Question

Water is filling a cylindrical water tank. As it is filled the radius of the tank expands at 0.2cm/s and the height decreases at the rate of 0.5cm/s. Find the rate at which the volume of the tank is changing at the instant the radius equals 1.8cm and height equals 65 cm. What is the volume of water at this point? Show all working. Volume, V, of the cylinder is given by V = ϖr2h.

Recall that V is a function of both r (radius of the cylinder) and h (the height of the cylinder). This is a partial differentiation problem. The volume is also changing with time.

Answer 1

141.94 cm³/s

Explanation:

Since the volume of the tank V = πr²h, and both the height and radius of the tank change with time, we find the rate of change of the volume with time, dV/dt from

dV/dt = dV//dr × dr/dt + dV/dh × dh/dt

where dV/dr = 2πrh, dr/dt = + 0.2 cm/s (since the radius of the tank expands), dV/dh = πr² and dh/dt = -0.5 cm/s (since the height of the tank decreases)

So,

dV/dt = dV/dr × dr/dt + dV/dh × dh/dt

dV/dt = 2πrh × + 0.2 cm/s + πr² × -0.5 cm/s

dV/dt = 0.4πrh cm/s - 0.5πr² cm/s

dV/dt = πr(0.4h - 0.5r) cm/s

We now find the rate at which the volume is changing when r = 1.8 cm and h = 65 cm.

So,

dV/dt = π(1.8 cm)(0.4 × 65 cm - 0.5 × 1.8 cm) cm/s

dV/dt = π(1.8 cm)(26 cm - 0.9 cm) cm/s

dV/dt = π(1.8 cm)(25.1 cm) cm/s

dV/dt = π(45.18 cm²) cm/s

dV/dt = 141.94 cm³/s