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: Formula for the volume of a sphere.
The volume \(V\) of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3 \] where \(r\) is the radius of the sphere. 
Step 2: Given data.
We are given that the radius is increasing at the rate of \(\frac{dr}{dt} = \frac{1}{2} \, \text{cm/s}\), and we need to find the rate at which the volume is increasing, i.e., \(\frac{dV}{dt}\), when the radius \(r = 1 \, \text{cm}\). 
Step 3: Differentiating the volume formula with respect to time.
To find the rate of change of volume with respect to time, we differentiate both sides of the volume equation with respect to time \(t\): \[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) \] Using the chain rule, we get: \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \] Step 4: Substituting the known values.
We are given that \(\frac{dr}{dt} = \frac{1}{2} \, \text{cm/s}\) and \(r = 1 \, \text{cm}\), so we substitute these values into the equation: \[ \frac{dV}{dt} = 4 \pi (1)^2 \left( \frac{1}{2} \right) \] \[ \frac{dV}{dt} = 4 \pi \times \frac{1}{2} = 2\pi \, \text{cm}^3/\text{s} \] Final Answer: The volume of the bubble is increasing at a rate of \( 2\pi \, \text{cm}^3/\text{s} \) when the radius is 1 cm.