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PostPosted: Wed, 21 Apr 2004 19:58:39 UTC 
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There is a parabola y=x^2

Imagine one circle with radius 5, drop inside this parabola and "stuck there" (hence there is some area between the circle and parabola)

1) find the centre of this circle (should be easy)
hint: obviously x coordinate is 0

Imagine another circle, with radius 4, drop inside this parabola on top of the previous circle. (so now there are 2 circles in the parabola) Unlike the first circle, this smaller circle is against the right hand side of the parabola.

2) Find the centre of this circle (challenging)

Sorry for my poor description, but just imagine you have a plastic bag (with shape y=x^2 fixed). You drop one ball into this bag, then drop another.

Does anyone know whether i can attach some picture files in this forum? So that anyone can view the graph clearly.


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PostPosted: Thu, 22 Apr 2004 23:52:04 UTC 
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Large circle = CircleA

CircleA=(x-x_0)^2+(y-y_0)^2=25, x_0=0\Leftrightarrow{x^2+(y-y_0)^2=25
y=\sqrt{25-x^2}+y_0
\frac{dy}{dx}=-\frac{x}{\sqrt{25-x^2}}

We must have

-\frac{x}{\sqrt{25-x^2}}=2x
-\sqrt{25-x^2}=\frac{1}{2}
x^2=25-\frac{1}{4}
x=\pm{\sqrt{24.75}}\Longrightarrow{y=24.75}

These are the points of tangency for CircleA and the parabola. We then must have

2\sqrt{24.75}=\sqrt{100-(10-2h)^2},
where h is the vertical distance between the tangent points and the lowest point on the circle. So;

100-(10-2h)^2=99
(10-2h)^2=1
h=4.5

The center y coordinate then follows.

5-4.5=0.5
24.75+0.5=25.25

So the center of CircleA is at (0,25.5). I'll be back a little later to work more on this.

-Kwyjibo

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PostPosted: Fri, 23 Apr 2004 01:47:30 UTC 
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thank you very much for your reply!

the 2nd problem would be a lot harder... hope can hear good news from you! have a good day


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PostPosted: Fri, 23 Apr 2004 02:31:12 UTC 
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Heres a slightly different approach for the first one:

Let the center of the circle be (0,h). Then the circles equation is:
x^2 + (y-h)^2 = 25
y=x^2

y + y^2 - 2hy + h^2 - 25 = 0
y^2 + (1-2h)y + (h^2 - 25) = 0

The circle and parabola intersect at two points of equal height, so this equation has 2 real equal roots, so the discriminant is 0:

(1-2h)^2 - 4(h^2 - 25) = 0
4h^2 - 4h + 1 - 4h^2 + 100 = 0
-4h = -101
h = 101/4

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PostPosted: Fri, 23 Apr 2004 10:43:14 UTC 
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you guys' approaches are elegant :)

I have another approach as well.

Let D be the distance btw (0,h) and (x,x^2), (circle centre & any point on the parabola)

D^2 = x^2 + (y-h)^2
= x^2 + (x^2-h)^2

For the point it touches, the distance must be shortest (radius)

Derivative:
(D^2)' = 2x + (2)(2x)(x^2-h) = 0

then x^2 = h - 1/2

put back to the equation, where the distance is 5

25 = x^2 + (x^2-h)^2
= (h - 1/2) + (h - 1/2 -h)^2
= (h - 1/2) + 1/4

h=101/4

I have found out the solution for the 2nd circle, but somehow not very satisfy my solution. Hope you guys can get one better than mine. :P


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PostPosted: Sat, 24 Apr 2004 09:35:36 UTC 
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Hello kongkahloon,

So far I only found the solution for the first one, it's about the same as Kwyjibo's.

Indeed, the second one is not easy to solve.
I found an rough, graphical, approximation of the center;the goal is, of course, an exact solution!

But I'll keep on trying, I'am waiting for an inspiration!


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PostPosted: Sat, 24 Apr 2004 10:30:10 UTC 
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Thanks for trying, CDMP

Yup, the 2nd question (created by myself) quite hard. In fact i totally spent more than 10 hrs on this question.

:evil:

Actually the original question was finding the area between 2 circles and the parabola (the little bending triangle). But I guess the ppl who is able to find the small circle's centre is almost surely able to find the area.


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PostPosted: Sun, 25 Apr 2004 19:23:41 UTC 
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Hello kongkahloon,

Finally I managed to get an answer,though not an exact one, but a good approximation.....

I found the center of the circle with radius 4:
x=1.82167.....
y=34.06371....

Here I will give a short description of my method.

1)First of all, the distance from the center of the first circle (0,25.25) to the center of the second circle is 9, so the equation of the center of the second circle is:
y(x)= (81-x²)^½+25.25...............eq(1)

2)Now we have to find some point on this circle (eq(1)), this point should be the center the second circle, and we know the distance from this point , perpendicular to the tangent of the parabola, is 4.

In other words, we have to construct an intersecting point.
I will find this intersecting point by finding another parabola ,parallel to the given one.

3)The second parabola I found this way:
First of all I chose 3 points P,Q and R on the first parabola, with coordinates and slope:
P:x_1=5,y_1=25 and y_1'=10
Q:x_2=6,y_2=36 and y_2'=12
R:x_3=7,y_3=49 and y_3'=14

So the slopes at these 3 points is known here, and also the slope of their normals.

Using this I found the coordinates of 3 points at a distance of 4 from P,Q and R.
Now I can set up 3 equations with 3 unknowns:
y(x)=Ax²+BX+C.

I found y(x)=1.00473251x²+7.952522153x+16.24260649.........eq(2)

4)Now we have eq(1)-eq(2)=0.
This yields x=1.82167.....
y=34.06371.....

We also can check the distance to the first parabola:4.00001....


Last edited by CMDP on Fri, 14 May 2004 18:47:29 UTC, edited 1 time in total.

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PostPosted: Tue, 27 Apr 2004 03:09:58 UTC 
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CMDP wrote:

1)
y(x)= (81-x²)^½+25.25...............eq(1)

I found y(x)=1.00473251x²+7.952522153+16.24260649.........eq(2)

4)Now we have eq(1)-eq(2)=0.
This yields x=1.82167.....
y=34.06371.....


CMDP, how did you solve (1)=(2) ?


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PostPosted: Tue, 27 Apr 2004 05:00:11 UTC 
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I think I figured out a way to do it. You are going to need a bunch of equations though.

Equation of Top Circle
Equation of Bottom Circle
Derivative of Top Circle
Derivative of Bottom Circle

Now, we can set the two pairs of respective equations equal and solve somewhat. But, we still have four unknowns so we need 4 equations. My idea is that the length of the line segment connecting the two centers is going to be 9. Therefore, we can use the distance formula to produce a third equation. The fourth equation could possibly be produced by setting the opposite of the concavities equal (I don't think this is legal though b/c the circles are of different radii).

However, these equations quickly reach high exponents making them unsolvable by algebraic means.


Comments?


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PostPosted: Tue, 27 Apr 2004 06:08:23 UTC 
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Thanks for responding DaSweepa.

As you mentioned, we do need a few equations to get the circle centre. At first I tried using this method and i got to solve an 8-degree polynomial equation (i.e. solve ax^8 + bx^7 + ... + hx + i = 0) something like that.

Then I tried to find some "more efficient" equations. I managed to reduce the degree of polynomial to 6. Then 4. Few weeks ago I managed to reduce it to 3-degree. Then I applied cardano's formula to solve.

When I kept on to try to reduce 2-degree (i.e. quadratic, which means that anyone above grade 10 is able to solve it), suddenly a bad news came to me. I found that there was a mistake in my calculation (the cubic equation), so far I havent found out the reason. (but i am sure that the error is a concept error, not calculation error)

Means that the best result I could get so far is to reduce the equation to 4-degree. I am going to have a big exam on May 7, after this day I will concentrate on finding out the mistake i made.


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PostPosted: Tue, 27 Apr 2004 17:52:55 UTC 
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kongkahloon,

I used the Newton's Method program on my TI-83.

The description of my method contains a small typo:
Quote:
1)First of all, the distance from the center of the first circle (0,25.25) to the center of the second circle is 9, so the equation of the center of the second circle is:
y(x)= (81-x²)^½+25.25...............eq(1)


This should read:
1)First of all, the distance from the center of the first circle (0,25.25) to the center of the second circle is 9, so the equation of the circle on which the center of the second circle is located is:
y(x)= (81-x²)^½+25.25...............eq(1)


This explains the need for a second curve,parallel to the first parabola, as described in 3).


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