The lake is a hyperbola.

Actually, it's a section of a hyperbola on one side of the axis of symmetry, then the mirror image of that section on the other side.

This is good knowledge to have if you're building optical equipment. Light follows paths of stationary time, so light passing through a lens is very similar to this problem. Most lenses are built with edges that are sections of sphere, not hyperbolas. This is just because it's easier to manufacture them that way. Then if you're doing some fancy optics, you'll have to correct for the "spherical aberration" of your lens.

Proving the the hyperbola is straightforward, but algebra-y. Just set up a coordinate system where the middle of the lake is (0,0) and the edge of the lake is given by x(y). Then the edge of the make must be long enough to satisfy the condition

((d-x)

^{2}+ y

^{2})

^{.5}+ 2x = c

where c is some constant, equal to (d

^{2}+ (w/2)

^{2})

^{.5}.

I got this by letting "y" be the height at which you cross the lake. Then I counted the distance to the point where you enter the lake, and the twice the distance to the axis of symmetry (since you go twice as slow in water). This must be equal to the same number for all "y" by the specification of the problem. If it is, the rest of the problem (you're going from A to B, not A to the axis of symmetry) follows because the second half is the same as the first half backwards, and also takes time "c".

The equation you get is a hyperbola. This makes some sense. Consider the limit as the width of the lake is much greater than your initial distance from it. Then in that limit, the additional path length added by crossing the lake one unit higher is one unit (for a trip to the axis of symmetry). So in that same limit, the width of the lake must be directly proportional to height of the lake, so that it slows you down the right amount. That's what a hyperbola does - goes asymptotically to a line. Ah well, I just realized that random babbling probably doesn't get the idea across very well. But I also don't care that much because if you're reading this I'm sure you can figure it out for yourself anyway. I just thought it was neat.

Also, note that you can quickly get into situations where the lake extends back behind your starting point, so if the lake is too wide, it's impossible. It'll be faster to swim across. This happens when (w/2)

^{2}+ d

^{2}>4d

^{2}. If that happens, you'll need the start farther away from the lake, or make it out of something that's harder to swim through.

New Problem: Lifeguard

Today's and yesterday's problems were inspired by Feynman's popular book "QED", which in turn is based on these lectures.

You're a lifeguard who sees a swimmer drowning somewhere off shore. You need to get out to the swimmer by a combination of running on the beach and swimming in the water. See the picture:

One particular path is fastest. Assume you run at speed n

_{1}and swim at speed n

_{2}. You take the fastest path. What is the relationship between the angle of your path on the beach relative to the normal, and your angle into the water relative to the normal?

## 1 comment:

High school optics once again comes into mind. Snell's law? What is it, n1sin(theta1) = n2sin(theta2), where n1 and n2 is this "index of refraction." If I recall correctly, n = c/v (wait, or is it the other way around? I thought this was right because the index of refraction should be greater than 1 for other media such as glass).

But based on the defined variables of your problem we'd be using n1 and n2 to represent velocity, not index of refraction.

In that case, take the equation I had above and switch the n1 and n2 and I think that will give you the answer you want.

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