WEBVTT 00:00:00.000 --> 00:00:07.690 JOEL LEWIS: Hi. 00:00:07.690 --> 00:00:09.150 Welcome to recitation. 00:00:09.150 --> 00:00:12.840 In lecture you've learned how to compute derivatives 00:00:12.840 --> 00:00:14.900 of polynomials, and you've learned 00:00:14.900 --> 00:00:18.050 the relationship between derivatives and tangents lines. 00:00:18.050 --> 00:00:22.340 So let's do a quick example that puts those two ideas together. 00:00:22.340 --> 00:00:24.450 So here I have a question on the board: 00:00:24.450 --> 00:00:26.840 compute the tangent line to the curve y 00:00:26.840 --> 00:00:30.710 equals x cubed minus x at the point (2, 6). 00:00:30.710 --> 00:00:32.210 So why don't you take a minute, work 00:00:32.210 --> 00:00:34.559 on that yourself, pause the video, we'll come back 00:00:34.559 --> 00:00:35.600 and we'll do it together. 00:00:38.691 --> 00:00:39.190 All right. 00:00:39.190 --> 00:00:40.170 Welcome back. 00:00:40.170 --> 00:00:44.510 So we have this function, y equals x cubed minus x. 00:00:44.510 --> 00:00:49.270 Let's just draw a quick sketch of it. 00:00:49.270 --> 00:00:57.760 So looks to me like it has zeros at 0, 1, minus 1. 00:00:57.760 --> 00:01:02.860 And it sort of does something like this in between. 00:01:02.860 --> 00:01:05.070 Very rough sketch there. 00:01:05.070 --> 00:01:07.820 And way over-- well, OK. 00:01:07.820 --> 00:01:10.060 We'll call that the point (2, 6). 00:01:10.060 --> 00:01:13.560 Feel a little sketchy, but all right. 00:01:13.560 --> 00:01:14.060 OK. 00:01:14.060 --> 00:01:16.220 So we want to know what the tangent line 00:01:16.220 --> 00:01:18.270 to the curve at that point is. 00:01:18.270 --> 00:01:23.440 So in order to do that, we need to know what its derivative is, 00:01:23.440 --> 00:01:25.050 and then that'll give us the slope. 00:01:25.050 --> 00:01:26.799 And then with the slope, we have the slope 00:01:26.799 --> 00:01:28.930 and we have a point, so we can slap 00:01:28.930 --> 00:01:32.320 that into, say, your point-slope formula for a line. 00:01:35.120 --> 00:01:41.150 So, all right, so the derivative of this function is y prime. 00:01:41.150 --> 00:01:43.940 So, OK, so here we have a sum of two things, 00:01:43.940 --> 00:01:45.810 and they're both powers of x. 00:01:45.810 --> 00:01:48.060 And so we learned our rules for a power 00:01:48.060 --> 00:01:50.440 of x that the derivative of x to the n 00:01:50.440 --> 00:01:52.510 is n times x to the n minus 1. 00:01:52.510 --> 00:01:54.640 And so we also learned that the derivative 00:01:54.640 --> 00:01:58.470 of a sum of two things is the sum of the derivatives. 00:01:58.470 --> 00:02:01.230 So in this case, so the derivative of x cubed 00:02:01.230 --> 00:02:08.076 minus x is 3 x squared minus 1. 00:02:08.076 --> 00:02:08.576 OK. 00:02:08.576 --> 00:02:13.480 So this is the slope of the function in terms of x. 00:02:13.480 --> 00:02:15.840 But in order to compute the tangent line, 00:02:15.840 --> 00:02:20.035 we need the slope at the particular point in question. 00:02:20.035 --> 00:02:20.535 Right? 00:02:20.535 --> 00:02:22.120 This is really important. 00:02:22.120 --> 00:02:25.870 So we aren't going to use 3 x squared plus 1 as our slope. 00:02:25.870 --> 00:02:26.780 Right? 00:02:26.780 --> 00:02:29.750 We want the slope at the point x equals 2. 00:02:29.750 --> 00:02:30.250 Right? 00:02:33.530 --> 00:02:34.030 OK. 00:02:34.030 --> 00:02:37.100 So what we want for the slope of the tangent line 00:02:37.100 --> 00:02:38.640 is y prime of 2. 00:02:38.640 --> 00:02:39.410 Right? 00:02:39.410 --> 00:02:42.050 We want it at this point when x is equal to 2. 00:02:42.050 --> 00:02:47.155 So that's equal to, well, 3 times 2 squared is 12, minus 1 00:02:47.155 --> 00:02:47.655 is 11. 00:02:58.681 --> 00:03:01.180 So this is the slope, this is the slope of the tangent line. 00:03:01.180 --> 00:03:03.310 I just want to say this one more time for emphasis. 00:03:03.310 --> 00:03:05.690 This is a really common mistake that we 00:03:05.690 --> 00:03:10.500 see on lots of homework and exams when teaching calculus. 00:03:10.500 --> 00:03:12.460 You have to remember that when you compute 00:03:12.460 --> 00:03:15.390 the slope of the tangent line, you compute the derivative 00:03:15.390 --> 00:03:18.980 and then you need to plug in the value of x 00:03:18.980 --> 00:03:20.160 at the point in question. 00:03:20.160 --> 00:03:22.620 Or the value of x and y. 00:03:22.620 --> 00:03:24.970 You know, you need to plug in the values of the point 00:03:24.970 --> 00:03:27.380 that you have. 00:03:27.380 --> 00:03:31.410 So here, the derivative is 3 x squared minus 1, 00:03:31.410 --> 00:03:34.760 so the slope at the point (2, 6) is 11. 00:03:34.760 --> 00:03:36.240 It's just a number. 00:03:36.240 --> 00:03:39.940 The slope at that point is that particular number. 00:03:39.940 --> 00:03:41.910 OK, so now, to compute the tangent line, 00:03:41.910 --> 00:03:45.950 we have a point, (2, 6), and we have a slope, 11. 00:03:45.950 --> 00:03:48.400 So we can plug into point-slope form. 00:03:57.060 --> 00:04:11.572 So the equation of the-- on the-- tangent line is y minus 6 00:04:11.572 --> 00:04:16.107 is equal to 11 times x minus 2. 00:04:16.107 --> 00:04:16.870 Right? 00:04:16.870 --> 00:04:22.340 So it's y minus y_0 is equal to the slope times x minus x_0. 00:04:22.340 --> 00:04:24.530 If you like, some people prefer to write 00:04:24.530 --> 00:04:28.209 their equations of their lines in slope-intercept form. 00:04:28.209 --> 00:04:30.000 So if you wanted to do that, you could just 00:04:30.000 --> 00:04:34.500 multiply through by 11 and then bring the constants together. 00:04:34.500 --> 00:04:42.680 So we could say, or y equals 11x-- well, 00:04:42.680 --> 00:04:48.050 we get minus 22 plus 6 is minus 16. 00:04:48.050 --> 00:04:50.540 So either of those is a perfectly good answer 00:04:50.540 --> 00:04:51.720 to the question. 00:04:51.720 --> 00:04:53.410 So that's that.