Question

If the rate of change of volume of a sphere is twice the rate of change of its radius, then the surface area of the sphere is:

• A. 1 sq unit
• B. 2 sq units
• C. 3 sq units
• D. 4 sq units

Hint:

When dealing with rates of change, use the chain rule to relate derivatives of volume and surface area to the radius, and solve for the given condition.

Answer 1

The Correct Option is B

Let the radius of the sphere be \( r \) and the surface area be \( S \). The volume \( V \) of a sphere is given by: \[ V = \frac{4}{3} \pi r^3 \] The surface area \( S \) of a sphere is given by: \[ S = 4 \pi r^2 \] We are given that the rate of change of volume is twice the rate of change of the radius. The rate of change of volume with respect to time \( t \) is \( \frac{dV}{dt} \), and the rate of change of radius with respect to time is \( \frac{dr}{dt} \). According to the problem: \[ \frac{dV}{dt} = 2 \frac{dr}{dt} \] First, differentiate the volume \( V = \frac{4}{3} \pi r^3 \) with respect to \( t \) using the chain rule: \[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = \frac{4}{3} \pi \cdot 3 r^2 \cdot \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt} \] Given \( \frac{dV}{dt} = 2 \frac{dr}{dt} \), substitute the expression for \( \frac{dV}{dt} \): \[ 4 \pi r^2 \frac{dr}{dt} = 2 \frac{dr}{dt} \] Assuming \( \frac{dr}{dt} \neq 0 \) (since the radius is changing), we can divide both sides by \( \frac{dr}{dt} \): \[ 4 \pi r^2 = 2 \] Solving for \( r^2 \): \[ r^2 = \frac{2}{4 \pi} = \frac{1}{2 \pi} \] Now, substitute \( r^2 \) into the surface area formula \( S = 4 \pi r^2 \): \[ S = 4 \pi \cdot \frac{1}{2 \pi} = \frac{4 \pi}{2 \pi} = 2 \]