Question

If the edge of a cube increases at the rate of $60$cm per second, at what rate the volume is increased when the edge is $90$cm?
A. $48600$ cu cm per second
B. $1458000$ cu cm per second
C. $4374000$ cu cm per second
D. None of the above

Answer 1

Hint: Since the problem is based on applications of derivatives of a function and we know that, if the edge of a cube is increasing then, the volume will also increase as both are directly proportional to each other. Here we have to find the first derivative of volume with respect to time, $t$ to get an accurate answer.

Formula Used:
Volume of cube = $V = {(edge)^3} = {a^3}$
$\dfrac{d}{{dx}}({x^n}) = n.{x^{n - 1}}$

Complete step by step Solution:
Let the edge of a cube be $a$.
Rate of increase of edge $\dfrac{{da}}{{dt}} = 60cm/s$ (given) … (1)
Now, we know that Volume of a cube, $V = {(edge)^3} = {a^3}$ … (2)
Also, we know, $\dfrac{d}{{dx}}({x^n}) = n.{x^{n - 1}}$
Differentiate eq. (2) with respect to $t$, we get
$\dfrac{{dV}}{{dt}} = 3{a^2}.\dfrac{{da}}{{dt}}$
Substituting the value of $\dfrac{{da}}{{dt}}$ from equation (1) in the above expression we get
 $\dfrac{{dV}}{{dt}} = 3{a^2}.(60)$
$\dfrac{{dV}}{{dt}} = 180{a^2}$
At $a = 90cm$,
\[{\left( {\dfrac{{dV}}{{dt}}} \right)_{a = 90}} = 180{(90)^2}\]
\[{\left( {\dfrac{{dV}}{{dt}}} \right)_{a = 90}} = 1458000c{m^3}/s\]
Thus, the rate at which volume is increasing when the edge = $90cm$ is $1458000$ cu cm per second.

Hence, the correct option is B.

Note: In the problems based on the application of derivatives, apply the required properties(formula) to get the accurate solution to the problem. Sometimes, the problem seems to be complex while calculating the rate of change of quantities hence, it is advised to calculate the rate of change of quantities one by one carefully.