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March 6th, 2017, 06:59 PM  #1 
Senior Member Joined: Apr 2008 Posts: 194 Thanks: 3  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=1sin(u) and r=1+sin(u) without drawing their graphs. This is how I do it. I equate the two equations and get 1sin(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. 
March 6th, 2017, 07:47 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 
You get: $\displaystyle \sin(u)=\sin(u)$ which is indeed true for: $\displaystyle \sin(u)=0$ or $\displaystyle u=k\pi$ where $\displaystyle k\in\mathbb{Z}$ But we also need to consider points of intersection that occur for different parameters for the two functions...so we could consider the identity: $\displaystyle \sin(u+(2k1)\pi)=\sin(u)=\sin(u)$ Hence: $\displaystyle u+(2k1)\pi=u$ $\displaystyle u=\frac{\pi}{2}(12k)$ Since the period of the two functions is $2\pi$, we are interesting in: $\displaystyle 0\le u<2\pi$ $\displaystyle 0\le \frac{\pi}{2}(12k)<2\pi$ $\displaystyle 0\le 12k<4$ $\displaystyle 1\le 2k<3$ $\displaystyle 1\ge 2k>3$ $\displaystyle \frac{1}{2}\ge k>\frac{3}{2}$ Thus: $\displaystyle k\in\{1,0\}$ 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). Last edited by MarkFL; March 6th, 2017 at 07:50 PM. 
March 7th, 2017, 09:51 PM  #3 
Senior Member Joined: Apr 2008 Posts: 194 Thanks: 3 
Thanks a lot, MarkFL.


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