Question

The volume of a spherical balloon is increasing at the rate of \( 6 cm^3/sec\). The rate of change of its surface area when its radius 2 cm is:

• A. \( 3 cm^2/sec\)
• B. \(\frac32 cm^2/sec\)
• C. \( 6cm^2/sec\)
• D. \( 9cm^2/sec\)

Answer 1

The Correct Option is C

To determine the rate of change of the surface area of a spherical balloon when its volume is increasing at a specific rate, we can use the known formulas for the volume and surface area of a sphere. The volume \(V\) of a sphere with radius \(r\) is given by the formula:

\( V = \frac{4}{3}\pi r^3 \)

Given that \( \frac{dV}{dt} = 6 \, \text{cm}^3/\text{sec} \), we need to find \( \frac{dA}{dt} \) when \( r = 2 \, \text{cm} \). The surface area \(A\) of a sphere is given by:

\( A = 4\pi r^2 \)

To solve this, we'll differentiate both \(V\) and \(A\) with respect to time \(t\):

\(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)

\(\frac{dA}{dt} = 8\pi r \frac{dr}{dt}\)

First, solve for \( \frac{dr}{dt} \) using the volume differential equation:

\(6 = 4\pi (2)^2 \frac{dr}{dt}\)

\(6 = 16\pi \frac{dr}{dt}\)

\(\frac{dr}{dt} = \frac{6}{16\pi} = \frac{3}{8\pi}\)

Now, substitute \( \frac{dr}{dt} \) into the surface area differential equation:

\(\frac{dA}{dt} = 8\pi(2)\left(\frac{3}{8\pi}\right)\)

\(\frac{dA}{dt} = 16\pi \cdot \frac{3}{8\pi}\)

\(\frac{dA}{dt} = 6 \, \text{cm}^2/\text{sec}\)

The correct rate of change of its surface area when the radius is 2 cm is therefore \( 6 \, \text{cm}^2/\text{sec} \).