related rates

Question

A coffee filter has the shape of an inverted cone. The radius is 8 cm at the top and the filter's height is 10 cm. Water is drained from the filter at the rate of 25 cm^3/s. How fast is the depth of the water decreasing when it is 5 cm deep?

Answer 1

Let r = radius of the surface of the water at time t

     h = depth of water at time t

     V = volume of water at time t

 

V = (1/3)πr²h

 

By similar triangles, 10/h = 8/r.  So, 8h = 10r.  r = (4/5)h

 

Therefore, V = (1/3)π(16/25)h³ = (16/75)πh³

 

Given:  dV/dt = -25   Find:  dh/dt when h = 5

 

dV/dt = (16/25)πh²(dh/dt)

 

-25 = (16/25)π(5)²(dh/dt)

 

dh/dt = -25/(16π) cm/sec ≈ -0.497 cm/sec

 

The depth of water is decreasing at the rate of 0.497 cm/sec when the water is 5 cm deep.