Related Rates (calculus)

Question

A balloon is at a height of 50 meters, and is rising at the constant rate of 5 m/sec. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10 m/sec. How fast is the distance between the bicyclist and the balloon increasing 2 seconds later?

Answer 1

SOLUTION

We know at t=2 s:

y = 60 ft

x = 20 ft

d^2 = x^2 + y^2

d=20√10 ft

dy/dt = 5 ft/s

dx/dt = 10 ft/s

d^2 = x^2 + y^2

Therefore:

2d(dd/dt) = 2x(dx/dt) + 2y(dy/dt)

2(20√10)(dd/dt) = 2(20)(10) +2(60)(5)

(40√10)(dd/dt) = 1000

dd/dt = 1000/(40√10)

dd/dt = (5√10)/2 ft/s