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provide solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise  12.2 question 30

Answers (1)

Answer: (i) -2 cm/min       (ii) 2 cm2/min

Hint: We know that, p=2(x+y)

Given: The length of the rectangle is x cm and width is y cm.

Solution: As per the given criteria, the length is decreasing at the rate of 5 cm/min

(i) \frac{d x}{d t}=-5 \mathrm{~cm} / \mathrm{min}

And width is increasing at the rate of 4 cm/min

\frac{d x}{d t}=4 \mathrm{cm} / \mathrm{min}                                                                                

Differentiating both sides,

\frac{d p}{d x}=\frac{d(2(x+y))}{d t}=2\left[\frac{d x}{d t}+\frac{d y}{d t}\right]

So, \frac{d p}{d x}=2[-5+4]=-2\; cm/min

 

(ii) Let A  be the area of the rectangle, A=xy

 \begin{aligned} &\frac{d A}{d t}=\frac{d(x y)}{d t} \\\\ &\frac{d(u v)}{d x}=v \frac{d u}{d x}+u \frac{d v}{d x} \end{aligned}

So the above equation becomes,

\frac{d A}{d t}=y \frac{d x}{d t}+x \frac{d y}{d t}

Substituting the values from equation (i) and (ii)

\frac{d A}{d t}=y(-5)+x(4)

When x=8 cm and y=6 cm the above equation becomes,

\frac{d A}{d t}=6(-5)+8(4)=-30+32=2\: cm^{2}/min

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