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

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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.


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.


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