question is in the description

Question

a car leaves an intersection traveling west. Its position 4 sec later is 20 ft from the intersection. At the same time, another car leaves the intersection heading north so that its position t sec later is t^2+2t ft from the intersection. If the speed of the first car 4 sec after leaving the intersection is 9ft/sec, find the rate at which the distance between the two cars is changing at that instant of time.

Answer 1

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Answer 2

Ty,

 

This is a related rates question. I assume that the two cars leave the intersection at the same time. If the second car leaves 4 seconds after the first, my solution will be incorrect. Let x(t) denote the position of the car moving west and y(t) denote the position of the car moving north. What you are given from the problem statement is the following:

 

x(4) = 20

x′(4) = 9

y(t) = t² + 2y

 

The object is to find s′(4), where s(t) is the distance between the two cars.

 

To get started, you must find an equation relating the three distances, x, y, and s. This is simply the Pythagorean theorem:

 

s² = x² + y².

 

Take the derivative with respect to t implicitly:

 

2s(t)s′(t) = 2x(t)x′(t) + 2y(t)y′(t).

 

At time t = 4, you can find all of the missing quantities except s′(4).

 

x(4) = 20 (given)

y(4) = 24 (computed from y(t))

x′(4) = 9 (given)

y′(4) = 10 (calculated from y′(t) = 2t + 2)

s(4) = √[20² + (4² + 2·4)²] = √976

 

When you plug all of this into the equation above, you can solve for s′(4):

 

2(√976)s′(4) = 2(20)(9) + 2(24)(10)

 

s′(4) = 420/√976 ≈ 13.44 ft/sec

 

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