:52.370","Text":"B, and C. In Part A,"},{"Start":"06:52.370 ","End":"06:57.320","Text":"we were given to find out what the velocity is when t is 30,"},{"Start":"06:57.320 ","End":"07:01.955","Text":"so it\u0027s just a substitution to put t equals 30 in here and a calculator job."},{"Start":"07:01.955 ","End":"07:03.290","Text":"This is the answer."},{"Start":"07:03.290 ","End":"07:07.700","Text":"In Part B, we had a reverse question is that we don\u0027t know t,"},{"Start":"07:07.700 ","End":"07:10.925","Text":"but we know that the velocity is 5, what is t?"},{"Start":"07:10.925 ","End":"07:12.440","Text":"We just solve this equation,"},{"Start":"07:12.440 ","End":"07:16.405","Text":"just substitute and get this thing is equal to 5."},{"Start":"07:16.405 ","End":"07:19.440","Text":"Then just bring this to this side,"},{"Start":"07:19.440 ","End":"07:22.085","Text":"bring the 5 over to that side, divide,"},{"Start":"07:22.085 ","End":"07:24.605","Text":"take the logarithm of both sides,"},{"Start":"07:24.605 ","End":"07:28.580","Text":"and then multiply by minus 250."},{"Start":"07:28.580 ","End":"07:34.430","Text":"In the end, we get this thing times minus 250 is 72."},{"Start":"07:34.430 ","End":"07:36.830","Text":"Time is measured in seconds."},{"Start":"07:36.830 ","End":"07:40.775","Text":"Finally, we come to the last part about the asymptotic speed,"},{"Start":"07:40.775 ","End":"07:45.500","Text":"which is what the speed of the boat eventually at infinity becomes."},{"Start":"07:45.500 ","End":"07:50.150","Text":"We just need to take the limit as t goes to infinity of v(t),"},{"Start":"07:50.150 ","End":"07:52.655","Text":"which is the limit of this thing."},{"Start":"07:52.655 ","End":"07:55.145","Text":"Just replace v(t) by that."},{"Start":"07:55.145 ","End":"07:58.640","Text":"That is just equal to 10 because when t goes to infinity,"},{"Start":"07:58.640 ","End":"08:01.970","Text":"let\u0027s remember that e to the minus infinity is"},{"Start":"08:01.970 ","End":"08:06.725","Text":"0 and so this becomes e to the minus infinity, this becomes 0."},{"Start":"08:06.725 ","End":"08:08.660","Text":"We\u0027re left with just the 10."},{"Start":"08:08.660 ","End":"08:11.000","Text":"That completes the question,"},{"Start":"08:11.000 ","End":"08:13.620","Text":"all 3 parts and we\u0027re done."}],"ID":7857},{"Watched":false,"Name":"Exercise 28","Duration":"7m 5s","ChapterTopicVideoID":7784,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.830","Text":"Here we have another word problem in physics and will lead us to a differential equation."},{"Start":"00:04.830 ","End":"00:06.615","Text":"Newton\u0027s Law of Cooling,"},{"Start":"00:06.615 ","End":"00:10.650","Text":"it states that the rate of change of the temperature of an object is"},{"Start":"00:10.650 ","End":"00:12.720","Text":"proportional to the difference between"},{"Start":"00:12.720 ","End":"00:16.635","Text":"its own temperature and the temperature of its surroundings."},{"Start":"00:16.635 ","End":"00:19.620","Text":"A substance with a temperature of 150 degrees"},{"Start":"00:19.620 ","End":"00:24.455","Text":"centigrade is in a container which has the surrounding temperature of the air,"},{"Start":"00:24.455 ","End":"00:26.990","Text":"a constant 30 degrees centigrade."},{"Start":"00:26.990 ","End":"00:30.195","Text":"The substance cools in accordance with Newton\u0027s Law of Cooling,"},{"Start":"00:30.195 ","End":"00:32.000","Text":"and after half an hour,"},{"Start":"00:32.000 ","End":"00:35.945","Text":"its temperature drops to 70 degrees centigrade."},{"Start":"00:35.945 ","End":"00:38.225","Text":"Then we have two questions to answer."},{"Start":"00:38.225 ","End":"00:40.730","Text":"What is its temperature after an hour,"},{"Start":"00:40.730 ","End":"00:45.350","Text":"and after how long will its temperature be 40 degrees centigrade."},{"Start":"00:45.350 ","End":"00:48.590","Text":"In the physics questions I usually like to say which units we\u0027re using."},{"Start":"00:48.590 ","End":"00:53.424","Text":"Well here the only unit of time that\u0027s mentioned is the hour."},{"Start":"00:53.424 ","End":"01:00.200","Text":"We\u0027re going to measure time in hours and temperature in degrees centigrade."},{"Start":"01:00.200 ","End":"01:01.970","Text":"These are our two units."},{"Start":"01:01.970 ","End":"01:08.375","Text":"So that when we start solving and when I say temperature as a function of time,"},{"Start":"01:08.375 ","End":"01:14.540","Text":"I mean that little t is measured in hours and big T is measured in degrees centigrade."},{"Start":"01:14.540 ","End":"01:16.580","Text":"Now let\u0027s see what we are given."},{"Start":"01:16.580 ","End":"01:18.965","Text":"We\u0027re given that after half an hour,"},{"Start":"01:18.965 ","End":"01:20.950","Text":"the temperature is 70."},{"Start":"01:20.950 ","End":"01:26.660","Text":"We also know that the starting temperature at time zero is 150."},{"Start":"01:26.660 ","End":"01:29.690","Text":"Question a is asking us,"},{"Start":"01:29.690 ","End":"01:31.950","Text":"what does T(1) equal?"},{"Start":"01:31.950 ","End":"01:34.915","Text":"Somehow we\u0027re missing question b,"},{"Start":"01:34.915 ","End":"01:38.840","Text":"which says that when will it be 40 degrees centigrade?"},{"Start":"01:38.840 ","End":"01:44.325","Text":"In other words, T of what time will equal 40."},{"Start":"01:44.325 ","End":"01:46.905","Text":"This is part b."},{"Start":"01:46.905 ","End":"01:51.375","Text":"Continuing, I need more space here. Let\u0027s see."},{"Start":"01:51.375 ","End":"01:55.130","Text":"What we have according to Newton\u0027s Law of Cooling is that the rate"},{"Start":"01:55.130 ","End":"01:58.865","Text":"of change of the temperature is proportional."},{"Start":"01:58.865 ","End":"02:02.220","Text":"Proportional means it\u0027s some constant times,"},{"Start":"02:02.220 ","End":"02:05.030","Text":"that\u0027s the proportional bit to the difference"},{"Start":"02:05.030 ","End":"02:07.819","Text":"between its temperature and the surrounding temperature."},{"Start":"02:07.819 ","End":"02:10.820","Text":"It\u0027s temperature is the variable big T,"},{"Start":"02:10.820 ","End":"02:13.010","Text":"surrounding temperature is the constant 30."},{"Start":"02:13.010 ","End":"02:20.075","Text":"This is our differential equation and we have our initial condition is that T(0) is 150."},{"Start":"02:20.075 ","End":"02:23.930","Text":"Together, we should be able to solve this differential equation."},{"Start":"02:23.930 ","End":"02:26.585","Text":"This one seems to be separable."},{"Start":"02:26.585 ","End":"02:30.560","Text":"What I say is let\u0027s move T minus 30 over to the other side,"},{"Start":"02:30.560 ","End":"02:34.070","Text":"and also remember that T\u0027 is d"},{"Start":"02:34.070 ","End":"02:39.715","Text":"capital T over d small t. So that we end up with this equation."},{"Start":"02:39.715 ","End":"02:43.190","Text":"We can put an integral in front of each of them."},{"Start":"02:43.190 ","End":"02:50.655","Text":"Integral of this equals integral of this and the integral of 1 over this is a logarithm,"},{"Start":"02:50.655 ","End":"02:53.870","Text":"and the integral of a constant is constant times t,"},{"Start":"02:53.870 ","End":"02:56.860","Text":"but we have to add a constant of integration."},{"Start":"02:56.860 ","End":"03:00.440","Text":"Now we can put in the initial condition."},{"Start":"03:00.440 ","End":"03:04.910","Text":"The initial condition says that T(0) is 150."},{"Start":"03:04.910 ","End":"03:09.820","Text":"If we put in 0 here for little t,"},{"Start":"03:10.010 ","End":"03:14.510","Text":"for big T, a 150 minus 30 is 120."},{"Start":"03:14.510 ","End":"03:18.490","Text":"So we found that C is natural log of 120."},{"Start":"03:18.490 ","End":"03:21.110","Text":"Now we have another constant to solve for k,"},{"Start":"03:21.110 ","End":"03:23.150","Text":"which is the constant of proportionality."},{"Start":"03:23.150 ","End":"03:25.265","Text":"But we have another piece of information."},{"Start":"03:25.265 ","End":"03:29.630","Text":"We know that after half an hour the temperature was 70."},{"Start":"03:29.630 ","End":"03:31.610","Text":"So this gives us, let\u0027s see,"},{"Start":"03:31.610 ","End":"03:35.150","Text":"putting 70 here, 70 minus 30 is 40."},{"Start":"03:35.150 ","End":"03:39.950","Text":"Natural log of 40 is k times 1/2 plus C. But we know what C is,"},{"Start":"03:39.950 ","End":"03:43.315","Text":"because C is 120."},{"Start":"03:43.315 ","End":"03:47.030","Text":"If we subtract natural log of 40,"},{"Start":"03:47.030 ","End":"03:48.730","Text":"which is bringing it to the other side,"},{"Start":"03:48.730 ","End":"03:51.245","Text":"and remembering that when we subtract,"},{"Start":"03:51.245 ","End":"03:55.357","Text":"we have log of 120 minus log of 40."},{"Start":"03:55.357 ","End":"04:00.485","Text":"This is the same as the natural log of 120 over 40,"},{"Start":"04:00.485 ","End":"04:01.910","Text":"and this is 3,"},{"Start":"04:01.910 ","End":"04:04.955","Text":"and that\u0027s where we get to this bit from."},{"Start":"04:04.955 ","End":"04:10.760","Text":"Now we can also multiply by 2 and twice natural log of 3."},{"Start":"04:10.760 ","End":"04:16.040","Text":"First of all is natural log of 3 to the power of 2 in other law of logarithms."},{"Start":"04:16.040 ","End":"04:19.940","Text":"Ultimately we get k is minus natural log of 9,"},{"Start":"04:19.940 ","End":"04:21.995","Text":"which is log of 9^-1."},{"Start":"04:21.995 ","End":"04:29.330","Text":"So we have both C and k. Now I want to plug C and k. Let\u0027s see, where\u0027s our equation?"},{"Start":"04:29.330 ","End":"04:31.220","Text":"This is our equation,"},{"Start":"04:31.220 ","End":"04:33.170","Text":"I should really highlight it."},{"Start":"04:33.170 ","End":"04:36.950","Text":"Of course it\u0027s written with 2 constants, k and C,"},{"Start":"04:36.950 ","End":"04:40.850","Text":"but we now have k and C. So I\u0027m going to write it on"},{"Start":"04:40.850 ","End":"04:46.925","Text":"the next page with k equals 1/9 and C is natural log of 120."},{"Start":"04:46.925 ","End":"04:48.380","Text":"Hold on. Okay,"},{"Start":"04:48.380 ","End":"04:49.640","Text":"here we are, as I said,"},{"Start":"04:49.640 ","End":"04:52.610","Text":"with that equation with the constants C and k"},{"Start":"04:52.610 ","End":"04:56.885","Text":"thrown in and now we\u0027re going to continue solving it."},{"Start":"04:56.885 ","End":"05:03.270","Text":"What I\u0027m going to do is put the t inside the natural logarithm,"},{"Start":"05:03.270 ","End":"05:06.440","Text":"which makes it the natural log of 9 to the power"},{"Start":"05:06.440 ","End":"05:09.935","Text":"of minus t. The rest of it stays the same."},{"Start":"05:09.935 ","End":"05:11.690","Text":"Before we get rid of the logarithm,"},{"Start":"05:11.690 ","End":"05:15.020","Text":"let\u0027s write this using sum of logs is log of the product,"},{"Start":"05:15.020 ","End":"05:16.940","Text":"and now we can throw out the logarithms"},{"Start":"05:16.940 ","End":"05:19.920","Text":"because logarithms are equal and so are the numbers."},{"Start":"05:19.920 ","End":"05:22.410","Text":"We get to this point here."},{"Start":"05:22.410 ","End":"05:24.860","Text":"Just putting the 30 here over to the other side,"},{"Start":"05:24.860 ","End":"05:28.745","Text":"we\u0027ve isolated what the temperature is in terms of the time."},{"Start":"05:28.745 ","End":"05:30.560","Text":"I\u0027m also going to highlight this."},{"Start":"05:30.560 ","End":"05:32.240","Text":"This is the same as we had before,"},{"Start":"05:32.240 ","End":"05:34.475","Text":"but with all the constants spelled out."},{"Start":"05:34.475 ","End":"05:39.440","Text":"Question a asked us the temperature after an hour."},{"Start":"05:39.440 ","End":"05:41.480","Text":"In other words, what is T(1)?"},{"Start":"05:41.480 ","End":"05:45.735","Text":"Well, let\u0027s just substituting t=1 here."},{"Start":"05:45.735 ","End":"05:50.055","Text":"It\u0027s 30 plus 120 over 9."},{"Start":"05:50.055 ","End":"05:52.890","Text":"That\u0027s 30 plus 40 over 3,"},{"Start":"05:52.890 ","End":"05:54.030","Text":"which is 43 and 1/3."},{"Start":"05:54.030 ","End":"05:56.945","Text":"Anyway, it comes out to be this in degrees."},{"Start":"05:56.945 ","End":"06:01.040","Text":"I forgot to say degrees and I forgot to say wherever I copied it from,"},{"Start":"06:01.040 ","End":"06:02.915","Text":"degrees centigrade of course."},{"Start":"06:02.915 ","End":"06:05.920","Text":"But that\u0027s the only degree we\u0027re talking about."},{"Start":"06:05.920 ","End":"06:11.615","Text":"In part b, just a second I scroll and get some more space here."},{"Start":"06:11.615 ","End":"06:14.330","Text":"In part b, let\u0027s see."},{"Start":"06:14.330 ","End":"06:17.855","Text":"We\u0027re given that the temperature is 40 degrees,"},{"Start":"06:17.855 ","End":"06:21.515","Text":"I won\u0027t keep writing the centigrade and we have to find what t is."},{"Start":"06:21.515 ","End":"06:27.230","Text":"We just have to substitute in this equation that big T is 40."},{"Start":"06:27.230 ","End":"06:30.780","Text":"So we get this side equals 40,"},{"Start":"06:30.780 ","End":"06:34.280","Text":"and now we have to solve for little t. We just"},{"Start":"06:34.280 ","End":"06:38.000","Text":"isolate the 9^ minus t by throwing this to the other side,"},{"Start":"06:38.000 ","End":"06:39.680","Text":"you get 10 over 120,"},{"Start":"06:39.680 ","End":"06:44.000","Text":"which is 1/12 and then we\u0027ve got to take"},{"Start":"06:44.000 ","End":"06:48.560","Text":"the logarithm of both sides minus t natural log of 9 equals log of this,"},{"Start":"06:48.560 ","End":"06:54.125","Text":"and divide by natural log of 9 multiplied by minus 1."},{"Start":"06:54.125 ","End":"07:00.380","Text":"Do it on the calculator and we get the answer to part b is 1.13,"},{"Start":"07:00.380 ","End":"07:02.570","Text":"and that would be in hours."},{"Start":"07:02.570 ","End":"07:04.980","Text":"We are done."}],"ID":7858},{"Watched":false,"Name":"Exercise 29","Duration":"6m 16s","ChapterTopicVideoID":7785,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.980","Text":"Here we have another word problem in physics involving a spring."},{"Start":"00:04.980 ","End":"00:09.825","Text":"Let\u0027s read it. A spring of negligible weight is suspended vertically."},{"Start":"00:09.825 ","End":"00:12.975","Text":"A mass m is connected to its free end."},{"Start":"00:12.975 ","End":"00:16.620","Text":"If the mass is moving at a velocity of v"},{"Start":"00:16.620 ","End":"00:20.205","Text":"naught meters per second when the spring is not extended,"},{"Start":"00:20.205 ","End":"00:23.220","Text":"find the velocity v in meters per"},{"Start":"00:23.220 ","End":"00:28.125","Text":"second as a function of the spring\u0027s extension in meters."},{"Start":"00:28.125 ","End":"00:37.290","Text":"First of all, let me highlight this. Let\u0027s see."},{"Start":"00:37.290 ","End":"00:41.910","Text":"We have time which is measured in seconds."},{"Start":"00:41.910 ","End":"00:46.505","Text":"We have length which is measured in meters."},{"Start":"00:46.505 ","End":"00:49.850","Text":"We have velocity which is of course in"},{"Start":"00:49.850 ","End":"00:53.855","Text":"meters per second because that\u0027s length and that\u0027s time."},{"Start":"00:53.855 ","End":"00:58.745","Text":"We also have mass and the units are not indicated,"},{"Start":"00:58.745 ","End":"01:02.390","Text":"so let\u0027s say that it\u0027s in kilograms."},{"Start":"01:02.390 ","End":"01:09.620","Text":"The reason I say in kilograms is that in order for the law that force=mass"},{"Start":"01:09.620 ","End":"01:17.735","Text":"times the acceleration due to gravity mg to work then m should be in kilograms."},{"Start":"01:17.735 ","End":"01:19.915","Text":"These are our units."},{"Start":"01:19.915 ","End":"01:25.735","Text":"We also have to be familiar with 2 laws of physics, or possibly 3."},{"Start":"01:25.735 ","End":"01:29.270","Text":"Let\u0027s let x be the extension of the spring that"},{"Start":"01:29.270 ","End":"01:32.960","Text":"we call here extension from the non-extended position,"},{"Start":"01:32.960 ","End":"01:35.800","Text":"and x is measured downwards."},{"Start":"01:35.800 ","End":"01:38.690","Text":"One of the laws we need to know is Hooke\u0027s law,"},{"Start":"01:38.690 ","End":"01:43.400","Text":"which says that there\u0027s a force on the spring proportional to the extension,"},{"Start":"01:43.400 ","End":"01:45.980","Text":"and proportional means some constant times."},{"Start":"01:45.980 ","End":"01:49.835","Text":"Then there\u0027s the law of physics that the force of"},{"Start":"01:49.835 ","End":"01:55.160","Text":"gravity acts downwards and it\u0027s equal to mass times g,"},{"Start":"01:55.160 ","End":"01:57.125","Text":"which is a well-known constant,"},{"Start":"01:57.125 ","End":"02:00.335","Text":"9.81 meters per second squared."},{"Start":"02:00.335 ","End":"02:04.535","Text":"We also need to know Newton\u0027s second law that in general,"},{"Start":"02:04.535 ","End":"02:09.620","Text":"the force applied to a body is equal to its mass times its acceleration."},{"Start":"02:09.620 ","End":"02:12.050","Text":"In case you don\u0027t know what acceleration is,"},{"Start":"02:12.050 ","End":"02:17.990","Text":"it\u0027s the second derivative of the distance or extension in our case,"},{"Start":"02:17.990 ","End":"02:22.835","Text":"second derivative with respect to time or the first derivative of the velocity."},{"Start":"02:22.835 ","End":"02:26.225","Text":"That\u0027s all the theoretical background that we need."},{"Start":"02:26.225 ","End":"02:29.870","Text":"Now let\u0027s get to work on the solution."},{"Start":"02:29.870 ","End":"02:32.510","Text":"As I mentioned indirectly,"},{"Start":"02:32.510 ","End":"02:38.620","Text":"the velocity is the derivative with respect to time of the extension."},{"Start":"02:38.620 ","End":"02:43.940","Text":"We also know that the acceleration of the mass is the second derivative,"},{"Start":"02:43.940 ","End":"02:46.055","Text":"which is the derivative of v,"},{"Start":"02:46.055 ","End":"02:51.240","Text":"and that\u0027s d^2 x/dt^2."},{"Start":"02:51.240 ","End":"02:52.860","Text":"These are different Fs by the way."},{"Start":"02:52.860 ","End":"02:54.695","Text":"Each F is in its own law,"},{"Start":"02:54.695 ","End":"03:00.685","Text":"but the mass times the acceleration of the object is made up of 2 bits."},{"Start":"03:00.685 ","End":"03:04.280","Text":"There\u0027s a force, but the force here is the resultant force."},{"Start":"03:04.280 ","End":"03:05.540","Text":"It\u0027s the total force."},{"Start":"03:05.540 ","End":"03:09.019","Text":"In our case, it will be this force minus"},{"Start":"03:09.019 ","End":"03:13.040","Text":"this force if we\u0027re taking a positive direction downwards."},{"Start":"03:13.040 ","End":"03:16.925","Text":"We get mg, which is the force due to gravity,"},{"Start":"03:16.925 ","End":"03:22.845","Text":"and minus kx, which is the force in the opposite direction due to Hooke\u0027s law."},{"Start":"03:22.845 ","End":"03:25.580","Text":"Now there\u0027s a bit of, I won\u0027t say confusion here,"},{"Start":"03:25.580 ","End":"03:28.310","Text":"but there\u0027s 2 possible independent variables."},{"Start":"03:28.310 ","End":"03:31.939","Text":"We could use x or we could use time."},{"Start":"03:31.939 ","End":"03:35.030","Text":"We want to use the extension as"},{"Start":"03:35.030 ","End":"03:38.630","Text":"independent variable because the way the problem was stated,"},{"Start":"03:38.630 ","End":"03:44.630","Text":"we want to know the velocity in terms of the extension."},{"Start":"03:44.630 ","End":"03:47.810","Text":"We\u0027re going to have to do some rearranging in order"},{"Start":"03:47.810 ","End":"03:51.380","Text":"to get the derivatives to be with respect to x."},{"Start":"03:51.380 ","End":"03:54.200","Text":"The good thing about the d something"},{"Start":"03:54.200 ","End":"03:57.605","Text":"over d something notation is that it acts like a fraction."},{"Start":"03:57.605 ","End":"04:00.065","Text":"If we want dv by dt,"},{"Start":"04:00.065 ","End":"04:06.140","Text":"we can say this is dv by dx times dx by dt as if the dx is canceled and"},{"Start":"04:06.140 ","End":"04:12.575","Text":"the dx by dt is v. What we get is that the acceleration,"},{"Start":"04:12.575 ","End":"04:17.215","Text":"which is dv/dt=v times dv/dx,"},{"Start":"04:17.215 ","End":"04:20.840","Text":"and that\u0027s 1.5 the derivative with respect to"},{"Start":"04:20.840 ","End":"04:25.370","Text":"x(v^2) because if I take 1.5v^2 and differentiate it,"},{"Start":"04:25.370 ","End":"04:34.550","Text":"I get v. The next step is to replace dv/dt in this boxed formula by this thing here,"},{"Start":"04:34.550 ","End":"04:36.640","Text":"which we just found that it\u0027s equal to,"},{"Start":"04:36.640 ","End":"04:39.350","Text":"and so the right-hand side remains the same,"},{"Start":"04:39.350 ","End":"04:40.490","Text":"the m remains the same,"},{"Start":"04:40.490 ","End":"04:46.685","Text":"but we replace dv/dt by 1.5d/dx(v^2)."},{"Start":"04:46.685 ","End":"04:48.665","Text":"I need to scroll a bit."},{"Start":"04:48.665 ","End":"04:53.750","Text":"The next step is to divide by m and multiply by 2."},{"Start":"04:53.750 ","End":"04:55.490","Text":"I\u0027m just doing some algebra here."},{"Start":"04:55.490 ","End":"05:01.130","Text":"Then we\u0027ve got the derivative of v^2 with respect to x is equal to this expression."},{"Start":"05:01.130 ","End":"05:03.980","Text":"If I know the derivative with respect to x,"},{"Start":"05:03.980 ","End":"05:09.245","Text":"then I can find v^2 by taking the integral of this with respect to x."},{"Start":"05:09.245 ","End":"05:10.910","Text":"Now this is an easy integral."},{"Start":"05:10.910 ","End":"05:14.150","Text":"The integral of 2g, which is a constant is 2gx,"},{"Start":"05:14.150 ","End":"05:17.905","Text":"and the integral of x is x^2/2."},{"Start":"05:17.905 ","End":"05:19.590","Text":"The 2 cancels with the 2,"},{"Start":"05:19.590 ","End":"05:21.400","Text":"the k/m stays,"},{"Start":"05:21.400 ","End":"05:23.900","Text":"and we add the constant of integration."},{"Start":"05:23.900 ","End":"05:27.560","Text":"Now what we know is that when x is 0,"},{"Start":"05:27.560 ","End":"05:30.650","Text":"our initial velocity is v_0,"},{"Start":"05:30.650 ","End":"05:36.150","Text":"which means that v_0^2=c because look,"},{"Start":"05:36.150 ","End":"05:39.030","Text":"if I put x=0 here and here,"},{"Start":"05:39.030 ","End":"05:45.870","Text":"and all that\u0027s left is c. If I put x=0 here, v is v_0."},{"Start":"05:45.870 ","End":"05:47.390","Text":"This is what we get."},{"Start":"05:47.390 ","End":"05:55.815","Text":"Now, all that remains for us to do is to put c=v_0^2 here."},{"Start":"05:55.815 ","End":"05:58.370","Text":"If I have c=v_0^2 here,"},{"Start":"05:58.370 ","End":"06:00.890","Text":"this thing here is v^2."},{"Start":"06:00.890 ","End":"06:05.364","Text":"V is just the square root of the following."},{"Start":"06:05.364 ","End":"06:06.960","Text":"Taking the square root,"},{"Start":"06:06.960 ","End":"06:12.230","Text":"we\u0027ve got v is plus or minus the square root of this thing with c replaced by v_0^2."},{"Start":"06:12.230 ","End":"06:13.430","Text":"An that ugly formula,"},{"Start":"06:13.430 ","End":"06:14.675","Text":"but there it is,"},{"Start":"06:14.675 ","End":"06:17.070","Text":"and we are done."}],"ID":7859},{"Watched":false,"Name":"Exercise 30","Duration":"9m 54s","ChapterTopicVideoID":28175,"CourseChapterTopicPlaylistID":4239,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.260 ","End":"00:07.290","Text":"Today we\u0027re looking at a dynamical systems problem so let\u0027s proceed."},{"Start":"00:07.290 ","End":"00:10.964","Text":"We\u0027re being asked to consider the dynamical system"},{"Start":"00:10.964 ","End":"00:14.985","Text":"described by the following 2 differential equations here."},{"Start":"00:14.985 ","End":"00:19.755","Text":"We have dN_1 by dt is equal to Alpha N_1,"},{"Start":"00:19.755 ","End":"00:23.040","Text":"N_2 over N minus Lambda N_1,"},{"Start":"00:23.040 ","End":"00:27.420","Text":"and dN_2 by dt is equal to minus Alpha N_1,"},{"Start":"00:27.420 ","End":"00:31.410","Text":"N_2 over N plus Lambda N_1."},{"Start":"00:31.410 ","End":"00:39.677","Text":"The question says, using the fact that N_1 plus N_2 is equal to N at all times,"},{"Start":"00:39.677 ","End":"00:44.695","Text":"we need to write dN_1 by dt as a function of N_1 only."},{"Start":"00:44.695 ","End":"00:47.675","Text":"Some useful observations to make."},{"Start":"00:47.675 ","End":"00:52.700","Text":"The first 1 is that if we look at these 2 differential equations,"},{"Start":"00:52.700 ","End":"00:58.485","Text":"they\u0027re very similar in that this first term in dN_1 by dt,"},{"Start":"00:58.485 ","End":"01:03.485","Text":"and the first term in dN_2 by dt are the negatives of each other."},{"Start":"01:03.485 ","End":"01:06.860","Text":"The same can be said for this second term."},{"Start":"01:06.860 ","End":"01:09.000","Text":"This minus Lambda N_1 here,"},{"Start":"01:09.000 ","End":"01:12.145","Text":"and this plus Lambda N_1 here."},{"Start":"01:12.145 ","End":"01:18.005","Text":"The second thing that maybe we should just clarify is that N_1,"},{"Start":"01:18.005 ","End":"01:23.091","Text":"and N_2 are equations that are"},{"Start":"01:23.091 ","End":"01:29.615","Text":"related to t. Remember here we have dN_1 by dt, dN_2 by dt."},{"Start":"01:29.615 ","End":"01:34.535","Text":"It wouldn\u0027t be very interesting if they want in terms of t otherwise,"},{"Start":"01:34.535 ","End":"01:36.800","Text":"obviously this bit would just be 0."},{"Start":"01:36.800 ","End":"01:39.875","Text":"If we\u0027re looking at N_1 plus N_2,"},{"Start":"01:39.875 ","End":"01:44.200","Text":"then we have N_1 plus N_2,"},{"Start":"01:44.200 ","End":"01:47.045","Text":"which remember are functions of t."},{"Start":"01:47.045 ","End":"01:53.700","Text":"Maybe we can say that N is also a function of t as well."},{"Start":"01:53.890 ","End":"01:59.420","Text":"I think the first thing that we can do when we start this question is"},{"Start":"01:59.420 ","End":"02:04.485","Text":"maybe we should call this equation 1,"},{"Start":"02:04.485 ","End":"02:08.685","Text":"and call this equation 2."},{"Start":"02:08.685 ","End":"02:10.902","Text":"So this is 1,"},{"Start":"02:10.902 ","End":"02:13.660","Text":"and this is 2."},{"Start":"02:13.660 ","End":"02:20.975","Text":"Then, let\u0027s see what happens when we add these 2 equations together."},{"Start":"02:20.975 ","End":"02:28.100","Text":"If we do 1 plus 2,"},{"Start":"02:28.100 ","End":"02:38.005","Text":"then that gives us dN_1 by dt plus dN_2 by dt."},{"Start":"02:38.005 ","End":"02:43.370","Text":"Remember what we can do when we are adding derivatives with respect to the same variable."},{"Start":"02:43.370 ","End":"02:47.660","Text":"Well, we can just put them both in the same bracket."},{"Start":"02:47.660 ","End":"02:52.705","Text":"So we have d over dt( N_1 plus N_2)."},{"Start":"02:52.705 ","End":"02:57.720","Text":"Remember the question says the N_1 plus N_2 is equal to N at all time."},{"Start":"02:57.720 ","End":"03:05.790","Text":"We can replace this N_1+ plus N_2 with an N. Then we get dN by dt."},{"Start":"03:05.790 ","End":"03:07.745","Text":"Now, what\u0027s this equal to?"},{"Start":"03:07.745 ","End":"03:12.800","Text":"Well, this first term here is going to cancel with this first term here."},{"Start":"03:12.800 ","End":"03:19.205","Text":"Similarly, this minus Lambda N_1 term is going to cancel with this plus Lambda N_1 term."},{"Start":"03:19.205 ","End":"03:29.690","Text":"We can say that dN by dt is equal to 0 for all times t. Why is this useful?"},{"Start":"03:29.690 ","End":"03:33.335","Text":"Because if we integrate both sides of this,"},{"Start":"03:33.335 ","End":"03:34.535","Text":"so let\u0027s write that down."},{"Start":"03:34.535 ","End":"03:42.155","Text":"If we do the integral of dN by dt with respect to t,"},{"Start":"03:42.155 ","End":"03:46.244","Text":"and then we integrate the other side as well with respect to t,"},{"Start":"03:46.244 ","End":"03:50.640","Text":"then that tells us that N,"},{"Start":"03:50.640 ","End":"03:53.985","Text":"which we said before may be a function of t,"},{"Start":"03:53.985 ","End":"03:56.615","Text":"is actually just equal to some constant,"},{"Start":"03:56.615 ","End":"04:02.470","Text":"which we can call K, which is of course just some real number."},{"Start":"04:02.470 ","End":"04:06.380","Text":"This means that whatever our N_1 and N_2 are,"},{"Start":"04:06.380 ","End":"04:08.105","Text":"which may be functions of t,"},{"Start":"04:08.105 ","End":"04:09.710","Text":"when we add them together,"},{"Start":"04:09.710 ","End":"04:12.335","Text":"we get a constant number."},{"Start":"04:12.335 ","End":"04:17.895","Text":"Let\u0027s call this equation here, star."},{"Start":"04:17.895 ","End":"04:19.340","Text":"N is equal to K,"},{"Start":"04:19.340 ","End":"04:21.200","Text":"which is some constant number."},{"Start":"04:21.200 ","End":"04:25.380","Text":"Now you might be wondering how do we have 2 functions that"},{"Start":"04:25.380 ","End":"04:30.540","Text":"are functions of t that then we add them together we get some constant number."},{"Start":"04:30.540 ","End":"04:38.100","Text":"Let\u0027s for example, take N_1(t) is equal to say,"},{"Start":"04:38.100 ","End":"04:42.210","Text":"10 over 10 plus t^2,"},{"Start":"04:42.210 ","End":"04:44.970","Text":"which obviously does depend on t,"},{"Start":"04:44.970 ","End":"04:49.335","Text":"and say N_2( t)"},{"Start":"04:49.335 ","End":"04:55.565","Text":"is equal to t^2 over 10 plus t^2."},{"Start":"04:55.565 ","End":"05:00.105","Text":"Then if we do N_1 plus N_2,"},{"Start":"05:00.105 ","End":"05:06.960","Text":"then that gives us 10 plus t^2 over 10 plus t^2,"},{"Start":"05:06.960 ","End":"05:08.565","Text":"which is just equal to 1,"},{"Start":"05:08.565 ","End":"05:12.750","Text":"which again is just a real number."},{"Start":"05:12.750 ","End":"05:16.340","Text":"This is one example of infinite number of cases"},{"Start":"05:16.340 ","End":"05:20.360","Text":"where we can have N_1 and N_2 which are functions of t. But,"},{"Start":"05:20.360 ","End":"05:22.880","Text":"when we take that addition, or their sum,"},{"Start":"05:22.880 ","End":"05:25.985","Text":"we get just a real valued number."},{"Start":"05:25.985 ","End":"05:29.795","Text":"This fact where N is a real number,"},{"Start":"05:29.795 ","End":"05:33.920","Text":"is about to come in useful later, as we will see."},{"Start":"05:33.920 ","End":"05:36.140","Text":"Now what we\u0027re going to do is,"},{"Start":"05:36.140 ","End":"05:39.740","Text":"remember what the question wants is dN_1 by dt."},{"Start":"05:39.740 ","End":"05:45.170","Text":"So we\u0027re going to use the facts now that N is a real number"},{"Start":"05:45.170 ","End":"05:51.545","Text":"to rewrite N_1 plus N_2 equals N and then we\u0027re going to sub that back into 1."},{"Start":"05:51.545 ","End":"05:53.720","Text":"Let\u0027s see what that gives us."},{"Start":"05:53.720 ","End":"05:56.675","Text":"We\u0027ve just tidied up what we had before,"},{"Start":"05:56.675 ","End":"05:58.340","Text":"and remember what we said."},{"Start":"05:58.340 ","End":"06:01.460","Text":"We said the N is a real number."},{"Start":"06:01.460 ","End":"06:05.750","Text":"We can use blue star to rewrite this part,"},{"Start":"06:05.750 ","End":"06:14.390","Text":"and we can say that N_1 plus N_2 is equal to K. Then we can"},{"Start":"06:14.390 ","End":"06:20.973","Text":"replace N_2 is equal"},{"Start":"06:20.973 ","End":"06:25.410","Text":"to K minus N_1 with just a little bit of rearranging."},{"Start":"06:25.410 ","End":"06:27.895","Text":"What we\u0027re going to do now is,"},{"Start":"06:27.895 ","End":"06:30.980","Text":"remember we want an equation just in terms of N_1."},{"Start":"06:30.980 ","End":"06:36.035","Text":"So we\u0027re going to put that into our equation 1 here."},{"Start":"06:36.035 ","End":"06:40.725","Text":"Putting into 1,"},{"Start":"06:40.725 ","End":"06:49.590","Text":"we get dN_1 by dt is equal to Alpha N_1."},{"Start":"06:49.590 ","End":"06:57.290","Text":"Then remember we\u0027ve just got this new expression for N_2, which is K-N_1."},{"Start":"06:57.290 ","End":"07:00.510","Text":"Then this is all over N. But,"},{"Start":"07:00.510 ","End":"07:02.925","Text":"remember we said that N was equal to K,"},{"Start":"07:02.925 ","End":"07:06.375","Text":"we may as well write that in here."},{"Start":"07:06.375 ","End":"07:09.830","Text":"If we tidy this up a little bit now,"},{"Start":"07:09.830 ","End":"07:18.010","Text":"then we get dN_1 by dt is equal to Alpha N_1."},{"Start":"07:18.010 ","End":"07:19.640","Text":"Because if you think about it,"},{"Start":"07:19.640 ","End":"07:22.085","Text":"these K\u0027s are going to cancel here."},{"Start":"07:22.085 ","End":"07:29.720","Text":"Then we\u0027re going to have minus Alpha over KN_1^2."},{"Start":"07:29.720 ","End":"07:37.515","Text":"This is expanding this and then dividing through by K. Let\u0027s try again,"},{"Start":"07:37.515 ","End":"07:40.125","Text":"tidy this up a little bit more."},{"Start":"07:40.125 ","End":"07:49.730","Text":"Don\u0027t forget, we need to remember to minus this Lambda N_1 term as well in both of them."},{"Start":"07:49.730 ","End":"07:55.010","Text":"If we then tidy this up and we bring terms together,"},{"Start":"07:55.010 ","End":"07:57.095","Text":"then let\u0027s see what that gives."},{"Start":"07:57.095 ","End":"08:00.425","Text":"We can actually write this a bit nicer."},{"Start":"08:00.425 ","End":"08:06.195","Text":"Let a equal Alpha minus Lambda,"},{"Start":"08:06.195 ","End":"08:11.265","Text":"and b equal Alpha over K. Because remember,"},{"Start":"08:11.265 ","End":"08:12.690","Text":"Alpha, Lambda,"},{"Start":"08:12.690 ","End":"08:15.450","Text":"and K are just real valued numbers."},{"Start":"08:15.450 ","End":"08:19.430","Text":"Therefore, a and b here would be real valued numbers."},{"Start":"08:19.430 ","End":"08:21.560","Text":"Why do we write it in this form?"},{"Start":"08:21.560 ","End":"08:24.995","Text":"Because then you\u0027ll see we get quite a nice equation and we\u0027ll get"},{"Start":"08:24.995 ","End":"08:29.685","Text":"dN_1 by dt is equal to,"},{"Start":"08:29.685 ","End":"08:31.864","Text":"remember a is Alpha minus Lambda."},{"Start":"08:31.864 ","End":"08:34.055","Text":"So you\u0027ve got aN_1,"},{"Start":"08:34.055 ","End":"08:36.860","Text":"and then we\u0027ve got minus Alpha over K,"},{"Start":"08:36.860 ","End":"08:38.180","Text":"which we said is b,"},{"Start":"08:38.180 ","End":"08:41.300","Text":"so we\u0027ve got minus bN_1^2."},{"Start":"08:42.580 ","End":"08:46.655","Text":"Then if we factorize this out,"},{"Start":"08:46.655 ","End":"08:48.710","Text":"then we can take on N_1,"},{"Start":"08:48.710 ","End":"08:57.285","Text":"and then we get a minus bN_1 and then obviously that\u0027s times by the factored out N_1."},{"Start":"08:57.285 ","End":"09:00.650","Text":"Why would we have a right in a simplified form?"},{"Start":"09:00.650 ","End":"09:03.200","Text":"Because then what we can do is,"},{"Start":"09:03.200 ","End":"09:10.156","Text":"we can say this means that 1 over N_1,"},{"Start":"09:10.156 ","End":"09:16.020","Text":"a minus bN_1^N_1 is equal to 1."},{"Start":"09:16.020 ","End":"09:20.010","Text":"Then we can actually just or equal to dt."},{"Start":"09:20.010 ","End":"09:24.350","Text":"Remember we had this dN_1 here and this dt here."},{"Start":"09:24.350 ","End":"09:28.170","Text":"We can actually then integrate both sides."},{"Start":"09:28.170 ","End":"09:30.340","Text":"Even though the question didn\u0027t ask for that,"},{"Start":"09:30.340 ","End":"09:33.310","Text":"it might be useful to understand why it would want"},{"Start":"09:33.310 ","End":"09:36.670","Text":"us to do this and it\u0027s so we can solve for N_1."},{"Start":"09:36.670 ","End":"09:39.130","Text":"Here, if we took partial fractions,"},{"Start":"09:39.130 ","End":"09:42.815","Text":"then we could work out what N_1 is in terms of t,"},{"Start":"09:42.815 ","End":"09:45.425","Text":"and we can see exactly what that is."},{"Start":"09:45.425 ","End":"09:48.175","Text":"But we\u0027ve answered the question because remember,"},{"Start":"09:48.175 ","End":"09:54.760","Text":"all we wanted to do was write dN_1 by dt as a function of N_1 only."}],"ID":29411}],"Thumbnail":null,"ID":4239}]

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1.1

1