The edge of the cube is increasing at a rate of `2cm//hr`.How fast is the cube's volume changing when its edge is `sqrt2`cm in length ?

Options

A. `6cm^3//hour`
B. `12cm^3//hour`
C. `3sqrt2cm^3//hour`
D. `6sqrt2`

Answer

To solve the problem step by step, we will find how fast the volume of the cube is changing when its edge is \(\sqrt{2}\) cm long, given that the edge is increasing at a rate of \(2 \text{ cm/hr}\).

### Step-by-Step Solution:

1. **Understand the relationship between edge and volume of a cube:**
The volume \(V\) of a cube with edge length \(x\) is given by the formula:
\[ V = x^3 \]

2. **Differentiate the volume with respect to time:**
To find how fast the volume is changing with respect to time, we differentiate the volume \(V\) with respect to time \(t\):
\[ \frac{dV}{dt} = \frac{d}{dt}(x^3) = 3x^2 \frac{dx}{dt} \]
Here, \(\frac{dx}{dt}\) is the rate at which the edge length \(x\) is increasing.

3. **Substitute the known values:**
We know that \(\frac{dx}{dt} = 2 \text{ cm/hr}\) and we need to find \(\frac{dV}{dt}\) when \(x = \sqrt{2} \text{ cm}\).
Substitute these values into the differentiated equation:
\[ \frac{dV}{dt} = 3(\sqrt{2})^2 \cdot 2 \]

4. **Calculate \((\sqrt{2})^2\):**
\[ (\sqrt{2})^2 = 2 \]
So, we can substitute this back into the equation:
\[ \frac{dV}{dt} = 3 \cdot 2 \cdot 2 \]

5. **Simplify the expression:**
\[ \frac{dV}{dt} = 12 \text{ cm}^3/\text{hr} \]

### Final Answer:
The volume of the cube is changing at a rate of \(12 \text{ cm}^3/\text{hr}\) when the edge length is \(\sqrt{2} \text{ cm}\).

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