Rates of Change

Question

If a hemispherical bowl of radius 6cm contains water to a depth of x cm the volume of the water is 1/3pix^2(18-x) cm^3. Water is poured into the bowl at a rate of 3cm^3/s. Find the rate at which the water level is rising when the depth is 2cm.

Answer 1

Daniel,

 

This is a related rates problem. You have the volume as a function of depth

V(x) = (1/3)πx²(18-x) cm³

 

We apply the chain rule to find the rate of change of volume

dV/dt = (dV/dx) (dx/dt)

 

You can solve this equation for dx/dt, the rate of change of height. The volume is increasing at a rate of dV/dt = 3cm³/s. Differentiate V(x) with respect to x and plug in.

 

dx/dt = (dV/dt)(1/(dV/dx))

 

The derivative you need to compute is

dV/dx = π(12x - x²)

 

I hope this helps get you started.