#### provide solution for RD Sharma maths class 12 chapter Derivative As a Rate Measure exercise  12.2 question 10

Answer: $\frac{d s}{d t}=0.4 m / \sec$

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

Given: A man 160 cm tall walks away from a source of light situated at the top of a pole 6m high, at the rate of 1.1 m/sec.

Solution:

Suppose $AB$  the lamp post and let $MN$  be the man of height 160 cm or 1.6 m.

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

Suppose length of the shadow $MS=s$

So, $\frac{\mathrm{d} \mathrm{I}}{\mathrm{dt}}=1.1 \mathrm{~m} / \mathrm{sec}$

Considering   $\Delta MNS$

$\tan \theta=\frac{M N}{M S}=\frac{1.6}{s}$.....…(i)

So considering $\Delta ASB,$

$\tan \theta=\frac{A B}{A S}=\frac{6}{I+s}$       ....… (ii)

Therefore, from equation (i) and (ii)

\begin{aligned} &\frac{6}{I+s}=\frac{1.6}{s} \\\\ &6 s=1.6(I+s) \\\\ &6 s=1.6 I+1.6 s \end{aligned}

\begin{aligned} &6 s-1.6 s=1.6 \\\\ &4.4 s=1.6I \\\\ &I=\frac{4.4}{1.6} s \\\\ &I=2.75 s \end{aligned}.................(iii)

By applying derivative with respect to time on both side

$\frac{d I}{d t}=\frac{d(2.75 s)}{d t}$                $I=2.75 s\left(\text { from } e q^{n} \text { iii }\right)$

\begin{aligned} &\frac{d I}{d t}=2.75 \frac{d s}{d t} \\\\ &1.1=2.75 \frac{d s}{d t} \end{aligned}                    $\ldots\left(\frac{d I}{d t}=1.1\right)(\text { given })$

$\frac{1.1}{2.75}=\frac{d s}{d t}$

$\frac{d s}{d t}=0.4 \mathrm{~m} / \mathrm{hr}$

Thus, the rate at which the length of his shadow increases by 0.4 m/hr.