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math puzzle: a rectangular field, a rope and a cow

T

tmc

Golden Member
There is a rectangular grass field of size 'a' by '2a'.

A cow is tied by a rope of length 'r' at the center of the field.

The cow is able to graze exactly half the area of the field.

What is the length of the rope (in terms of 'a')?

edit - this is not my HW and this is not that straight-forward (easy) as it looks.
 
It's a trick. The obvious answer leaves the cow grazing outside of the field. I am not motivated enough to find the right answer, though.

Got the equation but I have no interest in solving for r.

area/2 = 2*a*arcsin(a/(2r))*r^2 + r*a*sin(pi-arcsin(a/(2r)))/2

The first term is the area of the grazing area that isn't help up by the fence, and the second is the area of the grazing area that is limited by the fence.

edit for confusing
 
Originally posted by: LordMorpheus
It's a trick. The obvious answer leaves the cow grazing outside of the field. I am not motivated enough to find the right answer, though.
Click to expand...

I don't think it does. r = sqrt(3a/2pi) sub is say 5 for "a" and you get r = 1.545. That is less than 1/2 "a" so it will not go outside the field.

Unless i am missing something this is really easy.

Edit

Nevermind I fail. it is 2a^2 not 3a in the sqrt.

 
Originally posted by: SSSnail
rec = 3a
circle = 3a/2
length of rope = squareroot (1.5a/pie)
Click to expand...

You need to multiply a and 2a to get the area:

area of rec = a * 2a = 2a²

Half of area of rec = 1/2 * 2a² = a²

Area grazed = ¶r²

Solve for r: a² = ¶r²

a²/¶ = r²

sq root of a²/¶ = r
 
Originally posted by: tmc
Originally posted by: LordMorpheus
It's a trick.
Click to expand...

the answer comes to '0.583a' i think.
Click to expand...

The answer... in terms of a is as kthroyer said a * 1/sqrt(Pi), which is the same as your 0.583a. I fail to see how that wasn't obvious and "not as easy as it looks".

Aside from everybody thinking that a * 2a = 3a that is.
 
If the cow can graze outside the field, the answer is easy as shown above. If not then, r = 0.5828 * a.

Do I need to show work?
 
Originally posted by: SunnyD
Originally posted by: tmc
Originally posted by: LordMorpheus
It's a trick.
Click to expand...

the answer comes to '0.583a' i think.
Click to expand...

The answer... in terms of a is as kthroyer said a * 1/sqrt(Pi), which is the same as your 0.583a. I fail to see how that wasn't obvious and "not as easy as it looks".

Aside from everybody thinking that a * 2a = 3a that is.
Click to expand...

a/sqrt(pi) = 0.5642a

this would be correct if we were talking about a circle inside of a perfect square. In this case we aren't, so the correct answer is more complicated. In this case part of the grazing area you are attributing to the circle of radius 0.5642a is going to be limited by the fence (or simply being out of bounds).

No time to figure out the answer right now though...
 
Here is my work:

This picture will help.

The cow grazes an area of a^2. From my picture, that area is equal to a*b + 2*K, where K is the area of each of the two circle segments on the left and right. Thus:
[*]a^2 = a*b + 2*K

The segment areas combined are (segment area):
[*]2K = r^2*(C - sin(C))
where C is in radians.

Thus,
[*]a^2 = a*b + r^2*(C - sin(C)) *

Simple trignometry gives:
[*]C = 2*asin(a/2r)
[*]b = 2*(r^2-(a/2)^2)^0.5

So, if we knew r, we could find C and b and thus we can solve the equation marked with an asterisk (*). I just plugged it into Excel and used Solver. Pretty much, it guesses 'r', then checks equation (*) and adjusts 'r' until it works. You can use any value for 'a' that you want in Excel, just remember that 'r' is scaled by it.
 
Originally posted by: dullard
Here is my work:

This picture will help.

The cow grazes an area of a^2. From my picture, that area is equal to a*b + 2*K, where K is the area of each of the two circle segments on the left and right. Thus:
[*]a^2 = a*b + 2*K

The segment areas combined are (segment area):
[*]2K = r^2*(C - sin(C))
where C is in radians.

Thus,
[*]a^2 = a*b + r^2*(C - sin(C)) *

Simple trignometry gives:
[*]C = 2*asin(a/2r)
[*]b = 2*(r^2-(a/2)^2)^0.5

So, if we knew r, we could find C and b and thus we can solve the equation marked with an asterisk (*). I just plugged it into Excel and used Solver. Pretty much, it guesses 'r', then checks equation (*) and adjusts 'r' until it works.
Click to expand...

yes, i believe this is correct.
 
Originally posted by: Fritzo
Will this let me know if the fly will hover if the car is moving?
Click to expand...
The cow takes off, as long as it is at least 9.99999 lbs and has worked out sufficiently on a conveyor belt.
 
Originally posted by: SunnyD
Originally posted by: tmc
Originally posted by: LordMorpheus
It's a trick.
Click to expand...

the answer comes to '0.583a' i think.
Click to expand...

The answer... in terms of a is as kthroyer said a * 1/sqrt(Pi), which is the same as your 0.583a. I fail to see how that wasn't obvious and "not as easy as it looks".

Aside from everybody thinking that a * 2a = 3a that is.
Click to expand...

because, like i said, this was a homework problem for him and he was trying to get us to do it for him.

i remember these exact homework/quiz problems when i was in school.
 
Originally posted by: tmc
Originally posted by: dullard
Here is my work:

This picture will help.

The cow grazes an area of a^2. From my picture, that area is equal to a*b + 2*K, where K is the area of each of the two circle segments on the left and right. Thus:
[*]a^2 = a*b + 2*K

The segment areas combined are (segment area):
[*]2K = r^2*(C - sin(C))
where C is in radians.

Thus,
[*]a^2 = a*b + r^2*(C - sin(C)) *

Simple trignometry gives:
[*]C = 2*asin(a/2r)
[*]b = 2*(r^2-(a/2)^2)^0.5

So, if we knew r, we could find C and b and thus we can solve the equation marked with an asterisk (*). I just plugged it into Excel and used Solver. Pretty much, it guesses 'r', then checks equation (*) and adjusts 'r' until it works.
Click to expand...

yes, i believe this is correct.
Click to expand...

On the contrary... you're neglecting the length of rope needed for the loop and knots to secure the cow. 🙂

(And for some reason I keep thinking it was 0.058-0.06 whatever. Only an order of magnitude off...)
 
turns out that Mitt Romney has been living in the hollowed out ribcage of your cow. the answer is zero. forever.
 
Originally posted by: nkgreen
I'm going to call all the math teachers at Bob Jones HS and see if this is really HW. 😉
Click to expand...

Oh man, you're cruel! Might want to call Harvest-Monrovia as well just in case.
 
i guess too many ppl asking for HW solutions in ATOT.

anyways, it is not that hard to see i wasn't one of them (hint: profile, name, google).
 
Originally posted by: tmc
i guess too many ppl asking for HW solutions in ATOT.

anyways, it is not that hard to see i wasn't one of them (hint: profile, name, google).
Click to expand...

owned.
 
Wouldn't it be like:

Find area of rectangle:

A x 2A = 2A^2

Find half area of rectangle:

(2A^2)/2 = A^2

State that the area of the circle = half area of the rectangle:

- R^2 = A^2

Square root both sides to remove the squares: (Not sure about this one)

-^(1/2) x R = A

Divide both sides by the square root of Pi to leave R on one side and the answer on the other side in terms of A as requested.

R = A/-^(1/2)

I've never been good at math. If I'm wrong I'd like to know where though.
 
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