A sphere increases its volume at the rate of `pi cm^(3)//sec`. The rate at which its surface area increases when the radius is 1 cm is

Options

A. `2pisq.cm//s`
B. `pisq.cm//s`
C. `(3pi)/(2)sq. cm//s`
D. `(pi)/(2)sq.cm//s`

Answer

To solve the problem, we need to find the rate at which the surface area of a sphere increases when the radius is 1 cm, given that the volume of the sphere increases at a rate of \( \pi \, \text{cm}^3/\text{sec} \).

### Step 1: Understand 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: Differentiate the volume with respect to time
To find the rate of change of volume with respect to time, we differentiate both sides of the volume formula with respect to \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3} \pi r^3\right) \]
Using the chain rule, we have:
\[ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt} \]

### Step 3: Set up the equation with the given rate of volume change
We know that \( \frac{dV}{dt} = \pi \, \text{cm}^3/\text{sec} \). Therefore, we can set up the equation:
\[ 4\pi r^2 \frac{dr}{dt} = \pi \]

### Step 4: Solve for \( \frac{dr}{dt} \)
We can simplify the equation:
\[ 4r^2 \frac{dr}{dt} = 1 \]
Thus, we can express \( \frac{dr}{dt} \):
\[ \frac{dr}{dt} = \frac{1}{4r^2} \]

### Step 5: Find the rate of change of surface area
The surface area \( S \) of a sphere is given by:
\[ S = 4\pi r^2 \]
Differentiating both sides with respect to \( t \):
\[ \frac{dS}{dt} = \frac{d}{dt}(4\pi r^2) = 4\pi \cdot 2r \cdot \frac{dr}{dt} = 8\pi r \frac{dr}{dt} \]

### Step 6: Substitute \( r = 1 \, \text{cm} \) into the equation
Now, we substitute \( r = 1 \, \text{cm} \) into the equation:
\[ \frac{dS}{dt} = 8\pi (1) \frac{dr}{dt} \]
We already found \( \frac{dr}{dt} = \frac{1}{4(1^2)} = \frac{1}{4} \):
\[ \frac{dS}{dt} = 8\pi (1) \cdot \frac{1}{4} = 2\pi \]

### Conclusion
Thus, the rate at which the surface area increases when the radius is 1 cm is:
\[ \frac{dS}{dt} = 2\pi \, \text{cm}^2/\text{sec} \]