Related Rates — Question 3

Question 3 - Solution

We are given: $$\frac{dr}{dt} = 3 \ \text{cm/min}, \quad r = 10 \ \text{cm}$$

The volume of a sphere is given by: $$V = \frac{4}{3} \pi r^3$$

Differentiate both sides with respect to $t$: $$\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 4\pi r^2 \frac{dr}{dt}$$

Substitute the known values: $$\frac{dV}{dt} = 4\pi (10)^2 (3) = 4\pi (100)(3) = 1200\pi$$

$$\frac{dV}{dt} \approx 3769.91 \ \text{cm}^3/\text{min}$$

Answer: The volume is increasing at a rate of $\boxed{1200\pi \approx 3769.91 \ \text{cm}^3/\text{min}}$ when the radius is 10 cm.

Interpretation: When the balloon’s radius is 10 cm and growing at 3 cm per minute, the volume is expanding at approximately 3769.91 cubic centimeters per minute. This tells us how rapidly the space inside the balloon is increasing at that instant.