 Question 2

# The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Let x be the length of a side, V be the volume, and s be the surface area of the cube.

Then, V = x3 and S = 6x2 where x is a function of time t.

It is given that

\begin{align} \frac{dV}{dt} = 8 cm^3 / s \end{align}

Then, by using the chain rule, we have:

\begin{align} \frac{dV}{dt} = \frac{d}{dt} (x^3) . \frac{dx}{dt} = 3x^2 . \frac{dx}{dt} =8 \end{align}

\begin{align} \frac{dx}{dt} = \frac{8}{3 x^2} ……… (1) \end{align}

Now \begin{align} \frac{dS}{dt} = \frac{d}{dx}(6x^2).\frac{dx}{dt} [By Chain Rule] \end{align}

\begin{align} =12x .\frac{dx}{dt} =12x.(\frac{8}{3x^2}) = \frac{32}{x}  \end{align}

Thus, when x = 12 cm, \begin{align} \frac{dS}{dt} = \frac{32}{12} cm^2 / s = 8 cm^2 / s \end{align}

Hence, if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of \begin{align} \frac{8}{3} cm^2 / s \end{align}.