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Current time:0:00Total duration:3:35

0 energy points

AP.CALC:

CHA‑3 (EU)

, CHA‑3.D (LO)

, CHA‑3.D.1 (EK)

, CHA‑3.D.2 (EK)

two cars are driving towards an intersection from perpendicular directions the first cars velocity is 50 kilometers per hour and the second car's velocity is 90 kilometers per hour at a certain instant T sub zero the first car is a distance X sub T sub 0 or X of T sub zero of half a kilometer from the intersection and the second car is a distance Y of T sub 0 of 1.2 kilometers from the intersection what is the rate of change of the distance D of T between the cars at that instant so at t sub 0 which equation should be used to solve the problem and they give us a choice of four equations right over here so you could pause the video and try to work through it on your own but I'm about to do it as well so let's just draw what's going on that's always a healthy thing to do so two cars are driving towards an intersection from perpendicular directions so let's say that this is one car right over here and it is moving in the direct x direction towards that intersection which is right over there and then you have another car that is moving in the y direction so let's say it's moving like this so this is the other car I should have maybe done a top view well here we go this square represents the car and it is moving in that direction now they say at a certain instant T Sub Zero so let's draw that instant so the first car is a distance X of T Sub Zero of 0.5 kilometers so this distance right over here let's just call this X of T and let's call this distance right over here Y of T now how does the distance between the cars relate to X of T and Y of T well we could just use the distance formula which is essentially just the Pythagorean theorem to say well the distance between the cars would be the hypotenuse of this right triangle remember they're traveling from perpendicular directions so that's a right triangle there so this distance right over here would be X of T squared plus y t squared and the square root of that and that's just the Pythagorean theorem right over here this would be D of T or we could say that D of T squared is equal to X of T X of T squared plus y let me do any parentheses plus y of T squared so that's the relationship between D of T X of T and Y of T and it's useful for solving this problem because now we could take the derivative of both sides of this equation with respect to T and we'd be using various derivative rules including the chain rule in order to do it and then that would give us a relationship between the rate of change of D of T which would be D prime of T and the rate of change of X of T Y of T and X of T and Y of T themselves and so if we look at these choices right over here we in do we indeed see that D sets up that exact same relationship that we just did ourselves that it's it shows that the the distance squared between the cars is equal to that x distance from the intersection squared plus the Y distance from the intersection squared and then we can take the derivative of both sides to actually figure out this related rates question

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