Question

The radius of a cylinder is decreasing at a rate of 2 cm/s and the altitude is increasing at a rate of 3 cm/s. Find the rate of change of volume of this cylinder when its radius is 4 cm and altitude is 6 cm.

Hint:

To solve for rates of change involving volume, apply the product rule of differentiation when the volume formula involves multiple variables, and substitute the given values accordingly.

Answer 1

The formula for the volume \( V \) of a cylinder is \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. To determine the rate of change of volume with respect to time \( t \), we differentiate the volume formula with respect to \( t \): \[ \frac{dV}{dt} = \frac{d}{dt} \left( \pi r^2 h \right) \] Applying the product rule, we get: \[ \frac{dV}{dt} = \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right) \] Given the rates of change \( \frac{dr}{dt} = -2 \, \text{cm/s} \) and \( \frac{dh}{dt} = 3 \, \text{cm/s} \), and the dimensions \( r = 4 \, \text{cm} \) and \( h = 6 \, \text{cm} \), we substitute these values: \[ \frac{dV}{dt} = \pi \left( 2(4)(6)(-2) + (4)^2(3) \right) \] \[ \frac{dV}{dt} = \pi \left( -96 + 48 \right) = \pi (-48) \] Therefore, the rate of change of the cylinder's volume is: \[ \frac{dV}{dt} = -48\pi \, \text{cm}^3/\text{s} \]