If `V=(4)/(3)pir^(3)`, at what rate in cubic units is V increasing when r = 10 and `(dr)/(dt)` = 0.01?

Options

A. `pi`
B. `4pi`
C. `40pi`
D. `4pi//3`

Explanation

To solve the problem, we need to find the rate at which the volume \( V \) is increasing when the radius \( r = 10 \) and the rate of change of the radius \( \frac{dr}{dt} = 0.01 \).

### Step-by-Step Solution:

1. **Write the formula for volume**:
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]

2. **Differentiate the volume with respect to time**:
To find the rate at which the volume is changing with respect to time, we need to differentiate \( V \) with respect to \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3} \pi r^3\right) \]
Using the chain rule, we have:
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt} \]
The \( 3 \) from the differentiation of \( r^3 \) cancels with the \( \frac{1}{3} \) in front:
\[ \frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt} \]

3. **Substitute the known values**:
Now we substitute \( r = 10 \) and \( \frac{dr}{dt} = 0.01 \) into the equation:
\[ \frac{dV}{dt} = 4 \pi (10^2) \cdot 0.01 \]
Calculate \( 10^2 \):
\[ 10^2 = 100 \]
Substitute this back into the equation:
\[ \frac{dV}{dt} = 4 \pi \cdot 100 \cdot 0.01 \]

4. **Simplify the expression**:
\[ \frac{dV}{dt} = 4 \pi \cdot 1 = 4 \pi \]

5. **Final answer with units**:
Therefore, the rate at which the volume is increasing when \( r = 10 \) is:
\[ \frac{dV}{dt} = 4 \pi \text{ cubic units} \]

### Summary:
The volume \( V \) is increasing at a rate of \( 4 \pi \) cubic units when the radius \( r = 10 \) and \( \frac{dr}{dt} = 0.01 \).