Question

Surface area of a balloon (spherical), when air is blown into it, increases at a rate of 5 mm²/s. When the radius of the balloon is 8 mm, find the rate at which the volume of the balloon is increasing.

Hint:

To solve problems involving related rates, start by expressing the quantities in terms of the given variables and then differentiate using the chain rule.

Answer 1

For a spherical balloon: - Surface area \( A = 4\pi r^2 \) - Volume \( V = \frac{4}{3}\pi r^3 \) Given that the surface area increases at a rate of 5 mm²/s, we have: \[ \frac{dA}{dt} = 5 \, \text{mm}^2/\text{s} \] Differentiating the surface area with respect to time yields: \[ \frac{dA}{dt} = 8\pi r \frac{dr}{dt} \] Substituting \( \frac{dA}{dt} = 5 \) and \( r = 8 \) mm into this equation: \[ 5 = 8\pi (8) \frac{dr}{dt} \] Solving for \( \frac{dr}{dt} \) gives: \[ \frac{dr}{dt} = \frac{5}{64\pi} \] Subsequently, we aim to determine the rate of volume increase, \( \frac{dV}{dt} \). Differentiating the volume formula results in: \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \] Substituting \( r = 8 \) mm and \( \frac{dr}{dt} = \frac{5}{64\pi} \): \[ \frac{dV}{dt} = 4\pi (8)^2 \times \frac{5}{64\pi} \] Simplification leads to: \[ \frac{dV}{dt} = \frac{4\pi \times 64 \times 5}{64\pi} = 20 \, \text{mm}^3/\text{s} \] Therefore, the rate of volume increase for the balloon is \( \boxed{20 \, \text{mm}^3/\text{s}} \).