Two points $P$ and $Q$ lies in the $xy$-plane. $P$ is moving from $(1,0)$ to $(0,0)$ along the $x$-axis and $Q$ is moving from $(0,0)$ to $(1,0)$ along the $y$-axis following the parametrized equations ; $$\begin{cases} P(t)=(\cos t,0)\\ Q(t)=(0,\sin t) \end{cases}$$ Let $L$ be the line segment $PQ$. What is the region that is swept by $L$ as $0\le t\le\frac\pi2$?
I think the curve in the first quadrant is $x^{\frac32}+y^{\frac32}=1$.
(1) Intuitively (using Calculus)
The lines $L(t)$ and $L(t;\Delta t)$ at time $t$ and $t+\Delta t$ can be expressed as \begin{align*} L(t)&:\frac x{\cos t}+\frac y{\sin t}=1\\ L(t+\Delta t)&:\frac x{\cos(t+\Delta t)}+\frac y{\sin(t+\Delta t)}=1\\ \end{align*} Let $R(t;\Delta t)$ be the intersection of $L(t)$ and $L(t+\Delta t)$ and let $R(t)$ be the limit of $R(t;\Delta t)$ as $\Delta t\to0$. $$ \begin{bmatrix} \frac1{\cos t} &\frac1{\sin t}\\ \frac1{\cos(t+\Delta t)} &\frac1{\sin(t+\Delta t)} \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = \begin{bmatrix}1\\1\end{bmatrix} $$ \begin{align*} D &=\frac1{\cos t}\cdot\frac1{\sin(t+\Delta t)}-\frac1{\sin t}\cdot\frac1{\cos(t+\Delta t)}\\ &=\frac{\sin t\cos(t+\Delta t)-\cos t\sin(t+\Delta t)}{\cos t\sin t\cos(t+\Delta t)\sin(t+\Delta t)}\\ &=\frac{\sin(t-(t+\Delta t))}{\cos t\sin t\cos(t+\Delta t)\sin(t+\Delta t)}\\ &=-\frac{\sin t}{\cos t\sin t\cos(t+\Delta t)\sin(t+\Delta t)} \end{align*} \begin{align*} R(t;\Delta t)=\begin{bmatrix}x\\y\end{bmatrix} &=\frac1D \begin{bmatrix} \frac1{\sin(t+\Delta t)} &-\frac1{\sin t}\\ -\frac1{\cos(t+\Delta t)} &\frac1{\cos t} \end{bmatrix} \begin{bmatrix}1\\1\end{bmatrix}\\ &= \begin{bmatrix} \frac{\cos t\cos(t+\Delta t)}{\sin h}\left(\sin(t+\Delta t)-\sin t\right)\\ -\frac{\sin t\sin(t+\Delta t)}{\sin h}\left(\cos(t+\Delta t)-\cos t\right) \end{bmatrix} \end{align*} $$R(t)=\lim_{\Delta t\to0}R(t;\Delta t) = (\cos^3 t,\sin^3 t)$$ We can (intuitively) think of $R(t)$ as the point at time $t$. So the curve becomes $x^{\frac32}+y^{\frac32}=1$.
(2) Rigorously? (using the Convex Optimization)
A point $R$ in the line segment $L$ at time $t$ can be expressed as a parameter $s$ ($0\le s\le 1$) \begin{align*} R(t,s) &=sP(t)+(1-s)Q(t)\\ &=(s\cos t,(1-s)\sin t)\\ &=(x,y). \end{align*}
Finding the curve in the first octant is to find the function that assign $x$ to $y$ $$f(x)=\max_{\substack{0\le t\le\frac\pi2\\0\le s\le1}}y.$$ That is, it reduces to the convex optimization problem; $$\text{Maximize} \quad\quad (1-s)\sin t$$ under the three constraints ; $$\begin{cases} s\cos t = x\\ 0\le t\le\frac\pi2\\ 0\le s\le1 \end{cases}$$
Is it possible to evaluate the exact equation $x^{\frac32}+y^{\frac32}=1$ using the convex optimization?
(3) Or does anyone can show me a rigorous way to find the curve?
