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need solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise 12.2 question 3

Answers (1)

Answer: The rate of increasing of the perimeter of the square will be 0.8 cm/sec

Hint: Here, the perimeter of the square at any time t will be p = 4x cm

Given: The side of square is increasing at the rate of 0.2 cm/sec.

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

Now according to the question,

The rate of side of the square is increasing is     \frac{d x}{d t}=0.2\: cm/sec                          … (i)

By applying derivative in perimeter,

\frac{d p}{d t}=\frac{d(4 x)}{d t}=\frac{d x}{d t} \times 4=4 \times 0.2=0.8\: cm/sec

 

 \frac{d x}{d t}=0.2\: cm/sec (from eqn 1) and p = 4x (given)

So, Thus the rate of increasing of the perimeter of the square will be 0.8 cm/sec.

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