Question

The volume of the cube is increasing at a rate of 10cm^3/min. How fast is the surface area increasing when the length of an edge is 90 cm?

Answer 1

Hint: In order to do this question, you need to first right the formula for the volume of a cube and area of a cube. Since in the question, you are given the rate of change of volume, you need to differentiate the volume formula to get the rate of change of volume. From there you can find the value of the rate of change in the side of the cube. Next you are asked to find the rate of change of area, so you need to differentiate the area formula and then substitute the required values and get the answer.

Complete step-by-step answer:
Here is the step wise solution.
The first step we need to do is to write the formula for the volume of a cube and differentiate it to get the rate of change of volume.
$\Rightarrow V = a^3$
$\Rightarrow \dfrac{dV}{dt} = 3a^{2}\dfrac{da}{dt}$
In the question, we are given the rate of change of volume as $10\dfrac{cm^{3}}{min}$. Therefore, we substitute the value in the above equation, to get the value of rate of change in the side of the cube.
$\Rightarrow \dfrac{dV}{dt} = 3a^{2}\dfrac{da}{dt} = 10$
$\Rightarrow \dfrac{da}{dt} = \dfrac{10}{3\left(90^2\right)}$
$\Rightarrow \dfrac{da}{dt} = \dfrac{1}{2430}$
Now to find the change in area we do the following
$\Rightarrow A = 6a^2$
$\Rightarrow \dfrac{dA}{dt} = 12a\dfrac{da}{dt}$
Now we substitute the required values.
$\Rightarrow \dfrac{dA}{dt} = 12\times 90\times \dfrac{1}{2430}$
$\Rightarrow \dfrac{dA}{dt} = \dfrac{4}{9}$
Therefore, we get the final answer of the question, as $\dfrac{4}{9} \dfrac{cm^2}{min}$.

Note: Whenever you are given or asked to find the change in rate of something, you first need to write down that formula and differentiate it. You also need to remember to write the dimensions at the end of the answer.