Question

If the volume of a solid hemisphere increases at a uniform rate, prove that its surface area varies inversely as its radius.

Hint:

The volume of a hemisphere of radius \(r\) is \[ V=\frac{2}{3}\pi r^3 \]

Answer 1

Let the radius of the solid hemisphere be \( r \). The volume \( V \) of a hemisphere is given by the formula:
\[ V = \frac{2}{3} \pi r^3 \] Also, the surface area \( A \) of the hemisphere is given by:
\[ A = 3 \pi r^2 \] Now, we are told that the volume of the hemisphere increases at a uniform rate. This means that the rate of change of volume with respect to time is constant. Let \( \frac{dV}{dt} = k \), where \( k \) is a constant.
Differentiating the volume formula with respect to time \( t \), we get:
\[ \frac{dV}{dt} = \frac{d}{dt}\left( \frac{2}{3} \pi r^3 \right) = 2 \pi r^2 \frac{dr}{dt} \] Since \( \frac{dV}{dt} = k \), we have:
\[ 2 \pi r^2 \frac{dr}{dt} = k \] Solving for \( \frac{dr}{dt} \), we get:
\[ \frac{dr}{dt} = \frac{k}{2 \pi r^2} \] This shows that the rate of change of the radius with respect to time is inversely proportional to the square of the radius.
Next, we need to show that the surface area varies inversely as the radius. From the surface area formula, we know:
\[ A = 3 \pi r^2 \] Now, differentiate \( A \) with respect to time \( t \):
\[ \frac{dA}{dt} = \frac{d}{dt}\left( 3 \pi r^2 \right) = 6 \pi r \frac{dr}{dt} \] Substituting \( \frac{dr}{dt} = \frac{k}{2 \pi r^2} \) into this, we get:
\[ \frac{dA}{dt} = 6 \pi r \cdot \frac{k}{2 \pi r^2} = \frac{3k}{r} \] Hence, the rate of change of surface area with respect to time is inversely proportional to the radius. This implies that as the radius increases, the surface area increases at a slower rate, which shows that the surface area varies inversely with the radius.
Therefore, we have proven that if the volume of a solid hemisphere increases at a uniform rate, its surface area varies inversely as its radius.