# polar equations with trig intersection points

• Mar 1st 2008, 05:22 PM
stargirldrummer187
polar equations with trig intersection points
asd
• Mar 1st 2008, 05:38 PM
mr fantastic
Quote:

Originally Posted by stargirldrummer187
I am trying to find the intersecting points of two equations in a polor coordinate grid, by setting them equal.

In the equations x is meant to be theta.
And we are in degrees.

Our first equation is
r=1+5cos(x)
and the second is
r=3sec(x-30)

I have many trig identities and have tired using them any way I can.
If you can solve this:
1+5cos(x)=3sec(x-30)
Please let me know, it would be much appreciated.
Thank you,
from many very lost pre calc students(Speechless)

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 $r = 3 \sec (\theta - 30)$ is a line ...?
• Mar 1st 2008, 05:46 PM
stargirldrummer187
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 PM
stargirldrummer187
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 PM
mr fantastic
Quote:

Originally Posted by stargirldrummer187
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?

My only other thought at this stage is to get an equation with r.

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

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

$r = 1 + 5 \cos \theta$

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

From (2), $\sin \theta = \pm \frac{\sqrt{25 - (r - 1)^2}}{5} = \pm \frac{\sqrt{(6 - r)(r + 4)}}{5}$ .... (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 PM
stargirldrummer187
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 PM
mr fantastic
Quote:

Originally Posted by stargirldrummer187
I'm a little confused as to how you got that last line, but I'll work on it and get back to you.

From the Pythagorean Identity: $\sin^2 \theta = 1 - \cos^2 \theta$

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

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