Hard Area Question?

Question

A goat is grazing in a circular field with a rope attached to it of length r. The goat is free to eat all the grass as it moves around in the circular field and the area of the field is obviously pi.r^2.
Question - How long is another piece of the rope if it is now fixed to a point on the circumference of the circle and the goat can now only eat 50% of the grass in the original circle.

Answer 1

Let the first circle have radius R, and its equation in polar coordinates is
r = 2Rcosθ
Let the second circle have radius P, and its equation is
r = P
NOTE: the center of the second circle is on the circumference of the first. The circles are not concentric.

Both circles intersect at θ = +/- α
such that P = 2Rcosα
cosα = P/(2R)

A1 = The double integral of 2*r*dr*dθ
between the limits r = 0 to P and θ = 0 to α.
This will give us the part of the area enclosed between the two circles.

A2 = double integral of 2*r*dr*dθ
between the limits r = 0 and 2Rcosθ and &theta = α to π/2
This will give us the rest of the area between the two circles.

After performing the operations, we get
α * P^2 + πR^2 - 2αR^2 - R^2*sin2α = πR^2 / 2

Letting x = P/R ie fraction of first circle
α * (x^2 - 2) - sin2α + π/2 = 0

Now we need to solve numerically for x.
Writing α = arccos(x/2),
(x^2 - 2)*arccos(x/2) - x/2*sqrt(4-x^2) + π/2 = 0
we can solve for x using Newton's method. I'm getting
x = 1.158728

So the second circle must have radius 1.158728 times larger than the first in order to meet our required condition.

Answer 2

gregory_s19 is correct -- the first two answers are wrong. They didn't take into account that they goat is supposed to be able to eat 50% of the grase INSIDE the original circle.

I hope that you are in a calculus class. Here is a rough outline showing my train of thought, but I haven't worked out the details for you.

First, since we only care about the relative sizes of the radii, let the first circle be the unit circle centered at the origin:
x^2 + y^2 = 1

Then, for the sake of simplicity, since it doesn't matter where the center of the second circle is as long as it is on the circumference of the first, let the second circle be centered on the y-axis. So, letting r = the radius of the second circle, its formula is:
x^2 + (y-1)^2 = r

Obviously a radius of 1 would cover less than half of the original circle, while a radius of sqrt(2) would cover more than half of it. So the radius you want is somewhere between the original radius and sqrt(2) times the original radius.

Now, if you take the definite integral finding the area between the curves, you need the answer to come out as pi/2 so that its area is half of the original circle's area.

The first problem is how to solve for the points of interception of the two graphs (although they would be in the first and second quadrants and would be symmterical with respect to the y-axis. However, knowing this, you could integrate from 0 to the x-value of the point of interception in the first quadrant and look for where the integral is equal to pi/4.

Maybe someone else will actually figure this out for you, but I hope that I have at least given you a starting point.

Answer 3

The first two answers must be wrong, since a simple diagram demonstrates that a rope of length r allows the goat to eat far less than half his original circular plot. The new rope must be longer than the first -- just can't seem to find out how much longer....

Answer 4

just draw a nice locus diagram to get the ANSWER OUT!
dun need to stress so much!
do it sytematically!

Answer 5

it has to be r1 (r1 is radius of first rope, r2 radius of second)
but not sure what r2 is exactly

Answer 6

New area = (1/2)(pi.r^2) = pi.(r^2/2) = pi.(r/sqrt2)^2

New radius = r/sqrt2