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/sci/ - Science & Math

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2199636 No.2199636 [Reply] [Original]

None of you will get this.
A cow is tethered to any point on a circle with radius x. Find the length of a piece of rope in terms of x that would allow the cow to move in exactly half of the circle's area.

>> No.2199646 [DELETED] 

sqrt(2)*x

obvious hmw

>> No.2199652

calculus works

>> No.2199660

2(pi)x

Enjoy

>> No.2199662

A1=pi*x^2
A2=A1/2=(pi*x^2)/2=(pi*r^2)
(x^2)/2=r^2
SQRT
x/(2^-0.5)=r
And there we have it.

>> No.2199663

sqrt(1/2) * X

>> No.2199665

I HATE YOU GENIUSES AND YOUR GENIUS INSIDE JOKES

>> No.2199668

>>2199652

>calculus

doublefacepalm.jpg

>> No.2199670

OP here
>>2199646
>>2199660
you gotta aspain it bro

>> No.2199673

>>2199665
Your tears, they are delicious.

>> No.2199687

OP here again,
>>2199662
too easy, therefore wrong.
this really is hard. wasnt kidding when i said none of you would get it.

>> No.2199689

sqrt(2)*x. You're welcome

>> No.2199700

or you dumbasses could read the question

x=r

>> No.2199701
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2199701

>>2199687
It's not hard. You're just retarded, that's all.

>> No.2199709

All naga found was the radius of a circle with half the area of the original circle, doesn't the question ask for the radius for the cow to move inside the circle? (only the part of the new circle inside the old circle would count) otherwise the fact that the cow is tethered to a point on the circle would be meaningles. seems like calculus to me

>> No.2199712

This is just nontrivial enough for me to not care. Sorry, OP. I guess you win.

>> No.2199715

So the problem is to find the intersect of two circles of equal area that is half the area of the circles.

>> No.2199718

Draw it. That wouldn't be half the area.

>> No.2199729

take a rope that is 2x in length, and split the circle in half...

>> No.2199732

also x will be larger than r (to prove this draw a straight line through the circle and a new circle with a center on the original circle tangent to that line)

>> No.2199745 [DELETED] 

A = 1/2 r^2 (theta)
Half the circle's area => area of a half circle
theta = pi
A = (1/2)pi*r^2
A = pi*x^2
(1/2)r^2 = x^2
r = Sqrt(2)x

I just wanted to use another way of doing it. For kicks. Even though it is technically the same way as everyone else with just another definition of area.

>> No.2199750

>>2199718
Explain why it wouldn't be.
The cow can only move in HALF the circle.

>> No.2199751
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2199751

Where r is the radius of the circle and x is the length of the rope:

(1/2)(pi)r^2=((pi)r^2)-((pi)(r-x)^2)

First half of the above represents the whole circle's area divided by two.
Second half is the whole circle's area minus the area that can't be reached because of the rope (a circle of radius r-x).

x=r(1+(1/sqrt(2))

yes?

>> No.2199765

The problem is hard only if the cow has to stay inside the big circle. OP didn't state that.

>> No.2199773

This was posted on /sci/ a month ago.
The answer is "circle-circle intersection": http://mathworld.wolfram.com/Circle-CircleIntersection.html
No, OP, it isn't hard.

>> No.2199779

>>2199751
Learn to <span class="math">[/spoiler]

>> No.2199789
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2199789

>>2199751
Oh, wait, I misunderstood.
I worked it out for attaching at EVERY point around the circle. I see what the actual problem is now. My bad.

>>2199773
is correct, I think.