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  • explain solution RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 28 maths

explain solution RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 28 maths

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Answer: \frac{d s}{d t}=3.6\: cm^{2}/sec

Hint: We know that V=x^{3}

Given: The volume of a cube is increasing at the rate of 9  cm3/sec

Solution: Differentiating the equation,

\frac{d V}{d t}=\frac{d x^{3}}{d t}=3 x^{2} \frac{d x}{d t}........ …(i)

The cube is increasing at the rate of 9  cm3/sec

So the above equation becomes,

9=3 x^{2} \frac{d x}{d t}

\frac{d x}{d t}=\frac{9}{3 x^{2}}........(ii)

So, s=6 x^{2}

Again differentiating the above equation with respect to time is,

\frac{d s}{d t}=\frac{d\left(6 x^{2}\right)}{d t}=6 \times 2 x \frac{d x}{d t}

Substitute equation (i) in the above equation,

\frac{d s}{d t}=6 \times 2 x \times \frac{9}{3 x^{2}}=\frac{36}{x}

So, length 10 cm,

\frac{d s}{d t}=\frac{36}{x}=\frac{36}{10}=3.6\; cm^{2}/sec

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