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I am learning pre-calculus and I am not able to answer this question from the textbook:

Find the rectangular coordinates of all the points of intersection of the two polar curves ${\sqrt 3}\sin\theta=r$ and $\cos\theta=r$

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3 Answers 3

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Dealing with intersections of polar curves can be tricky. It would be awesome if we could just set the two equations equal to each other and solve for $\theta$. $$\sqrt3\sin\theta=\cos\theta\\ \tan\theta=\frac1{\sqrt 3}\\ \theta=\frac\pi6,\frac{7\pi}6\\ (r,\theta)=\left(\frac{\sqrt3}2,\frac\pi6\right),\left(\frac{-\sqrt3}2,\frac{7\pi}6\right)\\ (x,y)=\left(\frac34,\frac{\sqrt3}{4}\right)$$

because "both" of those polar coordinates are the same point (RED FLAG $1$). But let's double-check our work by graphing both of those polar curves in Desmos:

enter image description here

The curves also meet at the origin. Why didn't we catch that? Because the first curve goes through the origin at $(0,0)$ and $(0,\pi)$ and the second curve goes through the origin at $(0,\frac\pi2)$ and $(0,\frac{3\pi}2)$, and those are all the same point in polar coordinates (RED FLAG $2$).

The moral of the story is to always graph your polar curves to make sure that you're catching all of the intersection points.

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  • $\begingroup$ Can you please explain how did you calculated r, x and y? I was able only to get the angle θ. Sorry, I am beginner :) $\endgroup$
    – Dani Che
    Commented Sep 8, 2019 at 16:13
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    $\begingroup$ @MilanChe $r$ came from plugging $\theta$ into the two original curves, and your textbook probably talks about how $x=r\cos\theta$ and $y=4r\sin\theta$. $\endgroup$
    – user694818
    Commented Sep 8, 2019 at 19:44
  • $\begingroup$ Thank you very much :) $\endgroup$
    – Dani Che
    Commented Sep 9, 2019 at 8:06
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Use that $$\tan(\theta)=\frac{1}{\sqrt{3}}$$

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Hint: If the two of them intersect, then there is some $r,\theta$ satisfying the two equations simultaneously.

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    $\begingroup$ Or. unfortunately, some $(r,\theta),(r',\theta')$ satisfying $r'=-r$ and $\theta'=\theta+\pi$. $\endgroup$
    – user694818
    Commented Sep 8, 2019 at 13:08
  • $\begingroup$ @MatthewDaly Unfortunately. But if they find $r,\theta$ as described in my answer, they should then be able to generate such other pairs, no? $\endgroup$
    – Allawonder
    Commented Sep 8, 2019 at 13:28
  • $\begingroup$ Polar curves were a long time ago for me, but I recall that they got pathological as all heck. Not only do you need to check for negative radii, but you also have to do a separate check to see if the origin is in both graphs, because then you could have $r=0$ for any value of $\theta$. In the end, we always wound up just graphing the darned things to double-check that we didn't miss any intersections. Fortunately for this generation, there is Desmos. ^_^ $\endgroup$
    – user694818
    Commented Sep 8, 2019 at 13:49

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