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

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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.


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.


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