Variant of Sliding Ladder Related Rates Problem

Question

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 $\alpha = 45^{\circ}$?

Answer 1

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 $\frac v{\sqrt 2}$

Answer 2

Going with saulspatz's interpretation, let the top of the ladder be at $(0,y_1)$ and the bottom at $(x_2,0)$ at time $t.$ Then the middle of the ladder is at $$(x,y)=(\frac{x_2}{2},\frac{y_1}{2})$$.$$\dot s=\sqrt{\dot x^2+\dot y^2}$$ $$=\sqrt{(\frac {\dot x_2}{2})^2+(\frac {\dot y_1}{2})^2}$$ $$\frac {1}{2}\sqrt{\dot x_2^2+\dot y_1^2}$$ Since $\dot {x_2}$ is how fast the bottom of the ladder is sliding along the floor and $\dot{y_1}$ is how fast the top of the ladder is sliding down the wall, you should be able to take it from there.