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explain solution RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 8 maths

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Answer: $\frac{d s}{d t}=\frac{5}{2}=2.5\; km/hr$

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

Given: A man 2 meters high walk at a uniform speed of 5 km/hr away from lamp post 6 meters high.

Solution:

Suppose AB the lamp post and let MN be of the man of height 2 m.

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

Suppose length of the shadow MS=s

Given as man walk at the speed of 5 km/hr

$\frac{d I}{d t}=5\; km/hr$

So considering $\Delta ASB,$

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

Then considering $\Delta MNS,$

$\tan \theta=\frac{M N}{M S}=\frac{2}{s}$ ...................(ii)

From equation (i) and (ii)

$\begin{aligned} &\frac{6}{l+s}=\frac{2}{s} \\\\ &6 s=2(I+s) \\\\ &6 s=(2 I+2 s) \end{aligned}$

$\begin{aligned} &6 s-2 s=2 l \\\\ &4 s=2 l \\\\ &2 s=1 \end{aligned}$ ..............(iii)

By applying derivative, With respect to time on both side

$\frac{d l}{d t}=\frac{d(2 s)}{d t}$ $I=2 s\left(\text { from } e q^{n} \text { iii }\right)$

$\begin{aligned} &\frac{d l}{d t}=2 \frac{d s}{d t} \\\\ &5=2 \frac{d s}{d t} \end{aligned}$ $\left(\frac{d l}{d t}=5\right)(\text { given })$

$\begin{aligned} &\frac{5}{2}=\frac{d s}{d t} \\\\ &2.5=\frac{d s}{d t} \end{aligned}$

Thus, the rate at which the length of his shadow increases by 2.5 km/hr

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