Question

If the volume of a sphere is increasing at a constant rate, then the rate at which its radius is increasing is

• A. inversely proportional to its surface area
• B. proportional to the radius
• C. a constant
• D. inversely proportional to the radius

Hint:

When dealing with problems involving the volume of a sphere, remember that the rate of change of the radius is related to the rate of change of the volume by the equation \( \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \). This shows how the radius increases relative to the surface are(A)

Answer 1

The Correct Option is A

The radius of a sphere is increasing is related to the rate at which its volume is increasing. 1. The volume \( V \) of a sphere is given by: \[ V = \frac{4}{3} \pi r^3 \] 2. The surface area \( A \) of a sphere is given by: \[ A = 4 \pi r^2 \] Step 1: Differentiate the volume equation We are told that the volume of the sphere is increasing at a constant rate. Let the rate of change of the volume with respect to time be \( \frac{dV}{dt} \). Differentiating the volume equation with respect to time, we get: \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \] This equation relates the rate of change of the volume \( \frac{dV}{dt} \) to the rate of change of the radius \( \frac{dr}{dt} \). Step 2: Solve for \( \frac{dr}{dt} \) Since \( \frac{dV}{dt} \) is constant, we can solve for \( \frac{dr}{dt} \): \[ \frac{dr}{dt} = \frac{1}{4 \pi r^2} \frac{dV}{dt} \] Step 3: Interpretation The expression \( \frac{dr}{dt} \) shows that the rate at which the radius is increasing is inversely proportional to the square of the radius, which is proportional to the surface are(A) Hence, as the volume increases at a constant rate, the rate at which the radius increases depends on the surface area, not directly on the radius itself. Thus, the correct answer is \( \boxed{\text{inversely proportional to its surface area}} \).