Question

An edge of a variable cube is increasing at the rate of $10cm/s$. How fast the volume of the cube will increase when the edge is 5cm long?
1.$750c{m^3}/s$
2.$75c{m^3}/s$
3.$150c{m^3}/s$
4.$25c{m^3}/s$

Answer 1

Hint : we have to let the edge of the cube as x and denote the change in edge of the cube by $\dfrac{{dx}}{{dt}}$. We have to denote the volume of the cube by V and we will find a change in volume of the cube which will be denoted by $\dfrac{{dV}}{{dt}}$. Then, we have to differentiate the whole equation to find the solution.

Complete step-by-step answer :
According to the question,
Given: let the edge of the cube = x cm
Volume of cube $= V = {x^3}$
Change in edge of cube $= \dfrac{{dx}}{{dt}} = 10cm/s$-----(1)
Now we have to find change in volume of cube, so
$V = {x^3}$
Now to find change in volume we will differentiate this with respect to t
$\dfrac{{dV}}{{dt}} = 3{x^2}\dfrac{{dx}}{{dt}}$
We have to find change in volume when the value of edge is 5cm, so we will replace x with 5
$\dfrac{{dV}}{{dt}} = 3{(5)^2}\dfrac{{dx}}{{dt}}$
So, we know the change in edge of cube is $10cm/s$, so we will replace the value of $\dfrac{{dx}}{{dt}}$from 10 by using the equation (1)
$\dfrac{{dV}}{{dt}} = 3 \times 25 \times 10$
$\dfrac{{dV}}{{dt}} = 750c{m^3}/s$
So, the change in volume of the edge is $750c{m^3}/s$. Option (1) is the correct answer.
So, the correct answer is “Option 1”.

Note : Whenever change in anything is asked in any question it has to be found using a differentiation method as it tells the exact amount of change taking place at the point of time. Change is always to be found with respect to ‘t’ and generally the change is denoted by $\dfrac{d}{{dt}}$.