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 volume \( V \) of a cylinder is given by the formula: \[ V = \pi r^2 h \] where \( r \) is the radius and \( h \) is the height (altitude) of the cylinder. To find the rate of change of volume with respect to time, we differentiate both sides of the equation with respect to time \( t \): \[ \frac{dV}{dt} = \frac{d}{dt} \left( \pi r^2 h \right) \] Using the product rule: \[ \frac{dV}{dt} = \pi \left( 2r h \frac{dr}{dt} + r^2 \frac{dh}{dt} \right) \] We are given that \( \frac{dr}{dt} = -2 \, \text{cm/s} \), \( \frac{dh}{dt} = 3 \, \text{cm/s} \), \( r = 4 \, \text{cm} \), and \( h = 6 \, \text{cm} \). Substituting these values into the equation: \[ \frac{dV}{dt} = \pi \left( 2(4)(6)(-2) + (4)^2(3) \right) \] \[ \frac{dV}{dt} = \pi \left( -96 + 48 \right) = \pi (-48) \] Thus, the rate of change of the volume is: \[ \frac{dV}{dt} = -48\pi \, \text{cm}^3/\text{s} \]