#### Please solve RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 1 maths textbook solution

Answer: The area is increasing at the rate of 64 cm2/min

Hint: Here, we use the formula of square sheet in (cm) is, $A=x^{2} \mathrm{~cm}^{2}$

Given: The side of square sheet is increasing at the rate of 4 cm per minute, rate of area increasing when the side is 8 cm long. So Here, our x =8.

Solution:

Suppose the side of given square sheet be x cm at any instant time.

Now according to the question,

The rate of side of sheet increasing is $\frac{d x}{d t}=4\: cm/min$  cm/min

Now the area of square sheet at any time t will be $A=x^{2} \mathrm{~cm}^{2}$... (i)

On differentiating (i) with respect to time both side,

$\frac{d A}{d t}=\frac{d x^{2}}{d t}=2 x \frac{d x}{d t}=2 x \times 4=8 x$           [ $\frac{d x}{d t}=4$  From equation number (i)]

Therefore,  $\frac{d A}{d t}=8 x$

Let’s Put value of x = 8

$\Rightarrow \frac{d A}{d t}=8 \times 8=64 \quad(x=8\{\text { given }\})$

$d A / d t=64 \mathrm{~cm}^{2} / \mathrm{min}$

The area is increasing at the rate of $64\; cm^{2}/min$ when the side is 8 cm long.