Question

The surface area of a solid hemisphere is increasing at the rate of \( 8 \, \text{cm}^2/\text{sec} \) (retaining its shape). Then the rate of change of its volume (in \( \text{cm}^3/\text{sec} \)), when the radius is \( 5 \,\text{cm} \), is

• A. \( \frac{50}{3} \)
• B. \( \frac{20}{3} \)
• C. \( \frac{40}{3} \)
• D. \( \frac{25}{3} \)
• E. \( \frac{80}{3} \)

Hint:

In related rates, first find \( \frac{dr}{dt} \), then substitute into required derivative.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a related rates problem. We are given the rate of change of the surface area (\(\frac{dA}{dt}\)) and asked to find the rate of change of the volume (\(\frac{dV}{dt}\)) at a specific instant (when the radius r=5cm). The key is to find a relationship between V and A, or to find \(\frac{dr}{dt}\) as an intermediate step.
Step 2: Key Formula or Approach:
1. The surface area of a solid hemisphere is the sum of the curved surface area (\(2\pi r^2\)) and the area of the circular base (\(\pi r^2\)). So, \(A = 3\pi r^2\). 2. The volume of a hemisphere is \(V = \frac{2}{3}\pi r^3\). 3. Differentiate both A and V with respect to time `t` to find expressions for \(\frac{dA}{dt}\) and \(\frac{dV}{dt}\). 4. Use the given value of \(\frac{dA}{dt}\) to find \(\frac{dr}{dt}\) when r=5. 5. Use this value of \(\frac{dr}{dt}\) to find \(\frac{dV}{dt}\).
Step 3: Detailed Explanation:
1. Differentiate the Area formula. \[ A = 3\pi r^2 \] Differentiating with respect to time `t`: \[ \frac{dA}{dt} = 3\pi (2r) \frac{dr}{dt} = 6\pi r \frac{dr}{dt} \] 2. Find \(\frac{dr}{dt}\). We are given \(\frac{dA}{dt} = 8\) cm\(^2\)/sec and we need to find the rate at the instant when \(r=5\) cm. \[ 8 = 6\pi (5) \frac{dr}{dt} \] \[ 8 = 30\pi \frac{dr}{dt} \] \[ \frac{dr}{dt} = \frac{8}{30\pi} = \frac{4}{15\pi} \text{ cm/sec} \] 3. Differentiate the Volume formula. \[ V = \frac{2}{3}\pi r^3 \] Differentiating with respect to time `t`: \[ \frac{dV}{dt} = \frac{2}{3}\pi (3r^2) \frac{dr}{dt} = 2\pi r^2 \frac{dr}{dt} \] 4. Calculate \(\frac{dV}{dt}\). Now substitute \(r=5\) and the value of \(\frac{dr}{dt}\) we found: \[ \frac{dV}{dt} = 2\pi (5)^2 \left(\frac{4}{15\pi}\right) \] \[ \frac{dV}{dt} = 2\pi (25) \left(\frac{4}{15\pi}\right) = 50\pi \left(\frac{4}{15\pi}\right) \] The \(\pi\) terms cancel out. \[ \frac{dV}{dt} = \frac{50 \cdot 4}{15} = \frac{10 \cdot 4}{3} = \frac{40}{3} \] Step 4: Final Answer:
The rate of change of its volume is \(\frac{40}{3}\) cm\(^3\)/sec.