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I need to calculate the minimum value of the expression

**$ y = \frac{((x-a)^2 + p )^{\frac 12}}{v_1} + \frac{( (x-b)^2 + q )^{\frac12}} {v_2}$**

I’m calculating the first derivative , then calculting the corresponding value for $x$ . But the calculation is getting complex.

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Here $v_1 , v_2 , p$ and $q$ are positive integers greatest than $1$. I have also tried using $AM >= GM$ . but that too reduces finding maximum of a degree $4$ polynomial

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**It’s derivation of Snell’s Law from Fermat’s Principle**

Assume $a<b$,

\begin{align*}

T(x) &= \frac{\sqrt{(x-a)^2+p}}{v_1}+\frac{\sqrt{(x-b)^2+q}}{v_2} \\

T'(x) &=\frac{x-a}{v_1 \sqrt{(x-a)^2+p}}-

\frac{b-x}{v_2 \sqrt{(x-b)^2+q}} \\

0 &= \frac{\sin \theta_{1}}{v_{1}}-\frac{\sin \theta_{2}}{v_{2}} \\

s &= \sin \theta_{1} \\

\sin \theta_{2} &= \frac{v_{2} s}{v_{1}} \\

b-a &= \sqrt{p} \tan \theta_{1}+\sqrt{q} \tan \theta_{2} \\

b-a &=

\frac{s\sqrt{p}}{\sqrt{1-s^{2}}}+

\frac{v_{2} s\sqrt{q}}{\sqrt{v_{1}^{2}-v_{2}^2 s^2}}

\quad \cdots \cdots (*) \\

T_{\min} &=

\frac{\sqrt{p}}{v_{1}\cos \theta_{1}}+\frac{\sqrt{q}}{v_{2}\cos \theta_{2}} \\

&= \frac{\sqrt{p}}{v_{1}\sqrt{1-s^{2}}}+

\frac{v_{1} \sqrt{q}}{v_{2}\sqrt{v_{1}^{2}-v_{2}^2 s^2}}

\end{align*}

We seldom calculate the exact value of minimal $T$ but condition for minimum instead.

See also the link here.

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