Question

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

Hint:

In related rates problems, first identify the formula that connects the quantities involved (here, volume and radius). Then, differentiate the entire formula with respect to time (t), making sure to use the chain rule for any variable that is a function of t. Finally, substitute the given values to find the unknown rate.

Answer 1

Step 1: Understanding the Concept:
This is a related rates problem. We are given the rate of change of the radius of a sphere (\( \frac{dr}{dt} \)) and asked to find the rate of change of its volume (\( \frac{dV}{dt} \)) at a specific instant.
Step 2: Key Formula or Approach:
The volume (V) of a sphere with radius (r) is given by the formula:
\[ V = \frac{4}{3}\pi r^3 \] To find the relationship between the rates, we differentiate this formula with respect to time (t) using the chain rule.
Step 3: Detailed Explanation or Calculation:
First, differentiate the volume formula with respect to time, t:
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) \] Applying the chain rule:
\[ \frac{dV}{dt} = \frac{4}{3}\pi \cdot (3r^2) \cdot \frac{dr}{dt} \] \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \] We are given the following information:
- The rate of increase of the radius: \( \frac{dr}{dt} = \frac{1}{2} \) cm/s
- The specific radius at which to find the rate of volume increase: \( r = 1 \) cm
Now, substitute these values into the differentiated equation:
\[ \frac{dV}{dt} = 4\pi (1)^2 \left(\frac{1}{2}\right) \] \[ \frac{dV}{dt} = 4\pi (1) \left(\frac{1}{2}\right) \] \[ \frac{dV}{dt} = 2\pi \] Step 4: Final Answer:
The volume of the bubble is increasing at a rate of \( 2\pi \) cm\(^3\)/s when the radius is 1 cm.