# Thread: Polar equation exercise

1. Convert to cartesian the following polar equation:

r = asec(θ)

If you can solve this, could you also explain your answer to me?  2.

3. Originally Posted by Arcosseno Convert to cartesian the following polar equation:

r = asec(θ)

If you can solve this, could you also explain your answer to me?
Forum policy does not permit us to do homework for students. We can help if you get stuck, though. What have you done to try to solve this?  4. I wonder why?

I tried using my knowledge of cartesian equations (which is high school level) but I know nothing of polar equations, so I believe the problem is there.  5. Can give you hints though, what is the relationship between x, y, r and theta? And how would you write sec as a relationship using cos or sin.  6. That helped, but not enough. I'll have to find other texts on polar equations. I know r² = x² + y², but I can't see the bigger picture with what was presented to me (a single book)  7. Updating on the situation, I found out that x = rcosθ and y = rsenθ; and that secθ= 1/cosθ. But this was still not enough to find my answer. I also read on the Homework topic that updating on my situation was a good thing to do; so, that's what I'm doing.  8. You pretty much have all the parts you need now, r = a/cos(theta) is the same as r*cos(theta) = a. Do we have a different label for r*cos(theta)?  9. Use r=sqrt(x^2+y^2) and cos(theta)=x/sqrt(x^2+y^2).  10. @mathman you can use LaTeX to write those equations in here. @river_rat I'll try now, even though I still don't have an answer for y. Gonna find it, hopefully.  11. I have r = xsecθ, but this still isn't a cartesian equation. Any ideas of how to find y? Thank you for your help. I tried liking your post but it won't let me.  12. Originally Posted by Arcosseno I have r = xsecθ, but this still isn't a cartesian equation. Any ideas of how to find y? Thank you for your help. I tried liking your post but it won't let me.
Not sure how you got there, r*cos(theta) = a is the same as x = a. so a vertical line through a.  13. @arcosseno I am thinking on what you would do if you were asked something about toroidal and spherical coordinates...  14. @river_rat coordinates always (x,0)? I mean, even if it could be drawn that way, r is still xsecθ, secθ would probably define a y.

@unknown_artist I don't know yet  15. Originally Posted by Arcosseno @river_rat coordinates always (x,0)? I mean, even if it could be drawn that way, r is still xsecθ, secθ would probably define a y.
I'm not sure what you are trying to convey here sorry.  16. You said the function would be a vertical line through a, right? but there is a secθ in the way of that.  17. Originally Posted by Arcosseno @mathman you can use LaTeX to write those equations in here. @river_rat I'll try now, even though I still don't have an answer for y. Gonna find it, hopefully.
By trying all possibilities I knew of, I found how latex works here - the key is [tex]. My original post becomes: Use and .  18. Originally Posted by Arcosseno You said the function would be a vertical line through a, right? but there is a secθ in the way of that.
I think there is an error or two in your algebra, recheck that.  19. Start with . Substitute the expressions I previously posted and get or .  20. You think the answer is ?  21. is the answer.  22. If then y = what? The Question stands.  23. Originally Posted by Arcosseno If then y = what? The Question stands.
What is the equation for a vertical line on the x-y plane?  24. Originally Posted by Arcosseno If then y = what? The Question stands.
y=anything. It doesn't matter, since completely describes the function.  Bookmarks
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