What is the height of a man walking away from a street light 12 ft high, if his shadow is moving twice as fast as he is? The goal is to find h, the man's height. The height of the light post is given, 12ft. Also, we know that the change in position of the shadow is twice as much as the change in position as the man. From this, I got:
height (h) / 12 = Shadow Distance from Man (S) / Man's Dist. from light (M) + S
12s = h(M+S), so h= (12s) / (M+S)
This is where I was confused. I know I can derive with respect to time, with dh/dt = 0. However, there are three variables and we only know 1 of them: 2*dM/dt = dS/dt.
 A: The tip of the man's shadow has position $x+s=L$, so its rate of change with respect to time is: $\frac{d(x+s)}{dt}=\frac{dL}{dt}$. But it is given that this rate of change is twice that of the man, so we have the relationship: $\frac{d(x+s)}{dt}=\frac{dL}{dt}= 2\frac{dx}{dt}$, since his rate of change is $\frac{dx}{dt}$. 
To find the man's height, note that his speed is constant, so we have $\frac{dx}{dt}=v$ for some constant $v$. So then his position is $x(t)=vt +c_0$, where $c_0$ is his starting point. I assume the man starts at $0$, i.e. at the lamp, though this isn't mentioned in the problem. Then we have $x(t)=vt$. Now $\frac{dx}{dt}=\frac{ds}{dt}$ so $s(t)=x(t)+c=vt+c$ for some constant $c$. $c=0$, since if the man starts at $0$, at this time he casts no shadow as he is directly beneath the lamp, so we have $s(t)=x(t)$.By similar triangles: $\frac{12}{x+s}=\frac{h}{s}$, so $h=\frac{12s}{2s}=6$, so the man is $6$ feet tall.
A: 
Proportion in similar triangles:
$$ \frac{h}{12} = \frac{s}{s+m}= \frac{\dot s}{\dot s+\dot m} = \frac{\dot s/\dot m}{\dot s/\dot m +1 } = \frac{2}{2+1}= \frac{2}{3}\rightarrow h= 24/3 = 8^{\prime} $$
The ratio is constant, so numerator and denominator are differentiated with respect to time by the Quotient Rule.
A: Assume that the man starts walking from the base of the lamppost, where he has no shadow. Some time after that, the man is at distance $x$ from the lamppost. The tip of his shadow, having traveled twice as fast during the same time, will be at distance $2x$ from the lampost. By similarity of the obvious triangles, $\frac{2x}{12}=\frac{2x-x}{h}$ therefore $h=6$.
A: Denote the heights of the lamppost and of the man by $h_l$ and $h_m$, and denote the distance of the man and of the tip of his shadow from the lamppost by $m$ and $s$. Then
$${s\over h_l}={s-m\over h_m}$$
and therefore
$$s={h_l\over h_l-h_m} m$$
at all times. The condition $\dot s=2\dot m$ leads to $h_m={\displaystyle{h_l\over2}}=6$ ft.
