Question

The radius of an air bubble is increasing at the rate of \(\frac{1}{2} \, \text{cm/s}\). At what rate is the volume of the bubble increasing while the radius is 1 cm?

Hint:

{Key Steps for Related Rates:}

• Write formula relating variables
• Differentiate with respect to time
• Substitute known values
• Solve for required rate

For sphere: \( V = \frac{4}{3}\pi r^3 \) → \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \)

Answer 1

Step 1: Identify given information Let \( r \) be the radius of the air bubble. \[ \frac{dr}{dt} = \frac{1}{2} \, \text{cm/s} \] We need to find \( \frac{dV}{dt} \) when \( r = 1 \) cm.
Step 2: Formula for volume of a sphere The volume of a spherical bubble is: \[ V = \frac{4}{3} \pi r^3 \]
Step 3: Differentiate with respect to time \[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt} \] \[ \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \]
Step 4: Substitute the given values When \( r = 1 \) cm and \( \frac{dr}{dt} = \frac{1}{2} \, \text{cm/s} \): \[ \frac{dV}{dt} = 4\pi (1)^2 \cdot \frac{1}{2} \] \[ \frac{dV}{dt} = 4\pi \cdot \frac{1}{2} = 2\pi \]
Step 5: Final answer \[ \boxed{2\pi \, \text{cm}^3/\text{s}} \] Interpretation: When the radius is 1 cm, the volume of the air bubble is increasing at the rate of \( 2\pi \) cubic centimeters per second (approximately \( 6.28 \, \text{cm}^3/\text{s} \)).