Geometry/calculus (I think) goat problem.
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Geometry/calculus (I think) goat problem.
A goat on a 10 foot rope is tied to a circular barn with a radius of 8 feet. What is the total area in which the goat can graze?
We did some of these in school, but we always used polygons.
We did some of these in school, but we always used polygons.
 jestingrabbit
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Re: Geometry/calculus (I think) goat problem.
What have you done so far? If the barn weren't there, what area of grass would the goat have access to? How does the barn change things?
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 scarecrovv
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Re: Geometry/calculus (I think) goat problem.
Can the rope slide around the barn, so that the goat can reach any point within 10 feet of the barn, or is the rope fixed to one point on the barn? Because one of those problems is much easier than the other.

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Re: Geometry/calculus (I think) goat problem.
the field is a euclidean plane, and the rope is fixed at one point. I have absolutely no clue how to solve it, but there should be a way.
 jestingrabbit
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Re: Geometry/calculus (I think) goat problem.
Okay, so if there was no barn, how would you describe the points that the goat can get to? What shape does it have? It might help to actually make a little physical model with a pin for the stake and a bit of string for the chain, or at least to think about that situation.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
 silverhammermba
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Re: Geometry/calculus (I think) goat problem.
Alternatively, think about the maximum distance that the goat can move from the barn in a certain direction.
Assume that when viewing the barn from above, the goat is tied at the rightmost point. Then clearly the goat can stretch the rope to its full 10 feet so long as he doesn't walk to the left (e.g. he can move 10 feet up, or right, or down). However, if the goat moves at all to the left, the rope starts wrapping around the barn so the goat can no longer move all 10 feet in that direction. Try to figure out this relationship between direction and distance.
Assume that when viewing the barn from above, the goat is tied at the rightmost point. Then clearly the goat can stretch the rope to its full 10 feet so long as he doesn't walk to the left (e.g. he can move 10 feet up, or right, or down). However, if the goat moves at all to the left, the rope starts wrapping around the barn so the goat can no longer move all 10 feet in that direction. Try to figure out this relationship between direction and distance.
 agelessdrifter
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Re: Geometry/calculus (I think) goat problem.
I came up with a solution but it is embarrassingly inelegant. Comes to [imath](1000/24)+50\pi+400tan^{1}(10/8)32\pi[/imath]≈508.7ft[imath]^2[/imath]
I have to run, but I'd really like to know if that's accurate, if anyone else has worked the problem and can verify. I'll write up how I got it when I have a chance so you can all laugh at me
I have to run, but I'd really like to know if that's accurate, if anyone else has worked the problem and can verify. I'll write up how I got it when I have a chance so you can all laugh at me

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Re: Geometry/calculus (I think) goat problem.
Isn't 508.7 square feet more area than the goat could graze if there were no barn?
I might have a go at an answer later, I've worked out parametric equations for the funny shaped bit where the rope is partly clinging to the barn. I keep thinking there must be a nicer way to do it.
I might have a go at an answer later, I've worked out parametric equations for the funny shaped bit where the rope is partly clinging to the barn. I keep thinking there must be a nicer way to do it.
Re: Geometry/calculus (I think) goat problem.
I believe there is a much nicer way to do it. Consider the area the goat could graze without the barn, then consider the area the goat can't graze because of the barn. I get around 234.16 square feet.
EDIT: No calculus was used in the making of this post.
EDIT: No calculus was used in the making of this post.
 skeptical scientist
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Re: Geometry/calculus (I think) goat problem.
Kirby wrote:I believe there is a much nicer way to do it. Consider the area the goat could graze without the barn, then consider the area the goat can't graze because of the barn. I get around 234.16 square feet.
EDIT: No calculus was used in the making of this post.
That's not exactly right, because the barn not only makes certain areas inaccessible to the goat, it also makes certain areas inaccessible for the rope.
The solution I got when I did this problem back in high school using a physicsstyle handwavy argument for the area was [math]50\pi+2\int_0^{10}(10x)/2\, dx = 50\pi+50 \approx 207.[/math] But, IIRC, I also did the problem a different way and got a different answer, and I never did figure out which was right.
One way to go about it would be to write a parametric equation for the path of the rope if the goat walks around the barn with it stretched to the max, and then figure out the area under that curve.
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Re: Geometry/calculus (I think) goat problem.
Well, I finished working through it using parametric equations and integration to get the funny shaped area. Got a final answer of 198.7ish
I wouldn't swear by it though, it all got a bit messy and I might have made a mistake somewhere. Be interesting to see if anyone else gets the same number.
I wouldn't swear by it though, it all got a bit messy and I might have made a mistake somewhere. Be interesting to see if anyone else gets the same number.
 skeptical scientist
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Re: Geometry/calculus (I think) goat problem.
greengiant wrote:Well, I finished working through it using parametric equations and integration to get the funny shaped area. Got a final answer of 198.7ish
I wouldn't swear by it though, it all got a bit messy and I might have made a mistake somewhere. Be interesting to see if anyone else gets the same number.
That's the answer I got too. (Or, to be more exact, 250/6+50π) But, I'm not sure why my physicsy handwavy integration failed.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
"With math, all things are possible." —Rebecca Watson
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 agelessdrifter
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Re: Geometry/calculus (I think) goat problem.
greengiant wrote:Isn't 508.7 square feet more area than the goat could graze if there were no barn?
I might have a go at an answer later, I've worked out parametric equations for the funny shaped bit where the rope is partly clinging to the barn. I keep thinking there must be a nicer way to do it.
Err... Yeah, I was uh, just making sure you were paying attention
Revised answer + explanation:
[imath]50\pi + 2[125/3 + 40tan^{1}(5/4)32][/imath]≈263ft[imath]^2[/imath]
Hopefully that's at least a more reasonably incorrect answer than the last one. I agree there must be some simple way of doing it  mine clearly wasn't that, and I'm not sure it's right at all.

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Re: Geometry/calculus (I think) goat problem.
agelessdrifter wrote:stuff
Feel like a bit of a git for pointing out a second problem, but looking at your diagram I think you might have misread the question and used a barn with a diameter of 8 (rather than radius).
skeptical scientist wrote:That's the answer I got too. (Or, to be more exact, 250/6+50π) But, I'm not sure why my physicsy handwavy integration failed.
Excellent. Pretty satisfying that it worked out right. Good to have an exact answer too, I cut a corner and did numerical integration.
I've been looking at your other attempt and I can't work out where you got that integral from. Never really done much physics so I'm guessing it's a trick I don't know. What's the rationale?
 agelessdrifter
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Re: Geometry/calculus (I think) goat problem.
You're right about my barn, green giant, but that was only for the drawing  I used the right dimension for the barn in my calculation, but they're still wayy off, and I see why, now. The problem is that the points along the green curve there in my drawing aren't defined by [imath](r( \theta ), \theta)[/imath] where [imath]r(\theta)=\sqrt{64+[108(\theta)]^2}[/imath] like I need them to be for the integral I used to make sense  they're defined by [imath](r(\theta),\theta+tan^{1}(\frac{108(\theta)}{8}))[/imath].I dunno how to fix that.
I dunno how to fix that. I'm trying to look at this in euclidean coordinates now and am getting nowhere.
I dunno how to fix that. I'm trying to look at this in euclidean coordinates now and am getting nowhere.
 scarecrovv
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Re: Geometry/calculus (I think) goat problem.
After an error pointed out below, I agree with everyone else that the answer is [imath]\frac{\pi 10^2}{2} + 2\frac{10^3}{6 \times 8} \approx 198.75[/imath] square feet. My reasoning:
Obviously, [imath]\frac{\pi 10^2}{2}[/imath] is the region where the rope doesn't intersect the barn. Half the area of a circle and all that. I think everybody agrees here.
However, the rest of the area is two funny shaped regions, each of area [imath]\frac{10^3}{6 \times 8}[/imath], and this is where you're all getting tripped up. The key thing to notice is that as the rope wraps around the barn, it gets shorter and shorter, but is always tangent to the barn. We're going to integrate area between [imath]\phi = 0[/imath] radians (the rope is tangent to the barn, but not at all wrapped around it) to [imath]\phi = \frac{10}{8}[/imath] radians (the rope is wrapped around the barn completely). With me so far?
Now, what is the area of the little pie slice [imath]d\phi[/imath] radians wide at a given angle [imath]\phi[/imath]? Well, at [imath]\phi[/imath], the length of the rope that's not wrapped around the barn is [imath]10  8 \phi[/imath], right? So the little slice of area is [imath]\frac{(10  8 \phi)^2}{2} d \phi[/imath], with the factor of one half in there because it's a triangle, not a rectangle. So the integral we need to evaluate is:
[math]\int\limits_{0}^{\frac{10}{8}}\frac{(10  8 \phi)^2}{2} d \phi[/math]
Solving this:
[math]\frac{1}{2}\int\limits_{0}^{\frac{10}{8}}(10  8 \phi)^2 d \phi[/math]
Because I took calculus too long ago...
[math]\frac{(8 \times \frac{10}{8}  10)^3}{2 \times 3 \times 8}  \frac{(8 \times 0  10)^3}{2 \times 3 \times 8}[/math]
[math]\frac{(10  10)^3}{48}  \frac{1000}{48}[/math]
[math]\frac{0}{48}  \frac{1000}{48}[/math]
[math] \frac{1000}{48}[/math]
[math]\frac{10^3}{6 \times 8}[/math]
Multiply by 2, since we have 2 of these funny shaped regions, and then add the semicircle, and you get
[math]\frac{\pi 10^2}{2} + 2\frac{10^3}{6 \times 8} \approx 198.75[/math]
I think.
Obviously, [imath]\frac{\pi 10^2}{2}[/imath] is the region where the rope doesn't intersect the barn. Half the area of a circle and all that. I think everybody agrees here.
However, the rest of the area is two funny shaped regions, each of area [imath]\frac{10^3}{6 \times 8}[/imath], and this is where you're all getting tripped up. The key thing to notice is that as the rope wraps around the barn, it gets shorter and shorter, but is always tangent to the barn. We're going to integrate area between [imath]\phi = 0[/imath] radians (the rope is tangent to the barn, but not at all wrapped around it) to [imath]\phi = \frac{10}{8}[/imath] radians (the rope is wrapped around the barn completely). With me so far?
Now, what is the area of the little pie slice [imath]d\phi[/imath] radians wide at a given angle [imath]\phi[/imath]? Well, at [imath]\phi[/imath], the length of the rope that's not wrapped around the barn is [imath]10  8 \phi[/imath], right? So the little slice of area is [imath]\frac{(10  8 \phi)^2}{2} d \phi[/imath], with the factor of one half in there because it's a triangle, not a rectangle. So the integral we need to evaluate is:
[math]\int\limits_{0}^{\frac{10}{8}}\frac{(10  8 \phi)^2}{2} d \phi[/math]
Solving this:
[math]\frac{1}{2}\int\limits_{0}^{\frac{10}{8}}(10  8 \phi)^2 d \phi[/math]
Because I took calculus too long ago...
[math]\frac{(8 \times \frac{10}{8}  10)^3}{2 \times 3 \times 8}  \frac{(8 \times 0  10)^3}{2 \times 3 \times 8}[/math]
[math]\frac{(10  10)^3}{48}  \frac{1000}{48}[/math]
[math]\frac{0}{48}  \frac{1000}{48}[/math]
[math] \frac{1000}{48}[/math]
[math]\frac{10^3}{6 \times 8}[/math]
Multiply by 2, since we have 2 of these funny shaped regions, and then add the semicircle, and you get
[math]\frac{\pi 10^2}{2} + 2\frac{10^3}{6 \times 8} \approx 198.75[/math]
I think.
Last edited by scarecrovv on Mon May 09, 2011 2:21 am UTC, edited 2 times in total.
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Re: Geometry/calculus (I think) goat problem.
greengiant wrote:I've been looking at your other attempt and I can't work out where you got that integral from. Never really done much physics so I'm guessing it's a trick I don't know. What's the rationale?
It's more or less what scarecrovv just posted (the difference is because I forgot to account for the factor of 1/8 radians/ft when the rope is wrapped). That answer is wrong, as it disagrees with the parametric method which is more reliable than the handwavy physicsy explanation justifying why that integral should give the area. (scarecrovv, if you doubt this, just do it out on graph paper  the number your method gives is off by a lot, so it shouldn't be hard to see that it's wrong using a rough estimate.) But I feel like scarecrovv's method should give the right answer, so I'd be interested if someone can explain why it doesn't.
Last edited by skeptical scientist on Mon May 09, 2011 3:17 am UTC, edited 1 time in total.
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 agelessdrifter
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Re: Geometry/calculus (I think) goat problem.
Is it because they're little arc segments rather than little triangles? Shouldn't it be [imath]\frac{1}{2}(108\phi)^2d\phi[/imath] instead?
 scarecrovv
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Re: Geometry/calculus (I think) goat problem.
skeptical scientist wrote:(scarecrovv, if you doubt this, just do it out on graph paper  the number you're method gives is off by a lot, so it shouldn't be hard to see that it's wrong using a rough estimate.) But I feel like scarecrovv's method should give the right answer, so I'd be interested if someone can explain why it doesn't.
Ok, I'll accept that I'm wrong. I too am curious about why.
agelessdrifter wrote:Is it because they're little arc segments rather than little triangles? Shouldn't it be [imath]\frac{1}{2}(108\phi)^2d\phi[/imath] instead?
Oh, shit, you're right! hang on...
Post above fixed. My answer is now 198.75. I was finding the arc length of the funny shaped area, not the area.
 agelessdrifter
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Re: Geometry/calculus (I think) goat problem.
At least I was able to say one right thing in this thread.
 skeptical scientist
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Re: Geometry/calculus (I think) goat problem.
scarecrovv wrote:agelessdrifter wrote:Is it because they're little arc segments rather than little triangles? Shouldn't it be [imath]\frac{1}{2}(108\phi)^2d\phi[/imath] instead?
Oh, shit, you're right! hang on...
Post above fixed. My answer is now 198.75. I was finding the arc length of the funny shaped area, not the area.
Ah, that explains it! And it doesn't matter that it's an arc segment and not a triangle; the area formula is the same. The point is that the width of the triangle/arc should be [imath](108\phi)d\phi[/imath], rather than [imath]d\phi[/imath].
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
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Re: Geometry/calculus (I think) goat problem.
From [url=http://www.wolframalpha.com/input/?i=100+pi%28%28integrate+2+sqrt%28100%288%2Bx%29^2%29+from+7%2F4+to+0%29%2B%28integrate+2+sqrt%2864x^2%29+from+8+to+7%2F4%29%29]WolframAlpha[/url]
The 7/4 comes from the intersection between the circles.
The 7/4 comes from the intersection between the circles.
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 agelessdrifter
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Re: Geometry/calculus (I think) goat problem.
skeptical scientist wrote:It doesn't matter that it's an arc segment and not a triangle; the area formula is the same. The point is that the width of the triangle/arc should be [imath](108\phi)d\phi[/imath], rather than [imath]d\phi[/imath].
Thanks, I was finding the distinction there sort of puzzling.
 colinrmitchell
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Re: Geometry/calculus (I think) goat problem.
I tried it using tangent vectors. Kind of cheating, though...
Code: Select all
len = 10; # Rope length.
rad = 8; # Barn radius.
d = [0:0.1:len];
x = zeros(1, length(d));
y = zeros(1, length(d));
cx = zeros(1, length(d));
cy = zeros(1, length(d));
p = zeros(2, length(d));
dp = zeros(2, length(d));
for i = 1:length(d)
# Find point on circle.
p = [ rad * cos( pi/2  d(i)/rad);
rad * sin( pi/2  d(i)/rad)];
cx(i) = p(1);
cy(i) = p(2);
# Find tangent vector.
dp = [ 1;
p(1) / sqrt(64  p(1)^2)];
# Normalize it.
dp = dp / norm( dp );
# Scale it to remaining rope length.
dp = ( len  d(i) ) * dp;
# Add it to point on circle.
new_p = p + dp;
x(i) = new_p(1);
y(i) = new_p(2);
end
area = (pi * len^2)/2 + 2 * (rad*len  trapz(x,y)  trapz(cx,cy))
Code: Select all
> area = 198.75
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