Find an equation for a moving rod The two endpoints of a 1-metre long rod have an initial position at $(0,0),(0,1).$ The rod slides continuously to the position $(1,0),(0,0)$ sweeping out a region in the positive quadrant. Determine the equation for the boundary of this region.

My attempt:
At any position, let $a$ denote the distance from the lower endpoint to the origin, $b$ the distance from the upper endpoint to the origin. Then we have
$$a^2+b^2=1,$$
$$\frac{x}{a}+\frac{y}{b}=1.$$
I tried to fix $x$ and use the method of Lagrange multiplier to determine the maximal $y.$ But I couldn't solve the equations. Any ideas?
 A: Firstly obtain the straight-line equation of the rod at any point in the slide. If $x_0$ and $y_0$ are the $x$-intercept and $y$-intercept, then by Pythagoras, $x_0^2 + y_0^2 = 1$.
So we have the equation of this line:
\begin{eqnarray*}
y &=& -\dfrac{y_0}{x_0}x + y_0 \\
&=& -\dfrac{y_0}{\sqrt{1 - y_0^2}}x + y_0
\end{eqnarray*}
With this equation we treat $x$ as a constant and try to maximise $y$ with respect to $y_0$. That is, for each $x$ we want the maximum $y$ these equations achieve. To do this we differentiate:
\begin{eqnarray*}
\dfrac{dy}{dy_0} &=& \dfrac{-x}{\sqrt{1 - y_0^2}} - \dfrac{y_0^2x}{\left(1 - y_0^2\right)^{\frac{3}{2}}} + 1 \\
&=& 1 - \dfrac{x}{\left(1 - y_0^2\right)^{\frac{3}{2}}} \\
\end{eqnarray*}
Setting this to $0$ gives $y_0 = \dfrac{1}{\sqrt{1 - x\frac{2}{3}}}$ and $x_0 = x^\frac{1}{3}$.
Substitute these back to find $y$:
\begin{eqnarray*}
y &=& \dfrac{-\sqrt{1 - x^\frac{2}{3}}x}{x^\frac{1}{3}} + \sqrt{1 - x^\frac{2}{3}} \\
&=& \left(1 - x^\frac{2}{3}\right)^\frac{3}{2}
\end{eqnarray*}
So the region you want is the region bounded by this curve and the $x$ and $y$ axes.
A: Partial answer: The motion 
$$
A(t) = (t,0), \quad \quad \quad 0 \leq t \leq 1 \\
B(t) = (0, \sqrt{1-t^2}), 0\leq t \leq 1
$$
seems to do the trick.
