Simple related rates derivative question Rafael is walking away from a $12$-ft-tall lantern at a constant speed.
If the tip of Rafael's shadow is moving twice as fast as he walks, how tall is Rafael? I'm confused on the step where $dL/dt = 2dx/dt$. If we're looking at how fast the shadow is compared to his walk wouldn't we be comparing the derivative of s to the derivative of $x$ because $L$ is the whole distance including how much Raphael walked plus his shadow length?


 A: Making the comment of @turkeyhundt more explicit, the tip of the 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 Rafeal, so we have the relationship: $\frac{d(x+s)}{dt}=\frac{dL}{dt}= 2\frac{dx}{dt}$, since Rafael's rate of change is $\frac{dx}{dt}$. So that's where that comes from. 
To find Rafeal's height, note that Rafeal's speed is constant, so we have $\frac{dx}{dt}=v$ for some constant $v$. So then Rafeal's position is $x(t)=vt +c_0$, where $c_0$ is his starting point. I assume Rafeal 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 Rafeal 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 Rafeal is $6$ feet tall.
