$A$ man,$2 \ m$ tall,walks at the rate of $1 \frac{2}{3} \ m/s$ towards a street light which is $5 \frac{1}{3} \ m$ above the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he is $3 \frac{1}{3} \ m$ from the base of the light?

(A) Let $AB$ be the street light post and $CD$ be the height of the man,i.e.,$CD = 2 \ m$.
Let $BC = x \ m$,$CE = y \ m$,and $\frac{dx}{dt} = -\frac{5}{3} \ m/s$ (since the man is moving towards the light post).
From $\Delta ABE$ and $\Delta DCE$,we see that $\Delta ABE \sim \Delta DCE$ by $AAA$ similarity.
Therefore,$\frac{AB}{DC} = \frac{BE}{CE} \Rightarrow \frac{16/3}{2} = \frac{x+y}{y}$.
$\Rightarrow \frac{16}{6} = \frac{x+y}{y} \Rightarrow \frac{8}{3} = \frac{x+y}{y}$.
$\Rightarrow 8y = 3x + 3y \Rightarrow 5y = 3x \Rightarrow y = \frac{3}{5}x$.
On differentiating both sides with respect to $t$,we get $\frac{dy}{dt} = \frac{3}{5} \cdot \frac{dx}{dt} = \frac{3}{5} \cdot \left(-\frac{5}{3}\right) = -1 \ m/s$.
Thus,the length of the shadow is decreasing at the rate of $1 \ m/s$.
Let $z = x + y$ be the distance of the tip of the shadow from the light post.
Now,differentiating both sides with respect to $t$,we get $\frac{dz}{dt} = \frac{dx}{dt} + \frac{dy}{dt} = -\frac{5}{3} - 1 = -\frac{8}{3} = -2 \frac{2}{3} \ m/s$.
Hence,the tip of the shadow is moving at the rate of $2 \frac{2}{3} \ m/s$ towards the light source.