Surface area of cube is increasing at rate of `3.6 cm^2/s` . Find the rate at which its volume increases when lengths of side a is 10 cm.

Options

A. 9
B. 10
C. 18
D. 20

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Understand the problem
We need to find the rate at which the volume of a cube is increasing given that the surface area is increasing at a rate of \(3.6 \, \text{cm}^2/\text{s}\) when the length of the side of the cube \(a\) is \(10 \, \text{cm}\).

### Step 2: Write the formula for the surface area of a cube
The surface area \(S\) of a cube with side length \(a\) is given by:
\[ S = 6a^2 \]

### Step 3: Differentiate the surface area with respect to time
To find the rate of change of surface area with respect to time, we differentiate \(S\):
\[ \frac{dS}{dt} = \frac{d}{dt}(6a^2) = 12a \frac{da}{dt} \]

### Step 4: Substitute the known values
We know that \(\frac{dS}{dt} = 3.6 \, \text{cm}^2/\text{s}\) and \(a = 10 \, \text{cm}\). Substituting these values into the differentiated equation:
\[ 3.6 = 12 \cdot 10 \cdot \frac{da}{dt} \]

### Step 5: Solve for \(\frac{da}{dt}\)
Now we can solve for \(\frac{da}{dt}\):
\[ 3.6 = 120 \cdot \frac{da}{dt} \]
\[ \frac{da}{dt} = \frac{3.6}{120} = 0.03 \, \text{cm/s} \]

### Step 6: Write the formula for the volume of a cube
The volume \(V\) of a cube with side length \(a\) is given by:
\[ V = a^3 \]

### Step 7: Differentiate the volume with respect to time
Now we differentiate \(V\):
\[ \frac{dV}{dt} = \frac{d}{dt}(a^3) = 3a^2 \frac{da}{dt} \]

### Step 8: Substitute the known values
Now we substitute \(a = 10 \, \text{cm}\) and \(\frac{da}{dt} = 0.03 \, \text{cm/s}\):
\[ \frac{dV}{dt} = 3 \cdot (10)^2 \cdot 0.03 \]
\[ \frac{dV}{dt} = 3 \cdot 100 \cdot 0.03 = 9 \, \text{cm}^3/\text{s} \]

### Conclusion
Thus, the rate at which the volume of the cube is increasing when the length of the side is \(10 \, \text{cm}\) is:
\[ \frac{dV}{dt} = 9 \, \text{cm}^3/\text{s} \]