#### need solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise 12.2 question 11

Answer: $\frac{d s}{d t}=0.5 \mathrm{~m} / \mathrm{sec}$

Hint: The rate at which the length of the man’s shadow increase will be $\frac{ds}{dt}$ .

Given: A man 180 cm tall walks at a rate of 2 m/sec away from a source of light that is 9m above the ground.

Solution: Suppose $AB$  the lamp post and let $MN$  be of height of man.

Suppose AM =I meter and $MS$  be the shadow of the man.

Given as man walk at the speed of 2 m/sec

So, $\frac{d I}{d t}=2$ m/sec      …(i)

So considering $\Delta ASB,$

\begin{aligned} &\tan \theta=\frac{A B}{A S} \\\\ &\tan \theta=\frac{9}{I+s} \ldots(\mathrm{ii}) \end{aligned}

Then$\Delta MNS,$ ,

$\tan \theta=\frac{M N}{M S}=\frac{1.8}{s}$

So by (i) and (ii)

$\frac{9}{I+s}=\frac{1.8}{s}$

$\begin{gathered} \Rightarrow 9 s=1.8 I+1.8 s \\\\ \Rightarrow 9 s-1.8 s=1.8 I \\\\ \Rightarrow 7.2 s=1.8 I \end{gathered}$

$\Rightarrow I=4 s$        .......(iii)

By applying derivative with respect to time on both side

\begin{aligned} &\frac{d I}{d t}=4 \frac{d s}{d t} \\\\ &2=4 \frac{d s}{d t} \end{aligned}                        $\ldots \ldots\left(\frac{d I}{d t}=2\right)(\text { given })$

\begin{aligned} &\frac{2}{4}=\frac{d s}{d t} \\\\ &\frac{d s}{d t}=0.5 \mathrm{~m} / \mathrm{sec} \end{aligned}

Thus, the rate at which the length of his shadow increases by 0.57 m/sec.