[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.63,0:00:02.15,Default,,0000,0000,0000,,- [Instructor] Two cars are driving
Dialogue: 0,0:00:02.15,0:00:06.29,Default,,0000,0000,0000,,towards an intersection from\Nperpendicular directions.
Dialogue: 0,0:00:06.29,0:00:08.37,Default,,0000,0000,0000,,The first car's velocity is 50 km/h
Dialogue: 0,0:00:09.52,0:00:13.10,Default,,0000,0000,0000,,and the second car's velocity is 90 km/h.
Dialogue: 0,0:00:13.10,0:00:15.33,Default,,0000,0000,0000,,At a certain instant, T sub zero,
Dialogue: 0,0:00:15.33,0:00:19.08,Default,,0000,0000,0000,,the first car is a\Ndistance, X sub T sub zero
Dialogue: 0,0:00:19.08,0:00:22.49,Default,,0000,0000,0000,,or X of T sub zero, of half a kilometer
Dialogue: 0,0:00:22.49,0:00:24.22,Default,,0000,0000,0000,,from the intersection and the second car
Dialogue: 0,0:00:24.22,0:00:27.08,Default,,0000,0000,0000,,is a distance, Y of T sub zero,
Dialogue: 0,0:00:27.08,0:00:29.97,Default,,0000,0000,0000,,of 1.2 kilometers from the intersection.
Dialogue: 0,0:00:29.97,0:00:32.33,Default,,0000,0000,0000,,What is the rate of\Nchange of the distance,
Dialogue: 0,0:00:32.33,0:00:35.22,Default,,0000,0000,0000,,D of T, between the cars at that instant?
Dialogue: 0,0:00:35.22,0:00:36.93,Default,,0000,0000,0000,,So, T sub zero.
Dialogue: 0,0:00:36.93,0:00:39.94,Default,,0000,0000,0000,,Which equation should be\Nused to solve the problem?
Dialogue: 0,0:00:39.94,0:00:43.15,Default,,0000,0000,0000,,And they give us a\Nchoice of four equations,
Dialogue: 0,0:00:43.15,0:00:45.53,Default,,0000,0000,0000,,right over here, so you\Ncould pause the video
Dialogue: 0,0:00:45.53,0:00:47.10,Default,,0000,0000,0000,,and try to work through it on your own
Dialogue: 0,0:00:47.10,0:00:49.79,Default,,0000,0000,0000,,but I'm about to do it, as well.
Dialogue: 0,0:00:49.79,0:00:51.44,Default,,0000,0000,0000,,So, let's just draw what's going on.
Dialogue: 0,0:00:51.44,0:00:53.10,Default,,0000,0000,0000,,That's always a healthy thing to do.
Dialogue: 0,0:00:53.10,0:00:55.04,Default,,0000,0000,0000,,So, two cars are driving towards
Dialogue: 0,0:00:55.04,0:00:57.71,Default,,0000,0000,0000,,an intersection from\Nperpendicular directions.
Dialogue: 0,0:00:57.71,0:01:01.28,Default,,0000,0000,0000,,So, let's say that this\Nis one car right over here
Dialogue: 0,0:01:02.14,0:01:05.10,Default,,0000,0000,0000,,and it is moving in the X direction
Dialogue: 0,0:01:05.10,0:01:07.76,Default,,0000,0000,0000,,towards that intersection,\Nwhich is right over there.
Dialogue: 0,0:01:07.76,0:01:09.13,Default,,0000,0000,0000,,And then you have another car
Dialogue: 0,0:01:09.13,0:01:11.15,Default,,0000,0000,0000,,that is moving in the Y direction.
Dialogue: 0,0:01:11.15,0:01:13.47,Default,,0000,0000,0000,,So, let's say it's moving like this.
Dialogue: 0,0:01:13.47,0:01:15.15,Default,,0000,0000,0000,,So, this is the other car.
Dialogue: 0,0:01:15.15,0:01:17.30,Default,,0000,0000,0000,,I should have maybe done a top view.
Dialogue: 0,0:01:17.30,0:01:18.58,Default,,0000,0000,0000,,Well, here we go.
Dialogue: 0,0:01:18.58,0:01:20.25,Default,,0000,0000,0000,,This square represents the car
Dialogue: 0,0:01:20.25,0:01:23.02,Default,,0000,0000,0000,,and it is moving in that direction.
Dialogue: 0,0:01:23.02,0:01:25.22,Default,,0000,0000,0000,,Now, they say at a certain instance.
Dialogue: 0,0:01:25.22,0:01:27.43,Default,,0000,0000,0000,,T sub zero, so let's draw that instant.
Dialogue: 0,0:01:27.43,0:01:29.75,Default,,0000,0000,0000,,So, the first car is a distance,
Dialogue: 0,0:01:29.75,0:01:33.96,Default,,0000,0000,0000,,X of T sub zero, of 0.5 kilometers.
Dialogue: 0,0:01:33.96,0:01:36.69,Default,,0000,0000,0000,,So, this distance, right over here,
Dialogue: 0,0:01:36.69,0:01:40.36,Default,,0000,0000,0000,,let's just call this X of T and let's call
Dialogue: 0,0:01:40.36,0:01:44.83,Default,,0000,0000,0000,,this distance, right over here, Y of T.
Dialogue: 0,0:01:44.83,0:01:47.16,Default,,0000,0000,0000,,Now, how does the distance between the
Dialogue: 0,0:01:47.16,0:01:50.84,Default,,0000,0000,0000,,cars relate to X of T and Y of T?
Dialogue: 0,0:01:50.84,0:01:52.83,Default,,0000,0000,0000,,Well, we could just use\Nthe distance formula,
Dialogue: 0,0:01:52.83,0:01:55.24,Default,,0000,0000,0000,,which is essentially just\Nthe Pythagorean theorem,
Dialogue: 0,0:01:55.24,0:01:57.11,Default,,0000,0000,0000,,to say, well, the\Ndistance between the cars
Dialogue: 0,0:01:57.11,0:02:00.18,Default,,0000,0000,0000,,would be the hypotenuse\Nof this right triangle.
Dialogue: 0,0:02:00.18,0:02:03.04,Default,,0000,0000,0000,,Remember, they're traveling\Nfrom perpendicular directions.
Dialogue: 0,0:02:03.04,0:02:04.76,Default,,0000,0000,0000,,So, that's a right triangle, there.
Dialogue: 0,0:02:04.76,0:02:09.76,Default,,0000,0000,0000,,So, this distance, right over\Nhere, would be X of T squared
Dialogue: 0,0:02:11.30,0:02:16.30,Default,,0000,0000,0000,,plus Y of T squared, and\Nthe square root of that,
Dialogue: 0,0:02:16.71,0:02:19.07,Default,,0000,0000,0000,,and that's just the Pythagorean\Ntheorem, right over here.
Dialogue: 0,0:02:19.07,0:02:23.17,Default,,0000,0000,0000,,This would be D of T, or we could say
Dialogue: 0,0:02:23.17,0:02:27.24,Default,,0000,0000,0000,,that D of T squared is equal to
Dialogue: 0,0:02:29.27,0:02:34.27,Default,,0000,0000,0000,,X of T squared plus Y,\Ntoo many parentheses.
Dialogue: 0,0:02:34.90,0:02:38.30,Default,,0000,0000,0000,,Plus Y of T squared.
Dialogue: 0,0:02:38.30,0:02:39.61,Default,,0000,0000,0000,,So, that's the relationship between
Dialogue: 0,0:02:39.61,0:02:43.87,Default,,0000,0000,0000,,D of T, X of T, and Y\Nof T, and it's useful
Dialogue: 0,0:02:43.87,0:02:45.74,Default,,0000,0000,0000,,for solving this problem because, now,
Dialogue: 0,0:02:45.74,0:02:47.68,Default,,0000,0000,0000,,we can take the derivative of both sides
Dialogue: 0,0:02:47.68,0:02:49.51,Default,,0000,0000,0000,,of this equation with respect to T.
Dialogue: 0,0:02:49.51,0:02:52.00,Default,,0000,0000,0000,,We'd be using various derivative rules,
Dialogue: 0,0:02:52.00,0:02:54.40,Default,,0000,0000,0000,,including the chain\Nrule, in order to do it,
Dialogue: 0,0:02:54.40,0:02:56.61,Default,,0000,0000,0000,,and then that would give us a relationship
Dialogue: 0,0:02:56.61,0:02:59.62,Default,,0000,0000,0000,,between the rate of change of D of T,
Dialogue: 0,0:02:59.62,0:03:01.37,Default,,0000,0000,0000,,which would be D prime of T,
Dialogue: 0,0:03:01.37,0:03:04.48,Default,,0000,0000,0000,,and the rate of change of X of T, Y of T,
Dialogue: 0,0:03:04.48,0:03:07.45,Default,,0000,0000,0000,,and X of T and Y of T, themselves.
Dialogue: 0,0:03:07.45,0:03:10.42,Default,,0000,0000,0000,,And so, if we look at these\Nchoices, right over here,
Dialogue: 0,0:03:11.77,0:03:14.83,Default,,0000,0000,0000,,we indeed see that D sets up that exact
Dialogue: 0,0:03:14.83,0:03:17.94,Default,,0000,0000,0000,,same relationship that\Nwe just did, ourselves.
Dialogue: 0,0:03:17.94,0:03:20.23,Default,,0000,0000,0000,,That it shows that the distance
Dialogue: 0,0:03:20.23,0:03:22.43,Default,,0000,0000,0000,,squared between the cars is equal to
Dialogue: 0,0:03:22.43,0:03:25.37,Default,,0000,0000,0000,,that X distance from the\Nintersection squared,
Dialogue: 0,0:03:25.37,0:03:27.88,Default,,0000,0000,0000,,plus the Y distance from\Nthe intersection squared,
Dialogue: 0,0:03:27.88,0:03:30.44,Default,,0000,0000,0000,,and then we can take the\Nderivative of both sides
Dialogue: 0,0:03:30.44,0:03:33.38,Default,,0000,0000,0000,,to actually figure out this\Nrelated, rates question.