Calculus 1 problem.

Question

What is poured through a chute at a rate of 10 ft cube /min and falls in a conical pile whose bottom radius is always half the altitude. How fast is the radius of the base increasing when the pile is 8 feet high?

 

i got 5/12pi ft/sec roughly 0.08ft/sec? I'm a little lost. Please help.

Answer 1

Let r = radius of the pile at time t 

     h = height of the pile at time t

     V = volume of the pile at time t

 

Given:  dV/dt = 10

 

Find:  dr/dt when h = 8

 

Solution:  V = (1/3)πr²h

 

       Since r = (1/2)h, h = 2r.  When h = 8, r = 4.

 

       So, V = (2/3)πr³

 

Differentiating implicitly with respect to t, we have dV/dt = 2πr²(dr/dt)

 

So, 10 = 2π(4)²(dr/dt).

 

      dr/dt = 10/(32π) = 5/(16π) ft/min