Question

A spherical balloon is being inflated such that its radius increases at a constant rate of \( 2 \,\text{cm/sec} \). The rate at which the volume is increasing when the radius is \( 5 \,\text{cm} \) is

• A. \( 200\pi \)
• B. \( 150\pi \)
• C. \( 300\pi \)
• D. \( 250\pi \)
• E. \( 100\pi \)

Hint:

Always differentiate first, then substitute values in related rates problems.

Answer 1

The Correct Option is A

Concept:
• This is a related rates problem.
• Volume of a sphere: \[ V = \frac{4}{3}\pi r^3 \]
• When a quantity depends on another changing quantity, we differentiate with respect to time \(t\): \[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) \]

Step 1: Write the given information clearly.

• Radius is increasing at: \[ \frac{dr}{dt} = 2 \,\text{cm/sec} \]
• Radius at the instant: \[ r = 5 \,\text{cm} \]
• We need to find: \[ \frac{dV}{dt} \]

Step 2: Differentiate volume formula with respect to time.
\[ V = \frac{4}{3}\pi r^3 \] Differentiate both sides w.r.t time \(t\): \[ \frac{dV}{dt} = \frac{4}{3}\pi \cdot \frac{d}{dt}(r^3) \] Now apply chain rule: \[ \frac{d}{dt}(r^3) = 3r^2 \frac{dr}{dt} \] Thus: \[ \frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \frac{dr}{dt} \]

Step 3: Simplify expression.
\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \] This is the standard related rates formula for a sphere.

Step 4: Substitute the given values.
\[ r = 5, \quad \frac{dr}{dt} = 2 \] \[ \frac{dV}{dt} = 4\pi (5)^2 (2) \]

Step 5: Perform calculations step-by-step.
\[ (5)^2 = 25 \] \[ 4 \times 25 = 100 \] \[ 100 \times 2 = 200 \] Thus: \[ \frac{dV}{dt} = 200\pi \]

Step 6: Interpretation.

• Volume is increasing at \(200\pi\) cubic cm per second
• Faster radius growth leads to much faster volume increase (since \(r^2\) term)

Step 7: Final Answer.
\[ \boxed{200\pi} \]