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need solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise 12.2 question 7

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Answer\frac{d V}{d t}=2 \pi \mathrm{cm}^{3} / \mathrm{sec}

Hint: As we know that the volume of bubble is V=\frac{4}{3} \pi r^{3} .

Given: The radius of air bubble is increasing at the rate of 0.5 cm/sec

Solution: Suppose the radius of the given air bubble be r cm and let V  be the volume of the air bubble at any instant time.

The rate of increasing in the radius of the air bubble is \frac{\mathrm{dr}}{\mathrm{dt}}=0.5 \mathrm{~cm} / \mathrm{sec}


By applying derivative on volume

\frac{d V}{d t}=\frac{d\left(\frac{4}{3} \pi r^{3}\right)}{d t}                    \therefore\left(v=\frac{4}{3} \pi r^{3}\right)                                                                 


Let’s differentiate  d\left(\frac{4}{3} \pi r^{3}\right)

 \begin{aligned} &\frac{d V}{d t}=\frac{4 \times 3}{3} \times \pi r^{2} \frac{d r}{d t} \\\\ &\frac{d V}{d t}=4 \pi r^{2} \frac{d r}{d t} \end{aligned}


 Let’s  put value of \frac{d r}{d t}=0.5    (given)

\frac{d V}{d t}=4 \pi r^{2} \times 0.5


Let’s put  value of r = 1

 \begin{aligned} &\frac{d V}{d t}=4 \pi \times 1^{2} \times 0.5 \\\\ &\frac{d V}{d t}=2 \pi \mathrm{cm}^{3} / \mathrm{sec} \end{aligned}


The rate is increasing is 2 \pi \mathrm{cm}^{3} / \mathrm{sec}  when the radius is 1  cm.

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