#### Please solve RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 5 maths textbook solution

Answer: The rate of increase of its surface area, when the radius is 7 cm is  $11.2\pi$ cm2/sec

Hint: The surface area of spherical soap bubble at any time t  will be $S=4\pi r^{2}$ cm2

Given: The radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec.

Solution: Suppose the radius of the given spherical soap bubble be $r$ cm at any instant time.

Now according to the question,

$\frac{dr}{dt}=0.2$ cm/sec                     ….... (i)

By applying derivative on surface area

$\frac{d s}{d t}=\frac{d\left(4 \pi r^{2}\right)}{d t}$

\begin{aligned} &\frac{d s}{d t}=\frac{4 \pi\left(d r^{2}\right)}{d t} \\\\ &\frac{d s}{d t}=4 \pi \times 2 r \times \frac{d r}{d t} \end{aligned}

$\frac{d s}{d t}=4 \pi \times 2 r \times 0.2$                                    $\left(\frac{d r}{d t}=0.2\right)\left(\text { from } e q^{n}(i)\right)$

$\frac{d S}{d t}=1.6 \pi r$

lets put value of r in above formula $r=7$(given)

$\frac{d S}{d t}=1.6 \pi \times 7=11.2 \pi$ cm/sec

Thus the rate of increase of its surface area, when the radius is 7  cm is $11.2\pi$ cm2/sec