Question

The volume of a spherical balloon is increasing at the rate of 25 c.c./s. Find the rate of change of its surface area at the instant when its radius is 5 cm.

Hint:

To find the rate of change of surface area or volume, differentiate the respective formulas and use related rates.

Answer 1

Step 1: Relation between volume and surface area. 
The volume \( V \) and surface area \( A \) of a sphere are given by the formulas: \[ V = \frac{4}{3} \pi r^3, \quad A = 4 \pi r^2 \] We are given that the rate of change of the volume is \( \frac{dV}{dt} = 25 \) c.c./s and need to find \( \frac{dA}{dt} \) when \( r = 5 \) cm. 
Step 2: Differentiating the volume equation. 
Differentiating \( V = \frac{4}{3} \pi r^3 \) with respect to \( t \), we get: \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \] Substituting \( \frac{dV}{dt} = 25 \), we have: \[ 25 = 4 \pi (5)^2 \frac{dr}{dt} \] \[ 25 = 100 \pi \frac{dr}{dt} \] \[ \frac{dr}{dt} = \frac{25}{100 \pi} = \frac{1}{4\pi} \, \text{cm/s} \] Step 3: Finding the rate of change of the surface area. 
Now, differentiate \( A = 4 \pi r^2 \) with respect to \( t \): \[ \frac{dA}{dt} = 8 \pi r \frac{dr}{dt} \] Substitute \( r = 5 \) and \( \frac{dr}{dt} = \frac{1}{4\pi} \): \[ \frac{dA}{dt} = 8 \pi (5) \left( \frac{1}{4\pi} \right) = 10 \, \text{cm}^2/\text{s} \] Step 4: Conclusion. 
Thus, the rate of change of the surface area when the radius is 5 cm is \( 10 \, \text{cm}^2/\text{s} \).