Question

An edge of a cube is increasing at the rate of $10$ cm/s. How fast does the volume of the cube increase when the edge of the cube is $5$ cm long?

Answer 1

Hint: Use the formula of the volume of cube and then find the explicit differentiation as volume and the radius both are increasing with the time.
Then substitute the values to get the desired result.

Complete step by step solution:
It is given that the edge of a cube is increasing at the rate of $10$cm/s.
We have to find the rate at which the volume of the cube is increasing when the edge of the cube is $5$ cm.
Assume that the edge of the cube is $\left( a \right)$, then we have given
Rate of increase in the edge of the cube $\left( {\dfrac{{da}}{{dt}}} \right) = 10$cm/s
Edge of the cube$ = 5$cm
We know that the volume of the cube having edge length $a$, is given as:
Volume of cube $\left( V \right) = {\left( a \right)^3}$
Differentiate the volume with respect to $t$, then we have
$\dfrac{{dV}}{{dt}} = \dfrac{d}{{dt}}\left( {{a^3}} \right)$
$\dfrac{{dV}}{{dt}} = 3{a^2}\dfrac{{da}}{{dt}}$
As $a$ is also the function of $t$, therefore we use explicit differentiation to find the derivative.
Substitute the values, $a = 5$ and$\dfrac{{da}}{{dt}} = 10$ into the equation:
$\dfrac{{dV}}{{dt}} = 3{\left( 5 \right)^2}\left( {10} \right)$
Now, simplify the equation and get the rate of increase in the volume.
$ \Rightarrow \dfrac{{dV}}{{dt}} = 3\left( {25} \right)\left( {10} \right)$
$ \Rightarrow \dfrac{{dV}}{{dt}} = 750$cm/sec
Therefore, the volume of the cube is increasing at the rate of $750$ cm/second.

Note: We can notice that the volume is increasing with the increase in its edge, it means that both edges of the cube and the volume of the cube are the function of time, so while finding the derivative we have to use the explicit differentiation.