## Minimum Time to Move Between Fixed Points Through a Point Moving Along x Axis

A point P is free to move along the
$x$
axis between points A and B.Line segments joint to point P to points R and S as shown, and the lines PR and PS make angles
$\theta$
and
$\phi$
with the positive and negative
$y$
axes respectively. How can you find the minimum time to move from R to Q?

$tan \theta =\frac{AB-x}{b} \rightarrow sec^2 \theta \frac{d \theta}{dx}=- \frac{1}{b} \rightarrow \frac{d \theta}{dx}=- \frac{cos^2 \theta}{b}$

$tan \phi =\frac{x}{a} \rightarrow sec^2 \phi \frac{d \phi}{dx}= \frac{1}{a} \rightarrow \frac{d \phi}{dx}= \frac{cos^2 \phi}{a}$

Then
$\frac{d \theta}{d \phi}=\frac{\frac{d \theta}{dx}}{\frac{d \phi}{dx}}=\frac{- \frac{cos^2 \theta}{b}}{ \frac{cos^2 \phi}{a}}=- \frac{a cos^2 \theta}{b cos^2 \phi}$

Let a particle move from R to P with speed
$v_1$
and from P to Q with speed
$v_2$
.
The time taken to move from R to P is
$t_1=\frac{b}{v_1 cos \theta}$
.
The time taken to move from P to Q is
$t_2=\frac{a}{v_2 cos \phi}$
.
The time taken to move from R to Q is then
$T=\frac{b}{v_1 cos \theta}+ \frac{a}{v_2 cos \phi}$
.
\begin{aligned} \frac{dT}{d \theta}&= \frac{b sec \theta tan \theta}{v_1 } + \frac{a}{v_2 sec \phi tan \phi} \frac{d \phi}{d \theta} \\ &= \frac{b sec \theta tan \theta}{v_1 } + \frac{a}{v_2 sec \phi tan \phi} (- \frac{b cos^2 \phi}{a cos^2 \theta}) \\ &= b sec^2 \theta (\frac{sin \theta}{v_1}-\frac{sin \phi}{v_2}) =0 \end{aligned}
.
Hence
$\frac{sin \theta}{v_1}-\frac{sin \phi}{v_2} =0 \rightarrow \frac{sin \theta}{sin \phi}=\frac{v_1}{v_2}$
is a condition for what is obviously a minimum time.