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 carefully distinguish between total surface area (\(3\pi r^2\)) and curved surface area (\(2\pi r^2\)) for a solid hemisphere. If it mentions an open bowl or hollow hemisphere, use \(2\pi r^2\), but for a solid object, the flat base counts!

Answer 1

Concept: This problem involves application of derivatives related to changing rates. For a solid hemisphere of radius \(r\):
• The volume \(V\) is given by the formula: \( V = \frac{2}{3}\pi r^3 \)
• The total surface area \(S\) includes the curved surface area and the flat circular base: \( S = 2\pi r^2 + \pi r^2 = 3\pi r^2 \)
• "Uniform rate" implies that the derivative of volume with respect to time \(t\) is a constant value, say \(k\): \( \frac{dV}{dt} = k \).

Step 1: Write expressions for volume and relate its time derivative.
Let \(r\) be the radius, \(V\) be the volume, and \(S\) be the total surface area of the solid hemisphere at any given time \(t\). The volume formula is: \[ V = \frac{2}{3}\pi r^3 \] Differentiating both sides with respect to time \(t\) using the chain rule: \[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{2}{3}\pi r^3\right) = \frac{2}{3}\pi \cdot \left(3r^2 \cdot \frac{dr}{dt}\right) \] Simplifying the expression by cancelling the factor of 3: \[ \frac{dV}{dt} = 2\pi r^2 \frac{dr}{dt} \]

Step 2: Utilize the uniform rate condition.
Since the volume increases at a uniform rate, we can set \( \frac{dV}{dt} = k \), where \(k\) is a positive constant value: \[ 2\pi r^2 \frac{dr}{dt} = k \quad \Rightarrow \quad \frac{dr}{dt} = \frac{k}{2\pi r^2} \quad \cdots (1) \]

Step 3: Differentiate the surface area formula with respect to time.
The total surface area of a solid hemisphere is given by: \[ S = 3\pi r^2 \] Differentiating both sides with respect to time \(t\): \[ \frac{dS}{dt} = \frac{d}{dt}(3\pi r^2) = 3\pi \cdot \left(2r \frac{dr}{dt}\right) \] \[ \frac{dS}{dt} = 6\pi r \frac{dr}{dt} \quad \cdots (2) \]

Step 4: Substitute the value of \( \frac{dr}{dt} \) from Equation (1) into Equation (2).
\[ \frac{dS}{dt} = 6\pi r \cdot \left(\frac{k}{2\pi r^2}\right) \] Simplifying the fractions and terms carefully: \[ \frac{dS}{dt} = \frac{6\pi \cdot k \cdot r}{2\pi \cdot r^2} = \frac{3k}{r} \] Since \(k\) is a constant, \(3k\) is also a totally new combined constant, let's say \(C = 3k\). Therefore: \[ \frac{dS}{dt} = \frac{C}{r} \] This proves that the rate of change of surface area \( \frac{dS}{dt} \) is inversely proportional to its radius \(r\). That means \( \frac{dS}{dt} \propto \frac{1}{r} \).