1 00:00:00,630 --> 00:00:02,150 - [Instructor] Two cars are driving 2 00:00:02,150 --> 00:00:06,290 towards an intersection from perpendicular directions. 3 00:00:06,290 --> 00:00:08,370 The first car's velocity is 50 km/h 4 00:00:09,520 --> 00:00:13,100 and the second car's velocity is 90 km/h. 5 00:00:13,100 --> 00:00:15,330 At a certain instant, T sub zero, 6 00:00:15,330 --> 00:00:19,080 the first car is a distance, X sub T sub zero 7 00:00:19,080 --> 00:00:22,490 or X of T sub zero, of half a kilometer 8 00:00:22,490 --> 00:00:24,220 from the intersection and the second car 9 00:00:24,220 --> 00:00:27,080 is a distance, Y of T sub zero, 10 00:00:27,080 --> 00:00:29,970 of 1.2 kilometers from the intersection. 11 00:00:29,970 --> 00:00:32,330 What is the rate of change of the distance, 12 00:00:32,330 --> 00:00:35,220 D of T, between the cars at that instant? 13 00:00:35,220 --> 00:00:36,930 So, T sub zero. 14 00:00:36,930 --> 00:00:39,940 Which equation should be used to solve the problem? 15 00:00:39,940 --> 00:00:43,150 And they give us a choice of four equations, 16 00:00:43,150 --> 00:00:45,530 right over here, so you could pause the video 17 00:00:45,530 --> 00:00:47,100 and try to work through it on your own 18 00:00:47,100 --> 00:00:49,790 but I'm about to do it, as well. 19 00:00:49,790 --> 00:00:51,440 So, let's just draw what's going on. 20 00:00:51,440 --> 00:00:53,100 That's always a healthy thing to do. 21 00:00:53,100 --> 00:00:55,040 So, two cars are driving towards 22 00:00:55,040 --> 00:00:57,710 an intersection from perpendicular directions. 23 00:00:57,710 --> 00:01:01,280 So, let's say that this is one car right over here 24 00:01:02,140 --> 00:01:05,099 and it is moving in the X direction 25 00:01:05,099 --> 00:01:07,760 towards that intersection, which is right over there. 26 00:01:07,760 --> 00:01:09,130 And then you have another car 27 00:01:09,130 --> 00:01:11,150 that is moving in the Y direction. 28 00:01:11,150 --> 00:01:13,470 So, let's say it's moving like this. 29 00:01:13,470 --> 00:01:15,150 So, this is the other car. 30 00:01:15,150 --> 00:01:17,300 I should have maybe done a top view. 31 00:01:17,300 --> 00:01:18,580 Well, here we go. 32 00:01:18,580 --> 00:01:20,250 This square represents the car 33 00:01:20,250 --> 00:01:23,020 and it is moving in that direction. 34 00:01:23,020 --> 00:01:25,220 Now, they say at a certain instance. 35 00:01:25,220 --> 00:01:27,430 T sub zero, so let's draw that instant. 36 00:01:27,430 --> 00:01:29,750 So, the first car is a distance, 37 00:01:29,750 --> 00:01:33,960 X of T sub zero, of 0.5 kilometers. 38 00:01:33,960 --> 00:01:36,690 So, this distance, right over here, 39 00:01:36,690 --> 00:01:40,360 let's just call this X of T and let's call 40 00:01:40,360 --> 00:01:44,830 this distance, right over here, Y of T. 41 00:01:44,830 --> 00:01:47,160 Now, how does the distance between the 42 00:01:47,160 --> 00:01:50,840 cars relate to X of T and Y of T? 43 00:01:50,840 --> 00:01:52,830 Well, we could just use the distance formula, 44 00:01:52,830 --> 00:01:55,240 which is essentially just the Pythagorean theorem, 45 00:01:55,240 --> 00:01:57,110 to say, well, the distance between the cars 46 00:01:57,110 --> 00:02:00,180 would be the hypotenuse of this right triangle. 47 00:02:00,180 --> 00:02:03,040 Remember, they're traveling from perpendicular directions. 48 00:02:03,040 --> 00:02:04,760 So, that's a right triangle, there. 49 00:02:04,760 --> 00:02:09,759 So, this distance, right over here, would be X of T squared 50 00:02:11,300 --> 00:02:16,300 plus Y of T squared, and the square root of that, 51 00:02:16,710 --> 00:02:19,070 and that's just the Pythagorean theorem, right over here. 52 00:02:19,070 --> 00:02:23,170 This would be D of T, or we could say 53 00:02:23,170 --> 00:02:27,240 that D of T squared is equal to 54 00:02:29,273 --> 00:02:34,273 X of T squared plus Y, too many parentheses. 55 00:02:34,900 --> 00:02:38,300 Plus Y of T squared. 56 00:02:38,300 --> 00:02:39,610 So, that's the relationship between 57 00:02:39,610 --> 00:02:43,870 D of T, X of T, and Y of T, and it's useful 58 00:02:43,870 --> 00:02:45,740 for solving this problem because, now, 59 00:02:45,740 --> 00:02:47,680 we can take the derivative of both sides 60 00:02:47,680 --> 00:02:49,510 of this equation with respect to T. 61 00:02:49,510 --> 00:02:52,000 We'd be using various derivative rules, 62 00:02:52,000 --> 00:02:54,400 including the chain rule, in order to do it, 63 00:02:54,400 --> 00:02:56,610 and then that would give us a relationship 64 00:02:56,610 --> 00:02:59,620 between the rate of change of D of T, 65 00:02:59,620 --> 00:03:01,370 which would be D prime of T, 66 00:03:01,370 --> 00:03:04,480 and the rate of change of X of T, Y of T, 67 00:03:04,480 --> 00:03:07,450 and X of T and Y of T, themselves. 68 00:03:07,450 --> 00:03:10,423 And so, if we look at these choices, right over here, 69 00:03:11,766 --> 00:03:14,830 we indeed see that D sets up that exact 70 00:03:14,830 --> 00:03:17,940 same relationship that we just did, ourselves. 71 00:03:17,940 --> 00:03:20,230 That it shows that the distance 72 00:03:20,230 --> 00:03:22,430 squared between the cars is equal to 73 00:03:22,430 --> 00:03:25,370 that X distance from the intersection squared, 74 00:03:25,370 --> 00:03:27,880 plus the Y distance from the intersection squared, 75 00:03:27,880 --> 00:03:30,440 and then we can take the derivative of both sides 76 00:03:30,440 --> 00:03:33,383 to actually figure out this related, rates question.