Question

The surface area of a sphere when its volume changes at the same rate as its radius is :

• A. \(4\pi \text{ sq. units}\)
• B. \(1 \text{ sq. unit}\)
• C. \(4 \text{ sq. units}\)
• D. \(\pi \text{ sq. units}\)

Hint:

Always write down the basic formulas for volume and surface area first.
The rate of change of volume \(\frac{dV}{dt}\) is always equal to the surface area \(S\) times \(\frac{dr}{dt}\).
Pay close attention to units in the final answer.

Answer 1

The Correct Option is B

Concept:
• Volume of a sphere of radius \(r\) is given by \(V = \frac{4}{3}\pi r^3\).
• Surface area of a sphere of radius \(r\) is given by \(S = 4\pi r^2\).
• Rate of change of volume with respect to time is \(\frac{dV}{dt}\).
• Rate of change of radius with respect to time is \(\frac{dr}{dt}\).

Step 1: Find the derivative of volume with respect to time
Using the chain rule for \(V = \frac{4}{3}\pi r^3\):
\[ \frac{dV}{dt} = \frac{d}{dr}\left(\frac{4}{3}\pi r^3\right) \cdot \frac{dr}{dt} \] \[ \frac{dV}{dt} = \left(\frac{4}{3}\pi \cdot 3r^2\right) \cdot \frac{dr}{dt} \] \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]

Step 2: Use the given condition to find a relationship
The question states that the volume changes at the same rate as its radius.
\[ \frac{dV}{dt} = \frac{dr}{dt} \] Substituting the expression from
Step 1:
\[ 4\pi r^2 \frac{dr}{dt} = \frac{dr}{dt} \] Assuming the radius is changing (\(\frac{dr}{dt} \neq 0\)), we can divide both sides by \(\frac{dr}{dt}\):
\[ 4\pi r^2 = 1 \]

Step 3: Identify the surface area
We know that the formula for the surface area \(S\) is \(4\pi r^2\).
From Step 2, we found that \(4\pi r^2 = 1\).
Therefore, \(S = 1\).
The surface area is \(1 \text{ sq. unit}\).