# Math Help - snell's law

1. ## snell's law

Jamie is at the point
J = (0, 8) offshore, needing to
reach the destination D = (12,3) on land as quickly
as possible. The lake shore is the x-axis. Jamie is in a
boat that moves at 10 uph, with a motor bike on board
that will move 20 uph once the boat reaches land. Your
job is to find the landing point P = (x, 0) that minimizes
the total travel time from J to D. Assume that
the trip from P to D is along a straight line.

Could someone pls take a look at my work?

Thanks

2. ## Re: snell's law

You have correctly found, using the Pythagorean theorem and the relation between time, distance and constant velocity:

$f(x)=\frac{\sqrt{x^2+8^2}}{10}+\frac{\sqrt{(12-x)^2+3^2}}{20}$

where f measures time in hours. You then began the process of differentiating f with respect to x to find the critical values.

$f'(x)=\frac{2x}{20\sqrt{x^2+8^2}}+\frac{2(12-x)(-1)}{40\sqrt{(12-x)^2+3^2}}=$

$\frac{x}{10\sqrt{x^2+8^2}}+\frac{x-12}{20\sqrt{(12-x)^2+3^2}}=$

$\frac{2x\sqrt{(12-x)^2+3^2}+(x-12)\sqrt{x^2+8^2}}{20\sqrt{\left(x^2+8^2\right) \left((12-x)^2+3^2\right)}}$

Noting the denominator is always positive, we then consider:

$2x\sqrt{(12-x)^2+3^2}+(x-12)\sqrt{x^2+8^2}=0$

$2x\sqrt{(12-x)^2+3^2}=(12-x)\sqrt{x^2+8^2}$

Observing both sides represent non-negative values, we may square:

$4x^2\left((12-x)^2+3^2\right)=(12-x)^2\left(x^2+8^2\right)$

$(6x)^2=(12-x)^2\left(8^2-3x^2\right)$

$36x^2=\left(144-24x+x^2\right) \left(64-3x^2\right)$

$36x^2=-3x^4+72x^3-368x^2-1536x+9216$

$3x^4-72x^3+404x^2+1536x-9216=0$

A graph of this polynomial shows only one real root for $0\le x\le 12$.

Using a root-finding technique such as Newton's method, we find:

$x\approx4.2194869338023206885$

The first derivative test confirms this is a minimum, i.e., f'(4) < 0 and f'(5) > 0.

To apply Snell's Law, we could write:

$\frac{\sin\theta_2}{\sin\theta_1}=\frac{20}{10}=2$

$\sin\theta_2=2\sin\theta_1$

$\frac{12-x}{\sqrt{(12-x)^2+3^2}}=\frac{2x}{\sqrt{x^2+8^2}}$

This leads to the same equation we have above.

3. ## Re: snell's law

Let's generalize a bit to say we want to find the quickest path across two media:

(sorry, still learning the idiosyncrasies of this board...don't yet know how to include an image directly into a post...)

We are traveling from point A to point B then to point C by the quickest path possible. The question is, where should point B be placed?

I have set up xy coordinate axes such that:

$O=\left(0,0\right)$

$A=\left(0,-W_1\right)$

$B=\left(B,0\right)$

$C=\left(L,W_2\right)$

The distance from A to B is:

$AB=\sqrt{\left(B-0\right)^2+\left(0-\left(-W_1\right)\right)^2}=\sqrt{B^2+W_1^2}$

The distance from B to C is:

$BC=\sqrt{\left(L-B\right)^2+\left(W_2-0\right)^2}=\sqrt{\left(L-B\right)^2+W_2^2}$

The speed through medium 1 is $v_1$ and the speed through medium 2 is $v_2$. We assume both velocities are positive values. Thus, using the relationship between distance, constant velocity and time, we find the total time as a function of B is:

$t\left(B\right)=\frac{\sqrt{B^2+W_1^2}}{v_1}+\frac {\sqrt{\left(L-B\right)^2+W_2^2}}{v_2}$

Differentiating with respect to B, we find:

$t'\left(B\right)=\frac{B}{v_1\sqrt{B^2+W_1^2}}+ \frac{\left(B-L\right)}{v_2\sqrt{\left(L-B\right)^2+W_2^2}}=$

$\frac{Bv_2\sqrt{\left(L-B\right)^2+W_2^2}+\left(B-L\right)v_1\sqrt{B^2+W_1^2}}{v_1v_2\sqrt{B^2+W_1^2 }\sqrt{\left(L-B\right)^2+W_2^2}}$

The denominator will always be positive, so we need only consider:

$Bv_2\sqrt{\left(L-B\right)^2+W_2^2}+\left(B-L\right)v_1\sqrt{B^2+W_1^2}=0$

$Bv_2\sqrt{\left(L-B\right)^2+W_2^2}=\left(L-B\right)v_1\sqrt{B^2+W_1^2}$

Let's take a moment to rewrite this:

$\frac{B}{\sqrt{B^2+W_1^2}}\cdot\frac{\sqrt{\left(L-B\right)^2+W_2^2}}{L-B}=\frac{v_1}{v_2}$

$\frac{\frac{B}{\sqrt{B^2+W_1^2}}}{\frac{L-B}{\sqrt{$$L-B$$^2+W_2^2}}}=\frac{v_1}{v_2}$

$\frac{\sin\theta_1}{\sin\theta_2}=\frac{v_1}{v_2}$

Thus, we have shown that Snell's law (or Descartes' law) is satisfied when t'(B) = 0.