Question

The volume of a cube is increasing at the rate of 8 cm$^3$per second. How fast is the surface area increasing when the length of an edge is 12 cm?

Answer 1

Hint- In order to solve this type of question, we must know the concept of differentiation. Also we must know the chain rule as we have to find the area with the given length.
$\Rightarrow$ Volume of a cube $= {x^3}$
$\Rightarrow$Surface area $= 6{x^2}$

Complete step-by-step answer:
Let x be the length of the side, v be the volume and s be the surface area of a cube.
Here we are given that
$\Rightarrow$ $\dfrac{{dv}}{{dt}} = 8c{m^3}$ per second.
Then by using the chain rule,
We get
$\Rightarrow 8 = \dfrac{{dv}}{{dt}} = \dfrac{{d{x^3}}}{{dt}} \times \dfrac{{dx}}{{dx}}$
Or 8 = $3{x^2} \times \dfrac{{dx}}{{dt}}$
$\Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{8}{{3{x^2}}}$ ……………………………………………(1)
Now, we will solve $\dfrac{{ds}}{{dt}}$,
$\Rightarrow \dfrac{{d\left( {6{x^2}} \right)}}{{dt}} = \dfrac{{d\left( {6{x^2}} \right)}}{{dt}} \times \dfrac{{dx}}{{dx}} \\\ = \dfrac{{d\left( {6{x^2}} \right)}}{{dx}} \times \dfrac{{dx}}{{dt}} \\\ = 12x \times \dfrac{{dx}}{{dt}} \\\$
We used chain rule to solve the above mentioned equation.
$\Rightarrow$12x $\times \dfrac{{dx}}{{dt}}$
$\Rightarrow$ 12x $\times \dfrac{8}{{3{x^2}}}$
$\Rightarrow$ $\dfrac{{32}}{x}$
Thus using the given condition,
When x = 12 cm
$\Rightarrow$ $\dfrac{{ds}}{{dt}}$ = $\dfrac{{32}}{{12}}c{m^2}$per second
$\Rightarrow$ $\dfrac{8}{3}$ $c{m^2}$ per second is the right answer.
$\therefore$ Hence if the length of the edge of the cube is 12 cm, then the surface area is increasing at the rate of $\dfrac{8}{3}$ $c{m^2}$per second.

Note – In order to solve this type of question we must know the concept of chain rule and the derivatives. Also one can get confused with the rate asked in the question so we must check if the answer comes negative then it will be decreasing rate otherwise positive answer shows increasing rate. Hence we will get the desired result.