Question

An edge of a variable cube is increasing at the rate of $3cm/s.$ How fast is the volume of the cube increasing when the edge is $10$ cm long?

Answer 1

Hint : Here first of all we will assume the edge of the cube to be “x” and volume “v” and then will frame the differentiation equation and apply identity for the chain rule and simplify placing the given values in the equation to get how fast volume increases.

Complete step-by-step answer :
Let us assume that “x” be the length of the side of the cube
And “V” be the volume of the cube
Thus, the volume of the cube can be expressed as –
$V = {x^3}$
Now, take differentiation with respect to “x” and apply chain rule and will take the derivative of the function and then the derivative of the angle.
Apply the identity $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
$\therefore \dfrac{{dV}}{{dt}} = 3{x^2}.\dfrac{{dx}}{{dt}}$
Given that the cube is increasing at the rate of $3cm/s.$ , $\dfrac{{dx}}{{dt}} = 3cm/s$ Place it in the above equation –
$\therefore \dfrac{{dV}}{{dt}} = 3{x^2}.3$
Also, given that edge of the cube $x = 10cm$
$\therefore \dfrac{{dV}}{{dt}} = 3{(10)^2}.3$
Simplify the above expression –
$\therefore \dfrac{{dV}}{{dt}} = 900c{m^3}/s$
Hence, the volume of the cube increases with $900c{m^3}/s$ when the edge is $10$ cm long

Note : Know the difference between the differentiation and the integration and apply its formula accordingly. Differentiation is defined as the rate of change of the function and the integration represents the sum of the function over the range. Differentiation and integrations are inverses of each other.