The Cowculus Problem

[merjet]

I received The Joy of Mathematics from The Teaching Company as a Christmas gift. One lecture includes the "cowculus" problem. A cow is 1 mile N of the X-Axis River which runs E-W. Her barn is 3 miles E and 1 mile N of her. She wishes to drink from the X-Axis River then walk to the barn, minimizing the distance of her walk. What spot on the X-Axis River would achieve her goal?

The more obvious way to solve this problem uses calculus. However, with a few moments' reflection , it could be solved without calculus. What method would that be?

I will defer posting the solutions for a couple days. If you think you have the solutions and can't wait, send me a private message.

Edited January 13, 2011 by Merlin Jetton

[kiaer.ts]

I received The Joy of Mathematics from The Teaching Company as a Christmas gift. One lecture includes the "cowculus" problem. A cow is 1 mile N of the X-Axis River which runs E-W. Her barn is 3 miles E and 1 mile N. She wishes to drink from the X-Axis River then walk to the barn, minimizing the distance of her walk. What spot on the X-Axis River would achieve her goal?

The more obvious way to solve this problem uses calculus. However, with a few moments' reflection , it could be solved without calculus. What method would that be?

I will defer posting the solutions for a couple days. If you think you have the solutions and can't wait, send me a private message.

I know the answer. Can simply visualize the origin and destination as nails and the spot on the river as a loose point moving along a rod and ask where would the point on the rod move to if all three were connected by a rubber band.

(To avoid further confusion, note that the problem as quoted in this post differs from Merlin's edited problem, which changed the location of the barn after I made this and the three posts following it.)

Edited January 13, 2011 by Ted Keer

[merjet]

I know the answer. Can simply visualize the origin and destination as nails and the spot on the river as a loose point moving along a rod and ask where would the point on the rod move to if all three were connected by a rubber band.

You may have the answer, but I can't know if it's correct unless you tell me the (x,y) coordinate (private message okay).

[kiaer.ts]

I know the answer. Can simply visualize the origin and destination as nails and the spot on the river as a loose point moving along a rod and ask where would the point on the rod move to if all three were connected by a rubber band.

You may have the answer, but I can't know if it's correct unless you tell me the (x,y) coordinate (private message okay).

The y coordinate is fixed at 0, the x coordinate is half the distance between the origin and the destination. The triangle with the shortest possible limbs will be the solution, which in the case where two points are fixed and one varies along a line parallel to them will be the unique isosceles triangle defined by the given constraints.

Again, just visualize the origin and destination as nails on a board with the spot on the river indicated by a peg which can move along a track representing the river. String a rubberband around all three. The peg on the track will come to rest equidistant between the two fixed points, formingf an isosceles triangle. Any position to the right or left would be stretching the rubberband more than necessary - i.e., would be an inoptimal solution.

You can do a complex minimax problem with calculus to prove the solution mathematically, but the intuitive answer works just fine.

[merjet]

The y coordinate is fixed at 0, the x coordinate is half the distance between the origin and the destination.

If you are saying the answer is (x, y) = (1.5, 0), that is incorrect.

[kiaer.ts]

The y coordinate is fixed at 0, the x coordinate is half the distance between the origin and the destination.

If you are saying the answer is (x, y) = (1.5, 0), that is incorrect.

If the cow walks in two straight lines, the isosceles triangle is the solution.

Edited January 12, 2011 by Ted Keer

[merjet]

If the cow walks in two straight lines, the isosceles triangle is the solution.

Yes, the cow walks in two straight lines. What isosceles triangle? What are the coordinates of the 3 vertexes?

P.S. I edited post #1, adding "of her" to make it clearer where the barn is. Maybe Ted misunderstood, and if so, I apologize.

Edited January 13, 2011 by Merlin Jetton

[kiaer.ts]

If the cow walks in two straight lines, the isosceles triangle is the solution.

Yes, the cow walks in two straight lines. What isosceles triangle? What are the coordinates of the 3 vertexes?

Um, where is the draw geometrical figure button?

If I understand you, the origin is in Washington State, the destination in Maine, and the river in Texas, not California or Florida.

Can you not visualize what I am saying? Am I misunderstanding you? I think the rubberband solution is eloquent.

[merjet]

Um, where is the draw geometrical figure button?

It's not needed. (x,y) coordinates will do the job.

Can you not visualize what I am saying?

Yes, I can visualize it, but that doesn't tell me the (x,y) coordinate.

Edited January 13, 2011 by Merlin Jetton

[Jon Letendre]

The other method is intuition, and mine tells me she should shoot for that spot on the river where it is a 90 degree left turn to the barn. X would be about 1.2.

[BaalChatzaf]

Hint. The angle of incidence = the angle of reflection. What is the shortest path for a light beam shot from the cow to the river and bounces back to the barn. The cow should follow that light beam.

Ba'al Chatazf

Edited January 13, 2011 by BaalChatzaf

[Mikee]

Imagine a "bank" shot (like in pool). The reflection across the x-axis is -3,-2. Find the slope (m) of the straight line between the reflection point and 0,1 (B). Solve for y=0 Sol: -1,0

[BaalChatzaf]

Imagine a "bank" shot (like in pool). The reflection across the x-axis is -3,-2. Find the slope (m) of the straight line between the reflection point and 0,1 (B). Solve for y=0 Sol: -1,0

Same solution as the reflected light beam analogy. You got it.

Ba'al Chatzaf

[Mikee]

Oops. I just realized I picked the -x direction for "East" and most normal people would pick the +x direction which would make the real correct answer 1,0. Sorry.

[Jon Letendre]

Right, 1, not more than 1 like I said. At 1, it’s a 90 degree turn left to the barn.

[BaalChatzaf]

The more obvious way to solve this problem uses calculus. However, with a few moments' reflection , it could be solved without calculus. What method would that be?

Fermat's principle of least time for light travel gives the solution. In the case of light, least time means least distance. Assume you have a mirror with a source of light L1 units above the mirror and a target for the light beam L2 units above the mirror. Shine the light from the source to the point on the mirror where the reflection of the target is seen. This will give the least time for the bounce and consequentially the least distance.

Ba'al Chatzaf

[kiaer.ts]

Angle of incidence equals angle of reflection is another way of saying that the path creates an isosceles triangle. Not that our resident contrarian ever gives credit when someone else says something right.

(Note that Jon's and Bob's answers above are based on the different problem in Merlin's corrected post. Originally Merlin has specified that both origin and destination were one unit north of the river.)

Edited January 13, 2011 by Ted Keer

[kiaer.ts]

Um, adding "of her" after the fact doesn't make the question clearer, it makes it an entirely different problem, Merlin!

In that case, the triangle is no longer isosceles, since the origin and the destination are not on a line parallel to the river, as you had originally implied.

The length of the descending path is now the square root of two and of the ascending path the square root of eight, and Jon's right angle solution, (45 degrees x 2) is right.

[merjet]

Originally Merlin has specified that both origin and destination were one unit north of the river.

No, I did not say the latter. That was your misinterpretation. My original sentence was "Her barn is 3 miles E and 1 mile N." I never said "1 mile N of the river." Since "3 miles E" was relative to the cow, the more plausible interpretation was that "1 mile N" was also relative to the cow.

After reading your post #4 and writing post #5 it dawned on me you may have misinterpreted it, so I added "of her" to the sentence.

Edited January 13, 2011 by Merlin Jetton

[Jon Letendre]

“The more obvious way to solve this problem uses calculus. However, with a few moments' reflection , it could be solved without calculus. What method would that be?”

There are three N-S miles to cover and three E-W miles to cover, so all travel should be equal parts N-S and E-W, i.e., at 45 degree angles, which gives a landing at the river at 1,0, (or, since there's no reason to assume the cow is at x=0, she'll land at c+1,0.)

Edited January 13, 2011 by Jon Letendre

[merjet]

“The more obvious way to solve this problem uses calculus. However, with a few moments' reflection , it could be solved without calculus. What method would that be?”

There are three N-S miles to cover and three E-W miles to cover, so all travel should be equal parts N-S and E-W, i.e., at 45 degree angles, which gives a landing at the river at 1,0.

Okay. What would be your similar reply if the barn were at (4,2) rather than (3,2)? I'm trying to glean your method.

[Jon Letendre]

In that case, she’ll need to cover 4 E-W miles in her 3 miles of N-S travel. Or, 1 1/3 E-W per one mile of N-S travel. Getting to the river is one N-S mile of travel, and if she’ll need to cover 1 1/3 E-W miles in that one mile of N-S travel, then she’ll land at 1 1/3, 0.

[merjet]

In that case, she’ll need to cover 4 E-W miles in her 3 miles of N-S travel. Or, 1 1/3 E-W per one mile of N-S travel. Getting to the river is one N-S mile of travel, and if she’ll need to cover 1 1/3 E-W miles in that one mile of N-S travel, then she’ll land at 1 1/3, 0.

Sounds good. Your method is not the one presented by The Teaching Company lecturer, which I'll reveal it later.

[kiaer.ts]

Originally Merlin has specified that both origin and destination were one unit north of the river.

That should be "had specified."

No, I did not say the latter. That was your misinterpretation. My original sentence was "Her barn is 3 miles E and 1 mile N." I never said "1 mile N of the river." Since "3 miles E" was relative to the cow, the more plausible interpretation was that "1 mile N" was also relative to the cow.

After reading your post #4 and writing post #5 it dawned on me you may have misinterpreted it, so I added "of her" to the sentence.

Take whatever comfort you like in calling it my misinterpretation. But I didn't misinterpret what you actually wrote. I simply misunderstood what you meant to say, but didn't.

[merjet]

The following reports what The Teaching Company lecturer did.

Let (x,0) designate the spot to be found on the X-Axis River. Each leg of the trip is the hypotenuse of a triangle. The bases are x and 3-x. So the total length of the trip is: f(x) = (x^2+1)^0.5+(x^2-6x+13)^0.5

To minimize this, find the derivative: f'(x) = (0.5)(x^2+1)^(-0.5)2x+(0.5)(x^2-6x+13)^(-0.5)(2x-6)

Set f'(x) = 0 and solve. It can't be done by isolating x, but it's easy to verify that x = 1 works. Therefore, (1,0) is the spot.

The non-calculus method he used was as follows. The barn is at (3,2), so its "reflected image" across the river is at (3,-2). (x,0) to (3,2) and (x,0) to (3,-2) are equivalent. The shortest path is one straight line from (0,1) to (3,-2). It crosses the x-axis at (1,0).

The lecturer did not say so, but the slope of the shortest path is (-2-1)/(3-0)= -1. Maybe this is much like what Jon Letendre was thinking when he said "three N-S miles to cover and three E-W miles to cover" in post #20.

Similarly, when I asked Jon if the barn were located at (4,2), the "reflected image" is at (4,-2). The cow has per Jon "4 E-W miles in her 3 miles of N-S travel", so the slope of the shortest path is (-2-1)/(4-0)= -3/4 and the shortest path crosses the x-axis at (4/3,0).

Mikee was the first to post the solution (with his correction) and used the "reflected image" method.

But I didn't misinterpret what you actually wrote. I simply misunderstood what you meant to say, but didn't.

You misunderstood what I did not say.

Edited January 13, 2011 by Merlin Jetton