The edge of a cube is equal to the radius of a sphere. If the edge and the radius increase at the same rate, then the ratio of the increases in surface areas of the cube and sphere is

Options

A. `2pi : 3`
B. `3 : 2pi`
C. `6 : pi`
D. none of these

Answer

To solve the problem of finding the ratio of the increases in surface areas of a cube and a sphere when the edge of the cube is equal to the radius of the sphere and both are increasing at the same rate, we can follow these steps:

### Step 1: Define Variables
Let:
- \( a \) = edge of the cube
- \( r \) = radius of the sphere

According to the problem, we have:
\[ a = r \]

### Step 2: Write the Surface Area Formulas
The surface area \( S_1 \) of the cube is given by:
\[ S_1 = 6a^2 \]

The surface area \( S_2 \) of the sphere is given by:
\[ S_2 = 4\pi r^2 \]

### Step 3: Differentiate the Surface Area Formulas
Now, we need to find the rates of change of these surface areas with respect to time \( t \).

For the cube:
\[ \frac{dS_1}{dt} = \frac{d}{dt}(6a^2) = 12a \frac{da}{dt} \]

For the sphere:
\[ \frac{dS_2}{dt} = \frac{d}{dt}(4\pi r^2) = 8\pi r \frac{dr}{dt} \]

### Step 4: Substitute \( a = r \) and \( \frac{dr}{dt} = \frac{da}{dt} \)
Since \( a = r \) and both are increasing at the same rate, we can substitute \( r \) with \( a \) and \( \frac{dr}{dt} \) with \( \frac{da}{dt} \) in the equation for \( \frac{dS_2}{dt} \):
\[ \frac{dS_2}{dt} = 8\pi a \frac{da}{dt} \]

### Step 5: Find the Ratio of the Rates of Change of Surface Areas
Now we can find the ratio of the increases in surface areas:
\[ \frac{\frac{dS_1}{dt}}{\frac{dS_2}{dt}} = \frac{12a \frac{da}{dt}}{8\pi a \frac{da}{dt}} \]

### Step 6: Simplify the Ratio
We can cancel \( a \) and \( \frac{da}{dt} \) from both the numerator and the denominator:
\[ \frac{dS_1/dt}{dS_2/dt} = \frac{12}{8\pi} = \frac{3}{2\pi} \]

### Conclusion
Thus, the ratio of the increases in surface areas of the cube and the sphere is:
\[ \frac{3}{2\pi} \]