Question

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

Hint:

Always clarify if "surface area" refers to the curved surface area (\( 2\pi r^2 \)) or the total surface area (\( 3\pi r^2 \)) for a solid.
In this case, the proportionality holds for both (\( 2k/r \) vs \( 3k/r \)), so the proof remains valid regardless of the specific definition used.

Answer 1

Concept:
• Volume of a solid hemisphere: \( V = \frac{2}{3}\pi r^3 \).
• Total Surface Area of a solid hemisphere: \( S = 3\pi r^2 \) (Curved surface area \( 2\pi r^2 \) + base area \( \pi r^2 \)).
• Uniform rate of increase means \( \frac{dV}{dt} = k \), where \( k \) is a constant.
• Chain Rule for related rates: \( \frac{dy}{dt} = \frac{dy}{dr} \cdot \frac{dr}{dt} \).

Step 1: Relate the rate of change of volume to the rate of change of radius
Given \( V = \frac{2}{3}\pi r^3 \).
Differentiating with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dr}\left(\frac{2}{3}\pi r^3\right) \cdot \frac{dr}{dt} \]
\[ \frac{dV}{dt} = \left(\frac{2}{3}\pi \cdot 3r^2\right) \cdot \frac{dr}{dt} = 2\pi r^2 \frac{dr}{dt} \]
Since the volume increases at a uniform rate \( k \):
\[ k = 2\pi r^2 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{k}{2\pi r^2} \]

Step 2: Find the rate of change of surface area
Given \( S = 3\pi r^2 \).
Differentiating with respect to time \( t \):
\[ \frac{dS}{dt} = \frac{d}{dr}(3\pi r^2) \cdot \frac{dr}{dt} \]
\[ \frac{dS}{dt} = 6\pi r \frac{dr}{dt} \]

Step 3: Substitute the expression for \( \frac{dr}{dt} \) into the surface area rate equation
Substitute \( \frac{dr}{dt} = \frac{k}{2\pi r^2} \) into the equation from
Step 2:
\[ \frac{dS}{dt} = 6\pi r \cdot \left( \frac{k}{2\pi r^2} \right) \]
\[ \frac{dS}{dt} = \frac{3k}{r} \]

Step 4: Conclusion
Since \( 3k \) is a constant, we have:
\[ \frac{dS}{dt} \propto \frac{1}{r} \]
This proves that the surface area varies inversely as its radius.