The cow and the silo.

[Uncwilly]

Here is one that I have heard and don't have the answer to, nor do I have the math-ablity to solve it. I figure that it would make a good puzzle.
My one request is that those that can do it, please, oh please try to explain it so that someone that hasn't gone through calculus can follow it.

The problem:
A farmer has a cow and no fences, so the cow is on a chain that is 100' long.
The farmer has attached the chain to the side of a circular silo that is 50' in diameter.
What is the area that the cow can graze?

Assume that the chain does not angle, and that the cow's mouth is at the tip of the chain. This is a 2D puzzle.

[Wacky]

Regretfully, I know how to solve the problem by using Calculus, but do not know how to do so without it.

Even using Calculus, the problem is not trivial.

If we tear down the silo, the cow could graze a complete circle from the attachment point. Assuming that there is nothing to graze where the silo stood, the cow's grazing area would be simply the difference in the areas of two circles.

For the sake of explanation, let's establish a coordinate system for the cow. The units will be feet. The origin will be the point of attachment and the center of the silo will be on the positive x axis at (25,0).

Now make the following observations:
The diagram is symetrical about the x axis. The area grazed is twice the area grazed when y > 0.

When x < 0, the area grazed is just that of a circle with radius 100 ft, centered at the origin. So the area in quadrant 2 (x<0,y>0) is pi*r^2/4 = 2500*pi sqft.

In quadrant 1, the chain starts wrapping around the silo. The arc that the cow can graze becomes shorter and the center moves along the edge of the silo.

Now, if the silo were much bigger, the cow would move further along the silo until it came to a point where the chain was wrapped against the silo and the cow had no area to graze from that point.

If the silo is only slightly bigger than specified in the problem, diameter 200/pi ~ 63.66 ft, the cow would reach this point just as it reached the x axis. The resulting pattern would be similar to a cardoid.

But in this case, the silo is smaller. There are some points to the right of the silo that the cow can reach by going either way around the silo. We must avoid counting them twice.

It is not difficult to express the perimeter of the cow's reach and integrate the area contained within it. However, can anyone do it without Calculus?

[Fusion_power]

Its a good one Uncwilly and like most good ones has been used before. Break it down into two sections, the part the cow can graze freely which is half of the circle its chain will allow.

PI * R squared is the formula

1/2 * 3.14 * 100 ^ 2 equals 15,700 square feet.

Now you have to figure out how much of the remaining area is restricted by the silo. You would think at first guess that you could just subtract the area of the silo from 15,700. This won't work because the cows chain has to wrap around the silo losing length to the curvature of the walls.

Sorry, I don't have time to do the rest right now, work calls. But an approximation is 12,500 more square feet.

Maybe more later!

[Orgasmic Troll]

What do you mean that the chain does not angle?

p.s. did you pick those numbers out of thin air or were they the ones given? this would be much easier to do if the silo had a radius of 50 instead of a diameter of 50

p.p.s. The degree to which I will be impressed if someone explains this without calculus is inversely proportional to the chances that I think someone could explain this without calculus. And I would be very impressed.

[Uncwilly]

You have to use calculus, I understand, but to explain it so that a person that has had calculus can follow the explanattion might be possible. I know that one can explain a derivative if the student already understands a reasonable amount of geometry and alg.

If someone wants to do the problem, but make the silo say 75' in diameter to solve the over lap. That is cool, but the challenge with the overlap remains.

[Orgasmic Troll]

Someone without calculus knowledge could set up the problem (you could even find the parametric equation of a circle involute *cough*mathworld*cough* to get everything ready for the calculus end of it)

And then it's just a matter of finding the area using calculus .. and explaining that would have nothing more to do with the problem and all to do with explaining how taking the appropriate integral would give you the area .. and if someone was attentive and patient enough, I suppose they could follow it (I haven't set up the integral, so I don't know how involved it will be .. it could require a LOT of patience and attentiveness) but I think the best you could expect is to say "Set up these regions, and then I'll use calculus to find the area, and the basic concept behind it is this" without going into the nitty gritty of it

[Wacky]

Originally Posted by Uncwilly

You have to use calculus, I understand, but to explain it so that a person that has had calculus can follow the explanattion might be possible. I know that one can explain a derivative if the student already understands a reasonable amount of geometry and alg.

OK, I'll give it a stab.

Traditionally, the greek letter theta is used to express an angle. But since I don't know any easy way to do math expressions in this forum, I'll use "T" to represent the angle of interest.

Now I will start with the formula for the area of a circular arc. A=(R^2)*T/2. This comes from considering a very tiny arc. The base of the arc is R*T and the area is essentially a triangle of height R. You then integrate (add together many many very small areas) to cover the entire arc. For a full circle, you integrate for 0<T<2*pi, and get the expected answer (R^2)*2*pi/2 = pi*R^2.

We will apply the same idea to the cow. Basically, the cow can graze every location that any part of the chain can reach.

I'll divide the problem into three parts.

Assume that the chain is the hand of a clock.

First, as in my previous posting, we will start the cow at the 9 o'clock position and have it move clockwise, always keeping the chain taught.

While the cow is between the 9 o'clock position and the noon position, the entire 100 ft of the chain as moving. By the above formula, the area covered is

integral pi<T<pi/2 (-1/2*R^2*dT) == pi/4*(100^2) = 2500pi sqft.


In the next message, I'll describe what happens as the cow goes from the noon position to somewhere around 4 o'clock.

[Wacky]

Part 2

When the cow reaches the noon position, the chain becomes tangent to the circle of the silo. As the cow progresses, the chain can no longer be a straight line from the anchor point to the cow because such a straight line will pass through the silo. What happens is that the chain wraps along the side of the silo until it reaches the point that the rest of the chain lies along another tangent to the circle. However, only a part of the chain now gets to sweep across the grass. The rest is snug against the building.

So the next part of the problem is to decide how much of the chain is against the building and how much is sweeping through the grass.

Observe that a circle has a nice property that if you draw the radius of the circle to the point of tangency, it always forms a right angle. Further, if you change the angle of the radius with respect to the coordinate system, the tangent changes by an angle.

When the cow is at noon, the point of tangency is at 9 o'clock. Moving the point of tangency to the 10 o'clock position changes the chain to the 1 o clock angle, etc.

So let's take any arbitrary angle T. The central angle of the silo between the anchor point and point of tangency is T and the chain extends in a direction T measured from the noon direction.

Now, the amount of chain wrapped against the silo is S*T where S is the radius of the silo (25 ft). So the remainder of the chain sweeping the grass is (100 - 25 * T). We can integrate 1/2 *(100 - 25 * T)^2 dT through the appropriate angle and have the area covered.

This portion is integral 625/2 * (16 -8*T + T^2) dt

or

625/2 * (16 * (Tf - T0) -4*(Tf^2-T0^2) +1/3*(Tf^3-T0^3))

but we will measure the angle clockwise from the 9 o'clock position

therefore this reduces to

625/2 * (16* Tf -4 * Tf^2 + 1/3*Tf^3)

All that remains is to determine Tf.

[c00ler]

Wackerbarth is right. I got the same results.

And I've found that Tf is the solution(the smallest positive one) of
(x-4)^2=tg(x)^2+sin(x)^2
So x is non-algebraic (Is not a root of polynomial with integer coefficients) number (I suppose it can be proved)
I think that sin(x) is non-algebraic too (Who can prove?).
I believe that the answer cannot be expressed using elementary functions and operations.

[Wacky]

Part 3

Since the chain is longer than 1/2 the circumference of the silo, the cow would be below the x axis if we wrap the chain until it's tangent is at the 6 o' clock position.

But when the chain is at the 3 o clock position, the cow is S (25 ft) above the axis. Therefore, for some angle between 3 oclock and 6 oclock, the cow would be exactly on the axis. Call that angle Tf.

Now we have a right triangle whose acute angle is pi -Tf and its base is S, the radius of the silo (25 ft).

Therefore the tangent part of the chain is 25 * tan (pi - Tf) and the amount wrapped around the silo is 25 * Tf.

But the total chain is 100 ft

25 * Tf + 25 * tan (pi -Tf) = 100

or Tf - tan (Tf) = 4

Which yields Tf = 2.043245

[Wacky]

Part 4:

Returning to the integral, its value is

625/2 * (16* Tf -4 * Tf^2 + 1/3*Tf^3) = 5886.22918

plus the part of the circle = 2500 * pi

for a total of 13740.2042 sq ft above the axis

With an equal amount below the axis, the cow can graze 27480.4084 sq ft

-30-

Richard