Related rates calculus

Question

A road perpendicular to a highway leads to a farmhouse located a = 1.7

 km away (See figure below).

An automobile travels past the farmhouse at a speed of v = 45

 km/h. How fast is the distance between the automobile and the farmhouse increasing when the automobile is 3.4 km past the intersection of the highway and the road? (Round your answer to three decimal places.)

Answer 1

c^2 = a^2 + b^2 where c=distance from car to farm house

b'=v=45 km/hr, a'=0, b=3.4 km, a=1.7 km

c = sqr(3.4^2 +1.7^2) = about 3.8013

take the derivative with respect to time

2cc' = 2aa' + 2bb'

divide by 2

cc' = aa' + bb'

3.8013c' = 1.7(0) + 3.4(45) = 153

c' = 153/3.8013

= 40.249 km/hr = rate of change of distance from car to farmhouse