related rates- rate a man's shadow changes as he walks past a lamp post (is a fixed distance away from it) A $186$ cm man walks past a light mounted $5$ m up on the wall of a building, walking at $2\ m/s$ parallel to the wall on a path that is $2$ m from the wall. At what rate is the length of his shadow changing when he is $4$ m past the point where the light is mounted? ($4$ m is the distance along the wall).
I have been doing related rates in my year 12 maths class and I know how to figure it out if the person is walking towards or away from the light but I've never come across a question like this where they are walking past the light, and have no idea where to begin. Some help getting started would be a massive help. (does it still involve similar triangles but in a 3D way?)
 A: The first thing to do with a problem like this is to draw a diagram:

The man (at $M$) is walking parrallel to the wall at $2\mbox{ m/s}$ and $2\mbox{ m}$ from the wall. 
The plan distance from the wall below the lamp and the man's feet is:
$$
p=\sqrt{2^2+x^2}
$$ 
where $x$ is the distance of the man past his closest point of approach to the lamp.
Then the triangle formed by the lamp, the man's head and the point on the wall below the lamp at the same height as the man is similar to triangle LSO and the triangle formed by the man's head, his feet and $S$. So the length of the shadow $s=\mbox{ MS}$ satisfies:
$$
\frac{(5-1.86)}{p}=\frac{1.86}{s}
$$
So the length of the shaddow is:
$$
s=\frac{1.86\sqrt{2^2+x^2}}{3.14}
$$
Now you are asked to find $\frac{ds}{dt}$ when $x=4$ and $\frac{dx}{dt}=2$
A: Let the distance along the wall of the man from the lamp be $x$ and let the length of the man's shadow, given $x$, be $\ell$.
Let $L$ be the point of the lamp, $P$ be the point at the bottom of the wall exactly under the lamp, $F$ be the point of the feet of the man, $H$ be the point of the top of the head of the man, and $E$ be the end of the man's shadow.

Consider that $EF=\ell$ and $FP=\sqrt{x^2+4}$. We draw the triangle $ELP$, including $F$ and $H$.

Here, by using similar triangles, $$\begin{align}&\frac{\ell}{1.86}=\frac{\ell+\sqrt{x^2+4}}{5}\\\to\ &5\ell=1.86\ell+1.86\sqrt{x^2+4}\\\to\ &3.14\ell=1.86\sqrt{x^2+4}\\\to\ &\ell=\frac{1.86}{3.14}\sqrt{x^2+4}\\\to\ &\ell=\frac{93}{157}\sqrt{x^2+4}\text.\end{align}$$
With the above equation, and considering that $\frac{dx}{dt}=v=2\ m/s$, we need to find $\frac{d\ell}{dt}$.
Consider that $$\ell^2=\frac{93^2}{157^2}x^2+\frac{4\cdot93^2}{157^2}\Rightarrow\frac{d\ell^2}{dx}=\frac{2\cdot93^2}{157^2}x\text.$$
It is clear that $$\frac{d\ell}{d\ell^2}=\frac1{2\ell}=\frac{157}{2\cdot93\sqrt{x^2+4}}\text.$$
By multiplying, we obtain $$\begin{align}\frac{d\ell}{dt}&=\frac{dx}{dt}\frac{d\ell^2}{dx}\frac{d\ell}{d\ell^2}\\&=2\cdot\frac{2\cdot93^2}{157^2}x\cdot\frac{157}{2\cdot93\sqrt{x^2+4}}\ m/s\\&=\frac{2\cdot93}{157}x\cdot\frac1{\sqrt{x^2+4}}\ m/s=\frac{186x}{157\sqrt{x^2+4}}\ m/s\text.\end{align}$$
Because we know that $x=4\ m$, we can substitute $4\ m$ for $x$:$$\begin{align}\frac{d\ell}{dt}&=\frac{186\cdot4}{157\sqrt{16+4}}\ m/s\\&=\frac{744}{314\sqrt5}\ m/s\\&=\frac{372\sqrt5}{157\cdot5}\ m/s\\&=\frac{372}{785}\sqrt5\ m/s\text.\end{align}$$
Therefore, the rate of change of the length of the man's shadow when the man is $4$ metres past the lamp is $\frac{372}{785}\sqrt5\ m/s$.
