We all know the related rates problem of a ladder sliding off a wall. 
In this we are told how fast the bottom or top is sliding and then asked to find how fast the other end of the ladder is moving at. My question is fairly similar.
A ladder of length $2l$ is sliding off of a wall. The bottom moves at a velocity $v$. How fast will the middle of the ladder (which is located a distance $l$ from the ends of the ladder) be sliding when the angle between the rod, wall, and floor is $\alpha = 45^{\circ}$?
Best Answer
Let the bottom of the ladder have position $(x(t),0)$.
Let the top of the ladder have position $(0,y(t))$.
We know that $x^2+y^2=4l^2$
Differentiating with respect to $t$ gives $2x \frac {dx}{dt}+2y \frac {dy}{dt}=0$.
We know that $\frac {dx}{dt}=v$, so $2xv+2y \frac {dy}{dt}=0 \Rightarrow \frac {dy}{dt}=-\frac {vx}y$.
The midpoint has position $(x_m,y_m)=(\frac x2, \frac y2)$
$\frac {dx_m}{dt}=\frac 12 \frac {dx}{dt}=\frac v2$
$\frac {dy_m}{dt}=\frac 12 \frac {dy}{dt}=\frac {vx}{2y}$
Speed is given by $\sqrt{\left(\frac {dx_m}{dt}\right)^2+\left(\frac {dy_m}{dt}\right)^2}$
Speed is $\sqrt{\frac {v^2}{4}+\frac {v^2x^2}{4y^2}}=\sqrt{\frac {v^2y^2+v^2x^2}{4y^2}}=\sqrt{\frac {v^2(y^2+x^2)}{4y^2}}=\sqrt{\frac {v^24l^2}{4y^2}}=\frac {vl}{y}$
When $\alpha = 45^\circ$, $x=y=l\sqrt2$
So speed of midpoint is $\frac v{\sqrt 2}$