#### provide solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise  12.2 question 2

Answer: The volume of the cube increasing at the rate of 900cm3/sec

Hint: Here, we use the formula of cube’s volume $V=x^{3} \mathrm{~cm}^{3}$

Given: An edge of a variable cube is increasing at the rate of 3 cm per second, the volume of the cube increasing when the edge is 10 cm long, So Here  x =10

Solution: Suppose the edge of the given cube be x cm at any instant time.

Now according to the question,

The rate of edge of the cube increasing is $d x / d t=3 \mathrm{~cm} / \mathrm{sec}$...                                                  … (i)

Now, the formula is $V=x^{3} \mathrm{~cm}^{3}$

By applying derivative,

$\frac{d V}{d t}=\frac{d x^{3}}{d t}=3 x^{2} \frac{d x}{d t}=3 x^{2} \times 3=9 x^{2}$            $\left(\frac{d x}{d t}=3\right) \text { from eq}^{n}(i)$

Lets put the value of x in above equation

\begin{aligned} &\frac{d V}{d t}=9 x^{2} \mathrm{~cm}^{3} / \mathrm{sec} \\\\ &\frac{d V}{d t}=9 \times 10 \times 10 \mathrm{~cm}^{3} / \mathrm{sec} \quad(x=10) \text { Given } \\\\ &\frac{d V}{d t}=900 \mathrm{~cm}^{3} / \mathrm{sec} \end{aligned}

Thus the volume of cube increasing at the rate of 900 cm3/sec.