1. ## Alg. Geom. Proofs

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

2. Originally Posted by Varitron

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
for 1.3..

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

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

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

for 1.4..

let $C(x,y) = (x-r)^2 + y^2 - r^2$ and let
$D(x,y) = (x-h)^2 + (y-k)^2 - R^2$

solve for $I_{(0,0)}(C,D)$ by taking the cases $k=0$ and $k\neq 0$

3. 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.

4. Originally Posted by Varitron
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.
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, $C(x,y) = (x-r)^2 + y^2 - r^2 = x^2 -2rx + y^2$
and $D(x,y) = (x-h)^2 + (y-k)^2 - R^2 = x^2 -2hx + h^2 + y^2 - 2ky + k^2 - R^2$ (in fact, $R^2 = k^2 + h^2$ (why?))

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

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

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

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

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

hope you can do it from here..