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- Mar 1st 2008, 05:22 PMstargirldrummer187polar equations with trig intersection points
asd

- Mar 1st 2008, 05:38 PMmr fantastic
It might be easier to switch to Cartesian coordinates. I'll explain soon - have to run and put a fire out right now.

Nope, false alarm. Same for the fire.

Any reason to think you have to get exact solutions to the equation?

Also, have you tried drawing the curves .... you realise that $\displaystyle r = 3 \sec (\theta - 30)$ is a line ...? - Mar 1st 2008, 05:46 PMstargirldrummer187
We have somewhat tired converting the equations to cartesian equations with conversion formulas, but it turned out..funny.

If you can do it, that would be wonderful.

We also know you can graph them as cartesian equations and get the intersections, but if we were to do that (according to our teacher) we have to have proof of why it works.

Thank you for your input, We'll look forward to more help.(Talking) - Mar 1st 2008, 06:16 PMstargirldrummer187
Yeah, we need exact..our teacher is a tad insane.

Good..but very insane.

Yes we know it's a line, but how is that going to help us solve the situation? - Mar 1st 2008, 06:38 PMmr fantastic
My only other thought at this stage is to get an equation with r.

$\displaystyle r = 3 \sec (\theta - 30)$

$\displaystyle \Rightarrow \sqrt{3} r \cos \theta + r \sin \theta = 6$ .... (1).

$\displaystyle r = 1 + 5 \cos \theta$

$\displaystyle \Rightarrow \cos \theta = \frac{r - 1}{5}$ .... (2).

From (2), $\displaystyle \sin \theta = \pm \frac{\sqrt{25 - (r - 1)^2}}{5} = \pm \frac{\sqrt{(6 - r)(r + 4)}}{5}$ .... (3).

Substitute (2) and (3) into (1) and re-arrange into a quartic equation in r. You might get r from this (I'll cop that I haven't actually tried .....) - Mar 1st 2008, 08:47 PMstargirldrummer187
I'm a little confused as to how you got that last line, but I'll work on it and get back to you.

- Mar 1st 2008, 09:06 PMmr fantastic
From the Pythagorean Identity: $\displaystyle \sin^2 \theta = 1 - \cos^2 \theta$

$\displaystyle 1 - \cos^2 \theta = 1 - \left( \frac{r - 1}{5} \right)^2 = 1 - \frac{r^2 - 2r + 1}{25}$

$\displaystyle = \frac{25 - r^2 + 2r - 1}{25} = \frac{-r^2 + 2r + 24}{25} = \frac{-(r^2 - 2r - 24)}{25}$ etc.

But on reflection I don't think you're gonna get anything even a little bit easy popping out of the resulting quartic equation in r .....

I'll keep it in the back of my mind.