Question

An edge of a variable cube is increasing at the rate of \(3 cm/s\). How fast is the volume of the cube increasing when the edge is \(10 cm\) long?

Answer 1

The correct answer is \(900 cm^3/s\) 
Let \(x\) be the length of a side and \(V\) be the volume of the cube. Then,
\(V = x^3.\)
\(\frac{dv}{dt}=3x^2.\frac{dx}{dt} ...\)(By chain rule)
It is given that,
\(\frac{dx}{dt}=3cm/s dv/dt=3x2(3)=9x2\)
Thus, when \(x = 10 cm,\)
\(\frac{dv}{dt}=9(10)^2=900cm^2/s\)
Hence, the volume of the cube is increasing at the rate of \(900 cm^3/s\) when the edge is \(10 cm\) long.

Answer 2

The formula for the volume of a cube:
\(V = s^3\)
Where:
s is the length of one edge of the cube
We are given that the edge of the cube is increasing at a rate of \(3 \, \text{cm/s}\). So, the rate of change of the edge with respect to time \(\frac{ds}{dt}\,  is\, 3\text{cm/s}\).

We are asked to find how fast the volume of the cube is increasing when the edge is 10cm long, i.e., when s = 10cm. We need to find \(\frac{dV}{dt}\), the rate of change of the volume with respect to time.
To find \(\frac{dV}{dt}\), we can differentiate the volume formula with respect to t:
\(\frac{dV}{dt} = \frac{d}{dt}(s^3)\)

Using the chain rule:

\(\frac{dV}{dt} = 3s^2 \cdot \frac{ds}{dt}\)

\(\frac{dV}{dt} = 3 \times (10)^2 \times 3\)

\(\frac{dV}{dt} = 3 \times 100 \times 3\)

\(\frac{dV}{dt} = 900 \, \text{cm}^3/\text{s}\)

So, the answer is \(900 \, \text{cm}^3/\text{s}\).