Differentiation - Chain Rule

Question

A cone-shaped container with a hole at its top vertex is filled with water. The cone has a 4cm radius and 16cm in height. The cone was turned upside down to the let its water flow out of container. The water is draining at the rate of 2 cm^3 per second. Find the rate of change in height at the instant where water-level reached 10cm.

Answer 1

(Volume of cone at time t) = V = (1/3)πr²h

 

r = radius of surface of the water at time t

 

h = depth of water at time t

 

Given:  dV/dt = -2

 

Find:  dh/dt when h = 10

 

Using similar triangles, 16 / 4 = h/ r

 

                                  So, 4h = 16r   r = (1/4)h

 

Therefore, V = (1/3)π((1/4)h)²h

 

                   = (1/48)πh³

 

dV/dt = (1/16)πh²(dh/dt)

 

So, when h = 10, we have:  -2 = (1/16)π(10)²(dh/dt)

 

                                       dh/dt = -8/(25π) cm / sec

 

When h = 10 cm, the height (depth of water) is decreasing at the rate of 8/(25π) cm/sec.