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

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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)


\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.


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