What distance will the ant travel?

Question

A transparent cylinder has height 1 and
length of circumfernce of the base 6 sqrt(3).
Ant is located at height 3/4, and a droplet
of syrop at height 1/4 and 140 degrees apart.

What minimum distance must he ant travel
to reach the syrop?

'plug into the original distance formula

√(1+x²) + 2r/√(1+x²)

But we know that the minimum this can be is when √(1+x²) = √(2r), and this minimum will be 2√(2r).'

This amounts to the following similar,
but more obvious case of flawed logic:

Problem:
find minumum y(x) = x^2 + 2exp(x+1) +1.
Solution:
y'(x) = 2x + 2exp(x+1) = 0.

Pug into the original formula exp(x+1) = -x and we have:
y(x) = y(x) = x^2 - 2x +1 = (x-1)^2.

Therefore y_min = 0 when x=1.

'approximately x=0.5773502718'

Note that x^2 = 0.3333333363475338

Answer 1

Pretend the cylinder is cut vertically on the far side from the ant-syrup combo, unrolled, and spread flat.

Then it's easy to see that the ant must travel a "vertical" distance of 1/2, and a horizontal distance of (140/360)*circumference = (7/18)(6√3) = 7/√3

Thus the minimum distance the ant can travel in this fashion is:
sqrt(49/3+1/4) = sqrt(199/12).


And now, a word from our sponsors, algebraic manipulation:
Since x² > 0 when x is not 0, we have that
x(√d+x) + d > d + x√d ==>
x + d/(√d+x) > √d ==>
x + d/(√d+x) > √d ==>
(√d + x) + d/(√d+x) > 2√d
Thus, y + d/y is minimized when y = √d (with no calculus!).


Ahem. However, if we have a clever ant and a solid cylinder, then the ant will climb to the top of the cylinder, make a beeline (antline?) on top, and then scuttle down to the syrup. Hurrah for the thinking ant!

Hang onto your antennas, here we go... One portion of the ant's distance will be covered while on top of the cylinder while the rest will be covered in getting to and from the top. We'll designate by x the horizontal portion of his travels while not on the top. Therefore, the non top contribution to his total distance is √(1+x²). You can see this the same way as with the previous analysis: unroll the cylinder. Only in this case, you should imagine the ant traveling up and then continuing up on an extension of the cylinder, so the total vertical distance covered is 1 inch, and the total horizontal distance covered, as we already said, is x.

OK, what's the distance covered in making the shortcut across the top? We'll call c the circumference of the cylinder, so c = 6√3. That makes the radius of the cylinder r = c/(2π) = 3√3/π. Also, let the horizontal distance between the initial position of the ant and the syrup be h. Now the angle θ that the ant cuts off will be ((h-x)/(2πr))*2π = (h-x)/r, so the chord due to this will be 2rsin(θ/2) = 2rsin((h-x)/2r).

Thus, the total distance the ant travels may be expressed as:
√(1+x²) + 2rsin((h-x)/2r)

We want to minimize this so we take a derivative with respect to x and set it equal to 0:
x/√(1+x²) = cos((h-x)/2r). Square both sides to get
x²/(1+x²) = cos²((h-x)/2r). Negate and add 1 to get sin² on the right:
1 - x²/(1+x²) = sin²((h-x)/2r). Simplify to get
1/√(1+x²) = sin((h-x)/2r).

Now plug into the original distance formula above to get what the minimum distance will be, when it is achieved:

√(1+x²) + 2r/√(1+x²)

But we know (see the word from our sponsors) that the minimum this can be is when √(1+x²) = √(2r), and this minimum will be 2√(2r). Thus, the ant could travel a minimum distance of 2√(2c/(2π)) = 2√(c/π) =
2√(6√3/π)

That amount is approximately 3.64 vs. the 4.07 of the worker ant in the first part, a savings of about 10%! By the way, if the cylinder is not solid but the ant can tunnel under the edges, then it could achieve the same thing by heading down instead of up.

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The asker has pointed out an error in the above analysis. Good catch, and very nice example. It's OK to plug in, but it is not legitimate to use the resulting equation to determine x!

So, (with h=7/√3 and r=3√3/π) I have manually determined the zero of
1/√(1+x²) - sin((h-x)/2r)
to be at approximately x=0.5773502718
which leads to a total distance of approximately 4.01949

Phooey! This ant had to do a lot of mental work just to save only about 1%.

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Boy, that Alexander is a pretty crafty guy.
Let's reconsider that equation we want to find the zero of:
1/√(1+x²) = sin((h-x)/2r)

At this point, if we were the ant king, we might notice that if x = 1/√3 then
1/√(1+x²) = 1/√(4/3) = √3/2

And we might also notice that
sin((h-x)/2r) =
sin ((7√3/3-√3/3)/(2*3√3/π)) =
sin (π/3) = √3/2

Fancy that, the minimum happens at x = 1/√3 and the minimum distance is thus:
√(1+x²) + 2r/√(1+x²) =
2/√3 + (2*3√3/π)√3/2 =
2/√3 + 9/π

The ant is mollified that there is a closed form solution, after all.

Answer 2

pythagorean theorem

sqrt((.5inch)^2+ 108)