Using the coordinate system sketched in Figure 3.8.13(a), the distance from $A$ to $C$ is $d_1\=\sqrt{x^2\+a^2}$, whereas the distance from $C$ to $B$ is $d_2\=\sqrt{{\left(L-x\right)}^2\+b^2}$. The travel times in the two media are therefore $d_1\/v_1$ and $d_2\/v_2$, so the total travel time is

$T\left(x\right)\=\frac{\sqrt{x^2\+a^2}}{v_1}\+\frac{\sqrt{{\left(L-x\right)}^2\+b^2}}{v_2}$ 

It is tempting to write $d_1\=x\/\sin \left({\mathrm{\θ}}_1\right)$ and $d_2\=\left(L-x\right)\/\sin \left({\mathrm{\θ}}_2\right)$. Unfortunately, the angles ${\mathrm{\θ}}_1$ and ${\mathrm{\θ}}_2$ are themselves functions of $x$, so the total travel time expressed in terms of these angles can't be differentiated with respect to $x$.$$

Using $d_1\=x\/\sin \left({\mathrm{\θ}}_1\right)$ and $d_2\=\left(L-x\right)\/\sin \left({\mathrm{\θ}}_2\right)$, rewrite the equation $T\prime \left(x\right)\=0$ as

$\frac{\sin \left({\mathrm{\θ}}_1\right)}{v_1}-\frac{\sin \left({\mathrm{\θ}}_2\right)}{v_2}\=0$ 

from which Snell's law then follows.

Note that if $d_1\/v_1\+d_2\/v_2$ is written as $\frac{x}{v_1 \sin \left({\mathrm{\theta }}_1\right)}\+\frac{L-x}{v_2 \sin \left({\mathrm{\θ}}_2\right)}$ and differentiated with respect to $x$, the incorrect $\frac{v_2}{v_1}\=\frac{\sin \left({\mathrm{\θ}}_1\right)}{\sin \left({\mathrm{\θ}}_2\right)}$ will result.