Calculus rates question

Question

A road perpendicular to a highway leads to a farmhouse located 5 mile away. An automobile traveling on the highway passes through this intersection at a speed of 50mph.

How fast is the distance between the automobile and the farmhouse increasing when the automobile is 6 miles past the intersection of the highway and the road?

The distance between the automobile and the farmhouse is increasing at a rate of ____ miles per hour.

Answer 1

if the distance from the intersection to the farm house = 5 miles

then the distance from the car to the farmhouse = sqr(25+36) = sqr61

d= about 7.81 miles

d = sqr(5^2 + 6^2)= sqr61

d^2 = 25 + c^2 where c= car's distance from the intersection = 6

take the derivative with respect to time

2dd' = 2cc'

dd' = cc'

(sqr61)(d') = 6(50) = 300

d' = 300/sqr61 = about 300/7.81 = 38.41 miles per hour

Answer 2

Note that the distance (s) from the auto to the house will be sqrt(25+v^2t^2) where v is 50mph and t=time. Find the derivative ds/dt and substitute 6 for vt