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

Step 1: Write the formula for volume of a hemisphere. 
The volume of a hemisphere of radius \(r\) is \[ V=\frac{2}{3}\pi r^3 \] Differentiate with respect to time \(t\): \[ \frac{dV}{dt}=2\pi r^2\frac{dr}{dt} \] Step 2: Use the given condition. 
The problem states that the volume increases at a uniform rate. Thus \[ \frac{dV}{dt}=\text{constant} \] Therefore \[ 2\pi r^2\frac{dr}{dt}=\text{constant} \] Step 3: Express $\dfrac{dr}{dt}$. 
\[ \frac{dr}{dt}=\frac{\text{constant}}{2\pi r^2} \] Thus \[ \frac{dr}{dt}\propto\frac{1}{r^2} \] Step 4: Surface area of a hemisphere. 
Surface area of a hemisphere is \[ S=2\pi r^2 \] Differentiate with respect to time: \[ \frac{dS}{dt}=4\pi r\frac{dr}{dt} \] Step 5: Substitute the value of $\dfrac{dr{dt}$.} 
Substitute \[ \frac{dr}{dt}\propto\frac{1}{r^2} \] Thus \[ \frac{dS}{dt}\propto r\left(\frac{1}{r^2}\right) \] \[ \frac{dS}{dt}\propto\frac{1}{r} \] Step 6: Conclusion. 
Hence the rate of change of surface area is inversely proportional to the radius. 
Final Result: \[ \boxed{\frac{dS}{dt}\propto\frac{1}{r}} \] Thus the surface area varies inversely as the radius.