Maths question! (1 Viewer)

[mathsgod]

I have a circular paddock with AREA: 1 unit^2. If I have a goat with a leash tied to the outer fence (circumference of paddock), how long must the leash, L, be to only allow the goat to eat half the grass in the paddock?

-sorry if my wording confuses anyone (im a left brainer)

 

[ascentyx]

Okay the area of the circle is 1unit^2

Therefore using area = pi r^2

1 = pi r^2

r1 = 1 / sqrt(pi) (the radius of the circular paddock)

Now we have to work out how long the leash should be so that half the paddock can be eaten. This is essentially another circle inside the first one where the area of the 2nd circle is exactly half the area of the first one.

Therefore finding the radius of the smaller circle:

1/2 = pi r^2

r2 = 1/ sqrt(2pi)

Therefore the length of the leash L should be L = r1 - r2 = 1 / sqrt(pi) - 1/ sqrt(2pi).

Now you can rationalise each radius and get a nicer answer of:

[2sqrt(pi) - sqrt(2pi)] / 2pi

This would make more sense with a diagram but i kinda cbfed. also im so tired so i've probably made some mistakes

 

[Cazic]

The paddock has a radius r_p given by:



Now, you want the goat to eat half a square unit of grass, or equivalently, to leave half a square unit of grass uneaten. Assuming the goat can traverse the entire circumference of the paddock, the uneaten grass will be a small circle in the centre of the paddock. Ok, we have a picture in mind now.

The circle whose area is half a square unit has radius r_u given by



We want the length of the lead to be:

 

[Cazic]

Ahh, too quick for me

Now you can rationalise each radius and get a nicer answer of:

[2sqrt(pi) - sqrt(2pi)] / 2pi

This would make more sense with a diagram but i kinda cbfed. also im so tired so i've probably made some mistakes

P.S. pi isn't rational.

 

[ascentyx]

Ahh, too quick for me



P.S. pi isn't rational.

hahha yeh force of habit using that term, i used to say rationalise when i meant realise in 4unit too anyways yeh ur answer is a nicer form i probly should have done that haha.

 

[mathsgod]

Thanks for the replies but I must have worded the question wrong. The anchor point of the leash is non moving, it must stay still. The goat however can move on the circumference of the invading circle as illustrated in my attachment.

 

[jet]

Can you reproduce the question from your source?

 

[mitchy_boy]

Is the answer 1/sqrt(pi)?

 

[Cazic]

If the point of attachment of the leash is fixed (so it can't move around the circumference of the paddock), then my understanding (not necessarily right) is that this problem reduces to finding how to overlap two circles (one small and one large) such that the large circle centred on the circumference of a smaller circle overlaps half of the smaller circle. Since the centre of the large circle is fixed and we don't care how much of the large circle is used, this is indeed a matter of finding the radius (leash length) of the larger circle, and that's a cool problem.

Mathsgod, does this sound right?

 

[Cazic]

Is the answer 1/sqrt(pi)?

I don't think so, because that would be the same size as the original paddock, and if you draw a two circles, one centred on the perimeter of the other, it won't cover half of the other circle.

 

[mathsgod]

If the point of attachment of the leash is fixed (so it can't move around the circumference of the paddock), then my understanding (not necessarily right) is that this problem reduces to finding how to overlap two circles (one small and one large) such that the large circle centred on the circumference of a smaller circle overlaps half of the smaller circle. Since the centre of the large circle is fixed and we don't care how much of the large circle is used, this is indeed a matter of finding the radius (leash length) of the larger circle, and that's a cool problem.

Mathsgod, does this sound right?

Yeah, this is the right way... there's just the small problem of finding the solution now...

 

[jet]

Yeah, this is the right way... there's just the small problem of finding the solution now...

Tried it via integration and got a rather difficult answer which could only be solved by Newton's method/Halving the interval, I'm quite sure. Are you positive this is a 2 unit question?

 

[mathsgod]

Tried it via integration and got a rather difficult answer which could only be solved by Newton's method/Halving the interval, I'm quite sure. Are you positive this is a 2 unit question?

Nah, I was given it as a teaser and I needed to ask you guys to solve it for me. Not sure what level of mathematics it would come under.

 

[gurmies]

Seen this question before too, tried for ages and got nowhere.

 

[jet]

It's missing some piece of information to make the question solvable.

 

[ascentyx]

I don't see what's so impossible about this question. Can't you just translate it to a cartesian plane with the big circle sitting on the origin and the smaller circle under it. Then you just find the equations of both circles and where they intersect. Then just integrate the area between them and find that in terms of l (because l will be in the equation of the smaller circle and hence the intersection points aswell). Once you have that can't u just equate that area to 1/2 and solve for l?

 

[jet]

I don't see what's so impossible about this question. Can't you just translate it to a cartesian plane with the big circle sitting on the origin and the smaller circle under it. Then you just find the equations of both circles and where they intersect. Then just integrate the area between them and find that in terms of l (because l will be in the equation of the smaller circle and hence the intersection points aswell). Once you have that can't u just equate that area to 1/2 and solve for l?

That's what I did. You get a term in L and a term in arcsin(pi/2 - L) which does not solve algebraically. You need newton's method/halving the interval.

 

[Cazic]

Bump. I still haven't had a go at this, but did anyone else get anywhere?

 

[jet]

It would involve approximate solutions. This doesn't seem like a 2 unit question.

 

[Cazic]

Bah, with talk like that you'll make a great engineer You might be right though, but I still want to have a go.