Find the rate of change of volume of a sphere with respect to its surface area when the radius is 2cm.

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the rate of change of volume of a sphere with respect to its surface area when the radius is 2 cm, we can follow these steps:

### Step 1: Write the formulas for volume and surface area of a sphere.
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
The surface area \( S \) of a sphere is given by the formula:
\[ S = 4 \pi r^2 \]

### Step 2: Differentiate the volume with respect to the radius \( r \).
To find the rate of change of volume with respect to the radius, we differentiate \( V \) with respect to \( r \):
\[ \frac{dV}{dr} = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \]

### Step 3: Differentiate the surface area with respect to the radius \( r \).
Next, we differentiate \( S \) with respect to \( r \):
\[ \frac{dS}{dr} = \frac{d}{dr} (4 \pi r^2) = 8 \pi r \]

### Step 4: Use the chain rule to find \( \frac{dV}{dS} \).
Using the chain rule, we can express \( \frac{dV}{dS} \) as:
\[ \frac{dV}{dS} = \frac{dV}{dr} \cdot \frac{dr}{dS} = \frac{dV}{dr} \cdot \frac{1}{\frac{dS}{dr}} \]
Substituting the derivatives we found:
\[ \frac{dV}{dS} = \frac{4 \pi r^2}{8 \pi r} = \frac{4r}{8} = \frac{r}{2} \]

### Step 5: Evaluate \( \frac{dV}{dS} \) at \( r = 2 \) cm.
Now, we substitute \( r = 2 \) cm into the equation:
\[ \frac{dV}{dS} \bigg|_{r=2} = \frac{2}{2} = 1 \]

### Step 6: Write the final answer with units.
Thus, the rate of change of volume with respect to surface area when the radius is 2 cm is:
\[ \frac{dV}{dS} = 1 \text{ cm}^3/\text{cm}^2 \]