help! - tethered cow problem
 

I need to know how to solve a problem about a cow tethered to one point on a silo. If the length of the rope is two meters and the radius of the silo is one meter, how much area can the cow graze on around the silo? I found the problem where the length of rope is half the circumference, but I'd rather not use polar coordinates if it's possible. I'm pretty sure this problem can be solved with more simple calculus, but any solution will help.

Thanks for your time and help!

are you stupid? You have two circles.


Take the area of the larger one

subtract the area of the smaller

Done.



No calculus, no polar mathematics...

What about the wall of the silo if he is teatherd to one side it will affect the distance you would end up with a half moon shape

My understanding is the Cow was tethered outside the silo, and it would be full two meters to one side, and the silo to the other... so the circles dont line up perfectly, but its still larger minus smaller.

[QUOTE=Unregistered]My understanding is the Cow was tethered outside the silo, and it would be full two meters to one side, and the silo to the other... so the circles dont line up perfectly, but its still larger minus smaller.[/QUOTE]The effective length of the rope changes as the rope wraps around the silo.

[QUOTE=Unregistered]are you stupid? You have two circles.[/QUOTE]

Except for the fact that you do NOT have two circles, your approach might work.

Consider tearing down the silo. Since the silo covered the area, there is no grass inside.

In this case, your approach is correct. The cow can reach any point in its circle but you must subtract the silo area because it is void of grass.

But look at the tether when the cow is grazing near the far side of the silo. Consider the tether stretched tightly from the anchor point. This is the farthermost point that the cow could reach. The tether stretches across the silo floor and back into the grass on the far side.

But if the silo is in the way, the cow would not be able reach that far. The tether will have to wrap around the outside wall. Since the resulting tether position is not a straight line, it cannot reach as far in that direction. So the perimeter is not a circle.

[QUOTE=Unregistered]are you stupid?
[/QUOTE]
You have no right to insult other people, even if you think (wrongly) that they ask silly questions. Please learn to behave, or stop posting here.

Sorry if I didn't explain the problem fully the first time. It's not two circles and subtracting them. I don't know where you get that idea. The silo is tall and the rope has to wrap around it. Say the diameter of the silo is 1M and the length of the rope is 2M. The rope is tethered to one point on the silo, so there is a half circle of grazing area that can be calculated very easily. Once the cow goes beyond this half circle into the the half that has the silo in it, the line of the circle begins to approach the silo resulting in the half moon shape that someone mentioned earlier. The cow will continue to curve into the silo and eventually hit the wall. Since the circumference is pi, and the length of the rope is 2, the wrapped around grazing areas overlap. So, once you find the area of the half moon shaped grazing area, you have to subtract the areas that overlap. So, I hope that clears is up a little bit. Sorry about the first post, i meant to put diameter instead of radius. Can anyone think of a way to find the area? Thanks!

Have you tryed geometric tools?
A compass and ruler may help you come up with more elegant ideas. :)

The calculus solution is to just use the limit series definition of area. For angles not involving the silo you can just use (Pi*r^2)*(angle in degrees/360). The limit is used to find an approximation for angles involving the silo.

If it was me, then I'd just set the angle of the first equation to 180 degrees. Write this as a constant since the limit of a constant is still a constant. It's nice to eliminate as many variable terms as possible. The area of half a circle is 0.5*Pi*r^2 of course.

The second part is to find the critical points on the silo's perimeter. The point where the cow is no able to move (y=0) is easy to calculate. A full circle is Pi*r^2 so if r=1/2, then the circumference is Pi/4. It turns out that for a 2-meter-long cord you can go around the silo more then once.
This doesn't sound right. If the radius is 1 then it makes more sense so that's what I'll use. :) The unit circle has a radius of 1 and a circumference of Pi. A 2-meter cord would go to 2/Pi meters around the silo on either side. This means there's an overlap that will have to be considered as another section. Once you have the point for y=0, then you need the point for y=2 which is not too hard to get using the same method. Sorry if I'm a little tired - lator I'll probally get the solution into 3 steps since it's not that hard. :P

<2nd try> The easiest way using a compass is to start with half a circle, then at every point on the silo, calculate the length (or use a piece of string and a coin with the same diameter as my drawing) of each point and draw circles. You end up with a group of circles. Hehe, get ready for working with some sets or summation notation no matter what you do. You have to find just the areas of the _extensions_ of the previously enclosed area and add those to the final sum. After a couple trys I'd come up with some formula that I could plug into a "E" type function on my calculator.

[QUOTE=Marc Bacani]I need to know how to solve a problem about a cow tethered to one point on a silo. If the length of the rope is two meters and the radius of the silo is one meter, how much area can the cow graze on around the silo? I found the problem where the length of rope is half the circumference, but I'd rather not use polar coordinates if it's possible. I'm pretty sure this problem can be solved with more simple calculus, but any solution will help.

Thanks for your time and help![/QUOTE]
The best way to approach such a problem is to be systematic about it.

If it were me, I'd define a polar coordinate system centered about the tie-down point. Find a function, r(theta) describing the farthest distance the cow can reach at every angle theta. Because of the point at which the chain begins to wrap around the silo, this will be a piecewise function. Then integrate the area r*dr*d(theta) from -pi to pi (pi being the 'back side' of the silo). When you're done, subtract the area of the silo.

If you do it this way, you won't have to subtract out any overlap areas either, because you'll only account for each region once.

Drew

I would adopt a similar approach except that I would avoid worrying about the polar co-ordinates, as such.

If we draw a line which passes through both the center of the silo and the attachment point, we will have an axis of symmetry. Therefore, we can compute the grasing area on one side of that line and double it. This will resolve the overlap that might occur when the tether is longer than the half-circumference of the silo. Assume that the origin is at the center of the silo and that the attachment point is to the right (0 degrees).

We now wish to define an orientation and for each orientation the amount of grazing area that is assigned to it. Then we can simply integrate over all orientations to determine the area.

Note that for any given orientation, we can stretch the tether in that direction. In the general case, a portion of the tether will wrap around the silo and the remainder will be along a tangent in that direction.

Now, draw the radius of the silo that passes through that point of tangency.
Let's call the angle that that radius makes with the axis of symmetry "theta". The tether will be stretched at an angle of (theta + pi/2).

Note that the area grazed is just the area swept by the portion of the tether that is not against the silo. We need to determine L(theta) for each angle and integrate to find the total area grazed. The incremental area grazed is 1/2 L^2 dtheta.

Now for the endpoints in the interation.
When theta < -pi/2, the cow would be grazing on the wrong side of the axis.
For -pi/2 < theta < 0, the tether is stretched to its maximum length. Call it "T". This grazing area is a quadrant of a circle and the area is pi*(T^2)/4.

Call the radius of the silo "R". If T/R is sufficiently small, the cow would graze around the silo until the tether is completely against the silo.

Note that for an angle theta, R*theta of the tether is against the silo.
The remainder of the tether (T-R*theta) is the length to be used in the integration. If T = pi*R, the cow would reach the end of its rope just as it gets to the intersection of the silo and the axis opposite the attachment point.

For any longer tether, at some angle (less than pi), the cow would reach the axis. Determining this angle is left as an exercise for the student.

If we stop the integration before theta = pi, we must also add the additional area. This incremental area can be determined by expressing the remaining length as a function of theta. However, it is probably easier to observe that there is a triangle formed by the tangency point where we stopped the integration, the location of the cow on the axis at that time and the center of the silo. Calculate the area of the triangle and subtract the portion that is within the silo.

Now, add up the three parts and you have your answer in terms of T and R.
Substituting the values given produces the final result.