A child is drowning.

[BMAD]

At Newport Beach, CA, the long, straight shoreline separates the clear water (blue) from the sand (tan). Mitch, the lifeguard, is at M, 80 meters from the water, and sees a drowning child C, 120 meters from the sand. The points A and B are 280 meters apart along the shoreline. Mitch can run only 4 m/sec but he can swim 8 m/sec. At what spot on the shoreline AB should Mitch aim in order to reach the child at C as soon as possible? Compute the minimum time to the nearest tenth of a second; compare it to the times for the paths: MC, MAC, MBC.

[bonanova]

Mitch is quite a swimmer - do you want him running twice as fast, instead?

Either way, there is an answer.

[BMAD]

Mitch is quite a swimmer - do you want him running twice as fast, instead?

Either way, there is an answer.

No it is hard to run in the sand

[Yoruichi-san]

If I'm reading this correctly...

...is 39.35m away from A, assuming A is the point along the shore that corresponds to Mitch's initial position.

The equation for the total time it takes Mitch to get to the child is: t=1/4*sqrt(x^2+80^2)+1/8*sqrt((280-x)^2+120^2), where x is the distance from A to the point Mitch hits the shoreline.

Taking a derivative and setting equal to zero gives the minimum, after some messy algebra.

[bonanova]

The distance x from point A that the lifeguard enters the water.

Following the path MAC, (x=0) his time to the swimmer is 58.08 seconds

Following the path MBC, (x=280) his time to the swimmer is 87.80 seconds.

Following the straight path MC, (x=112) his time to the swimmer is 60.22 seconds.

So there is a value of x, less than 112, that minimizes his time to the swimmer.

Snell's law says the path of least time occurs at interfaces of differing speeds

when the sines of the angles with the normal are in the same ratio as the speeds.

We can have some fun deriving that result here.

The distance run on sand increases with x as sin(a) where a is his angle to the perpendicular to the shore.

The distance swum in the water decreases with x as cos(b) where b is his angle with the shore.

Since the speed ratio is 2, the stationary condition is 2 sin (a) = cos (b).

If we had used angle with normal in the water, we would have 2 sin (a) = sin (b) which is Snell's law.

This occurs when x = 40, and coincidentally when a = b.

Thus tan (a) = 0.5 and we get a = 26.56^o.

The sand distance is 89.44 meters (22.36 seconds) and water is 268.33 m (33.54 seconds.)

So the minimum time to the swimmer of 55.90 seconds is achieved by entering the water 40 meters from point A.

[dark_magician_92]

The calculation part is a headache..........

[bonanova]

The calculation part is a headache..........

Sometimes the fun part challenge is to find the shortest an intuitive path through the messy calculations.

Y-san did the brute force calculation.

I tried all the trig approaches until I rediscovered Snell's law (usually used for optics).

It's a cute puzzle because it ends up with similar triangles on the sand and in the water.

I don't know whether that was because the speed ratio was 2.

[bonanova]

If I'm reading this correctly...

...is 39.35m away from A, assuming A is the point along the shore that corresponds to Mitch's initial position.

The equation for the total time it takes Mitch to get to the child is: t=1/4*sqrt(x^2+80^2)+1/8*sqrt((280-x)^2+120^2), where x is the distance from A to the point Mitch hits the shoreline.

Taking a derivative and setting equal to zero gives the minimum, after some messy algebra.

Verified x = 40.

But I didn't do all the messy stuff, I just plugged 40 into

2x sqrt((280-x)² + 120²) = (280-x) sqrt(x² + 80²)

How did you solve the quartic???

[Yoruichi-san]

Online equation solver . Okay, yes, I'm lazy .

You have a great knowledge of different mathematical principles, but I abhor memorizing things and hence, although the name "Snell's Law" sounds vaguely familiar, I have no memory of what it actually entails. For me, calculus is very intuitive and finding a max/min via derivative is second nature .

[bonanova]

Online equation solver . Okay, yes, I'm lazy .

You have a great knowledge of different mathematical principles, but I abhor memorizing things and hence, although the name "Snell's Law" sounds vaguely familiar, I have no memory of what it actually entails. For me, calculus is very intuitive and finding a max/min via derivative is second nature .

Our approaches are often complementary. I'm not ready to say that I abhor calculus, but I am old (and tired) enough to call in transferable concepts where they apply if it avoids re-deriving a result. But, this afternoon I did the whole d/dx on this problem, but surrendered when I saw x⁴ and x³ terms. So I didn't derive x=40 that way, I got it numerically from my Newton-Raphson program, then plugged 40 into the stationary condition just to verify the approaches agree.

I worked in optics long enough for Snell's law to have been like f = ma to me. It just shows the bend in light's path at interfaces that creates the least time path. Just as OP asks here. In optics it's refractive indices, in this puzzle it's running vs swimming, but it's the same effect. But my optics journey was 40 years ago and the connection didn't hit me at first. Full disclosure, Snell's law is an extension of the principle of least action, a fundamental underpinning in several disciplines.

Long version of the answer to a question that wasn't asked...