1 00:00:00,025 --> 00:00:08,290 [music] Here is a classic related rates problem. 2 00:00:08,290 --> 00:00:12,722 So here's the question. You're walking away from a light source, 3 00:00:12,722 --> 00:00:18,012 at some speed, and you don't know how fast is your shadow's length changing. 4 00:00:18,012 --> 00:00:23,030 There's a standard method to solving a related rates problem, four steps. 5 00:00:23,030 --> 00:00:27,395 So, step 1 is draw a picture, alright? You're told some sort of story, you should 6 00:00:27,395 --> 00:00:30,710 convert it into some diagram. Step 2, find an equation. 7 00:00:30,710 --> 00:00:34,010 Given that picture, you should label everything in the picture so you can write 8 00:00:34,010 --> 00:00:37,372 down some equation. Step 3 is to differentiate that equation 9 00:00:37,372 --> 00:00:39,633 which would probably involve the chain rule. 10 00:00:39,633 --> 00:00:43,766 And step 4 is to solve that equation for whatever you're interested in finding out. 11 00:00:43,766 --> 00:00:47,710 I mean, hopefully you're trying to find some rate of something so, you're going to 12 00:00:47,710 --> 00:00:50,021 try to find one of these derivatives in there. 13 00:00:50,021 --> 00:00:55,969 Let's draw a picture to represent this person moving away from the light source. 14 00:00:55,970 --> 00:01:00,350 Here's my picture, I've got a light source right here and I've got this person right 15 00:01:00,350 --> 00:01:02,876 here. And I imagine that person walking away 16 00:01:02,876 --> 00:01:05,507 from the light and I've drawn in their shadow. 17 00:01:05,508 --> 00:01:10,037 And I've drawn this beam of light just glancing across this person's head. 18 00:01:10,038 --> 00:01:15,070 Now we'll turn this picture into an equation using similar triangles. 19 00:01:15,070 --> 00:01:19,472 You might suppose that the original problem told us that the lamppost was 3 20 00:01:19,472 --> 00:01:22,760 meters tall, and the person was, say, two meters tall. 21 00:01:22,760 --> 00:01:26,250 And I can label the other relevant lengths on this diagram. 22 00:01:26,250 --> 00:01:32,079 I'll call x the distance from the bottom of the lamppost to the person's feet, and 23 00:01:32,079 --> 00:01:37,760 the distance from the feet to the tip of the shadow, I'll call that s for shadow. 24 00:01:37,760 --> 00:01:42,186 So, now I've got a nicely labeled diagram, and I can replace that nicely labeled 25 00:01:42,186 --> 00:01:46,139 diagram with a slightly more extract diagram, where the person is now a 26 00:01:46,139 --> 00:01:50,025 vertical line, the lamppost is now a vertical line, and I've got these 27 00:01:50,025 --> 00:01:52,967 triangles. The up shot here is that I've got similar 28 00:01:52,967 --> 00:01:55,833 triangles. This little triangle is similiar to this 29 00:01:55,833 --> 00:01:58,917 big triangle. Well, because the angles are all the same, 30 00:01:58,917 --> 00:02:01,665 right? Both of these triangles have a right angle 31 00:02:01,665 --> 00:02:05,960 here, they have the same angle here, consequently these angles are the same. 32 00:02:05,961 --> 00:02:10,667 They're two triangles that have the same angles, they're similar triangles. 33 00:02:10,668 --> 00:02:14,854 And as a result, I get this equation out of the diagram. 34 00:02:14,855 --> 00:02:19,600 This distance to this distance, x plus s to 3, must be the same as the 35 00:02:19,600 --> 00:02:24,592 corresponding ratio in to the other triangle, which is this distance to this 36 00:02:24,592 --> 00:02:28,354 distance s over 2. With an equation in hand we can now 37 00:02:28,354 --> 00:02:34,561 differentiate. Thinking of my position and the shadow's 38 00:02:34,561 --> 00:02:43,073 length as a function of t, I could rewrite this equation as say, x of t plus s of t 39 00:02:43,073 --> 00:02:48,134 divided by 3 equals s of t divided by 2, right? 40 00:02:48,134 --> 00:02:52,298 My position is a function of time and my show's length is a function of time, and I 41 00:02:52,298 --> 00:02:54,904 can differentiate both sides of this equation. 42 00:02:54,904 --> 00:03:03,224 The derivative of this is 1 3rd, the derivative of x plus the derivative of s, 43 00:03:03,224 --> 00:03:10,690 and the derivative of the other side is one half the derivative of s. 44 00:03:10,690 --> 00:03:17,590 And I could solve for the derivative of s. Let me first expand this out. 45 00:03:17,590 --> 00:03:26,215 I've got 1/3 derivative of x plus 1/3 derivative of s, is 1/2 derivative of s. 46 00:03:26,215 --> 00:03:37,596 I'll subtract 1/3 s prime from both sides. Got 1 3rd x prime is 1 6th s prime. 47 00:03:37,596 --> 00:03:42,633 'Because a half minus a 3rd is a 6th. And then I multiply both sides by 6. 48 00:03:42,633 --> 00:03:49,128 So I find out that s prime. This is the rate of change in the shadow's 49 00:03:49,128 --> 00:03:54,280 length is twice. X prime this is the speed for which the 50 00:03:54,280 --> 00:03:58,454 person is moving. There's a bit of a surprise to this 51 00:03:58,454 --> 00:04:02,280 answer. Take a look at this equation, It's telling 52 00:04:02,280 --> 00:04:08,260 you that the speed that your shadows length is changing is just twice the speed 53 00:04:08,260 --> 00:04:12,090 that your walking. But it doesnt matter where your standing. 54 00:04:12,090 --> 00:04:17,042 The value of x doesn't appear on this side only x prime appears on this side. 55 00:04:17,042 --> 00:04:20,913 So that's kind of interesting right? The speed with which your shadow is 56 00:04:20,913 --> 00:04:24,250 growing only has to do with how fast you're walking. 57 00:04:24,250 --> 00:04:28,345 In fact, the speed at which your shadow is growing is exactly twice the speed at 58 00:04:28,345 --> 00:04:31,842 which you are walking. Assuming of course, you are two meters 59 00:04:31,842 --> 00:04:34,839 tall and the lamp post is three meters tall. 60 00:04:34,840 --> 00:04:39,022 I think the important thing to take away from this example is that similar 61 00:04:39,022 --> 00:04:43,557 triangles are hugely helpful. If you see similar triangles, use them. 62 00:04:43,558 --> 00:04:47,054 There's also one thing that I think is very funny about this particular solution 63 00:04:47,054 --> 00:04:49,414 to the problem. Here's the challenge for you to think 64 00:04:49,414 --> 00:04:51,896 about. If your velocity is 90% the speed of 65 00:04:51,896 --> 00:04:54,988 light, so you're almost going the speed of light. 66 00:04:54,988 --> 00:05:04,435 Then this formula would be telling you that your shadow's length would be moving 67 00:05:04,435 --> 00:05:11,073 faster than the speed of light, is that really possible?