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March 6th, 2017, 06:59 PM   #1
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I can't find the third answer in the polar equations.

I am going to write down a problem below.

(ex) Find the points of intersection of r=1-sin(u) and r=1+sin(u) without drawing their graphs.

This is how I do it.

I equate the two equations and get 1-sin(u)=1+sin(u). Simplifying it gives

sin(u)=0. Solving for u gives 0 and pi. Then, the two points are (1 , pi) and (1 , 0).

They are correct. The answer key gives one more answer (0 , pi/2).

How do you find the third answer without graphing? Thanks for your help.
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March 6th, 2017, 07:47 PM   #2
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Math Focus: Calculus/ODEs
You get:

$\displaystyle \sin(u)=-\sin(u)$

which is indeed true for:

$\displaystyle \sin(u)=0$


$\displaystyle u=k\pi$ where $\displaystyle k\in\mathbb{Z}$

But we also need to consider points of intersection that occur for different parameters for the two we could consider the identity:

$\displaystyle \sin(u+(2k-1)\pi)=-\sin(u)=\sin(-u)$


$\displaystyle u+(2k-1)\pi=-u$

$\displaystyle u=\frac{\pi}{2}(1-2k)$

Since the period of the two functions is $2\pi$, we are interesting in:

$\displaystyle 0\le u<2\pi$

$\displaystyle 0\le \frac{\pi}{2}(1-2k)<2\pi$

$\displaystyle 0\le 1-2k<4$

$\displaystyle -1\le -2k<3$

$\displaystyle 1\ge 2k>-3$

$\displaystyle \frac{1}{2}\ge k>-\frac{3}{2}$


$\displaystyle k\in\{-1,0\}$

And so:

$\displaystyle u=\frac{\pi}{2},\,\frac{3\pi}{2}$

We find:

$\displaystyle r_1\left(\frac{\pi}{2}\right)=1-\sin\left(\frac{\pi}{2}\right)=0$

$\displaystyle r_2\left(\frac{3\pi}{2}\right)=1+\sin\left(\frac{3 \pi}{2}\right)=0$

And so the two functions intersect at the origin, but for different values of $u$.

We also find that:

$\displaystyle r_1\left(\frac{3\pi}{2}\right)=1-\sin\left(\frac{3\pi}{2}\right)=2$

$\displaystyle r_2\left(\frac{\pi}{2}\right)= 1+\sin\left(\frac{\pi}{2}\right)=2$

But, we observe that the first function corresponds with the Cartesian point (0,-2) while the second corresponds with (0,2).
Thanks from davedave and romsek

Last edited by MarkFL; March 6th, 2017 at 07:50 PM.
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March 7th, 2017, 09:51 PM   #3
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Thanks a lot, MarkFL.
Thanks from MarkFL
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