Let S be the position of source of light.
Let BD be the position of the man at a time t.
Let AB = x and BC = length of the shadow = y
Now
$\frac{d x}{d t}=100$
From the $\triangle ASC \sim \triangle BDC$
$\therefore \frac{A S}{B D}=\frac{A C}{B C}$
$\frac{30}{6}=\frac{x+y}{y}$
$\therefore 5 y=x+y$
$4 y=x$
$4 \frac{ dy }{ dt }=\frac{ dx }{ dt }$
$\therefore \frac{d y}{d t}=\frac{1}{4}(100)=25$
∴ The shadow of the man is lengthening at the rate 25 ft/min
The tip of shadow is at C. Let AC = z.
∴ AB = c
$\frac{A S}{B D}=\frac{A C}{B C}$
∴ BC = z – x
$\therefore \frac{30}{6}=\frac{z}{z-x}$
5z – 5x = z
4z = 5x
$4 \frac{ dz }{ dt }=5 \frac{ dx }{ dt }$
$4 \frac{ dz }{ dt }=5 \times 100$
$\frac{ dz }{ dt }=5 \times 25=125$
The tip of the shadow is moving at the rate 125 ft/min.
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$\log (1+\cos x)-x \tan \left(\frac{x}{2}\right)$
$\left(x^2+3 x y+y^2\right) d x-x^2 d y=0$
given by $\bar{d}=\lambda\left(\frac{a}{|\bar{b}|}+\frac{\bar{b}}{|\bar{b}|}\right)$
Question is modified
If $\overline{O A}=\bar{a}$ and $\overline{O B}=\bar{b}$ then show that the vector along the angle bisector of $\angle \mathrm{AOB}$ is
given by $\bar{d}=\lambda\left(\frac{\bar{a}}{|a|}+\frac{\bar{b}}{|\bar{b}|}\right)$
