Question

Let the volume of a metallic hollow sphere be constant. If the inner radius increases at the rate of 2 cm/s, find the rate of increase of the outer radius when the radii are 2 cm and 4 cm respectively.

Hint:

When solving related rate problems, remember to differentiate implicitly and use the given rates to find the unknown rate.

Answer 1

The volume \( V \) of a hollow sphere, defined by \( V = \frac{4}{3} \pi (R^3 - r^3) \) with outer radius \( R \) and inner radius \( r \), is constant. This constancy implies \( \frac{dV}{dt} = 0 \). Differentiating the volume equation with respect to time \( t \) yields: \[ \frac{d}{dt}\left(\frac{4}{3} \pi (R^3 - r^3)\right) = 0 \] Applying the chain rule, we get: \[ \frac{4}{3} \pi \left( 3R^2 \frac{dR}{dt} - 3r^2 \frac{dr}{dt} \right) = 0 \] Simplifying this equation results in: \[ R^2 \frac{dR}{dt} = r^2 \frac{dr}{dt} \] Given the values: - \( \frac{dr}{dt} = 2 \, \text{cm/s} \) (the rate of change of the inner radius),
- \( r = 2 \, \text{cm} \), and
- \( R = 4 \, \text{cm} \).
Substituting these values into the simplified equation: \[ (4)^2 \cdot \frac{dR}{dt} = (2)^2 \cdot 2 \] \[ 16 \cdot \frac{dR}{dt} = 4 \cdot 2 \] \[ 16 \cdot \frac{dR}{dt} = 8 \] Solving for \( \frac{dR}{dt} \): \[ \frac{dR}{dt} = \frac{8}{16} = \frac{1}{2} \, \text{cm/s} \] Consequently, the rate of increase of the outer radius is \( \frac{1}{2} \, \text{cm/s} \).