http://i33.tinypic.com/mai452.jpg

I'm not very good at thinking about these things... any help is appreciated, it's due 10/27.

This thread has problem 1.2 in it:

http://www.mathhelpforum.com/math-he...ns-origin.html

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- Oct 26th 2008, 02:25 PMVaritronAlg. Geom. Proofs
http://i33.tinypic.com/mai452.jpg

I'm not very good at thinking about these things... any help is appreciated, it's due 10/27.

This thread has problem 1.2 in it:

http://www.mathhelpforum.com/math-he...ns-origin.html - Oct 26th 2008, 09:21 PMkalagota
for 1.3..

let $\displaystyle x=r\cos \theta$ and $\displaystyle y = r\sin \theta$, thus $\displaystyle r^2 = x^2 + y^2$

try changing $\displaystyle \sin 3\theta$ into into something with $\displaystyle \cos \theta$ and $\displaystyle \sin \theta$ only..

after that, try multiplying $\displaystyle r^3$ on both sides of $\displaystyle r=\sin 3\theta$..

for 1.4..

let $\displaystyle C(x,y) = (x-r)^2 + y^2 - r^2$ and let

$\displaystyle D(x,y) = (x-h)^2 + (y-k)^2 - R^2$

solve for $\displaystyle I_{(0,0)}(C,D)$ by taking the cases $\displaystyle k=0$ and $\displaystyle k\neq 0$ - Oct 26th 2008, 09:38 PMVaritron
I got 1.3 actually, and I did exactly that. It worked out.

For 1.4, I'm still a bit confused. Why can you put (x-r^2)^2 instead of just x^2 for C, and how do we know that D even contains the origin?

EDIT: We know D contains the origin because the problem says it does! I overlooked that. I'm still having trouble computing I(C,D) at the origin though. Showing some steps would help though. I keep thinking that either r or R need to be 0 otherwise the origin isn't contained. I wound up having something like x(2r-2h) - R^2 on one side, and I'm just very confused. - Oct 27th 2008, 02:14 AMkalagota
r and R are the radii of C and D respectively.. so if either one is 0, then the curve is just a point..

now, $\displaystyle C(x,y) = (x-r)^2 + y^2 - r^2 = x^2 -2rx + y^2$

and $\displaystyle D(x,y) = (x-h)^2 + (y-k)^2 - R^2 = x^2 -2hx + h^2 + y^2 - 2ky + k^2 - R^2$ (in fact, $\displaystyle R^2 = k^2 + h^2$ (why?))

so $\displaystyle D(x,y) = x^2 -2hx + y^2 - 2ky $

if the center of $\displaystyle D$ is on the x-axis, that is, $\displaystyle k=0$, then

$\displaystyle I_{(0,0)}(x^2 -2rx + y^2, x^2 -2hx + y^2) = ...$

if the center is not on the x-axis, i.e., $\displaystyle k\neq 0$ then

$\displaystyle I_{(0,0)}(x^2 -2rx + y^2, x^2 -2hx + y^2 -2ky) = ...$

hope you can do it from here..