Application of Derivatives — Class 12 Maths Solution

Let us assume AB be the street light post and CD be the height of man i.e., $CD = 2$ m. ( as shown in the given figure)

Let $BC = xm,CE = ym$ and $\frac{{dx}}{{dt}} = \frac{{ - 5}}{3}\;{\rm{m}}/{\rm{s}}$

From $\Delta ABE$ and $\Delta DCE$, we see that
$\Delta ABE \sim \Delta DCE$ [by AAA similarity]

Therefore,$\frac{{AB}}{{DC}} = \frac{{BE}}{{CE}} \Rightarrow \frac{{\frac{{16}}{3}}}{2} = \frac{{x + y}}{y}$

$\Rightarrow$ $\frac{{16}}{6} = \frac{{x + y}}{y}$

$\Rightarrow$ $16y = 6x + 6y \Rightarrow 10y = 6x$

$\Rightarrow$ $y = \frac{3}{5}x$

On differentiating both sides w.r.t. $t$, we get

$\frac{{dy}}{{dt}} = \frac{3}{5} \cdot \frac{{dx}}{{dt}} = \frac{3}{5} \cdot \left( { - 1\frac{2}{3}} \right)$

[as man is moving towards the light post]

$= \frac{3}{5} \cdot \left( {\frac{{ - 5}}{3}} \right) = - 1\;{\rm{m}}/{\rm{s}}$

Let $z = x + y$

Now, differentiating both sides w.r.t. $t$, we get

$\frac{{dz}}{{dt}} = \frac{{dx}}{{dt}} + \frac{{dy}}{{dt}} = - \left( {\frac{5}{3} + 1} \right)$

$= - \frac{8}{3} = - 2\frac{2}{3}\;{\rm{m}}/{\rm{s}}$

Hence, the tip of shadow is moving at the rate of $2\frac{2}{3}\;{\rm{m}}/{\rm{s}}$ towards the light source and length of the shadow is decreasing at the rate of $1\;{\rm{m}}/{\rm{s}}$.