(Trigonometry) How to convert 2x - 3y = 5 to polar form?
@uri
Do you know Trig?
You're in which grade?
Going to be a Senior.
12th
Oh which group? I mean Commerce?
Sorry, I'm in High School.
Oh okay cool,I'm just 12 so no idea about it. :)
Man... Nobody can answer this...
@jim_thompson5910 Can :D
@jim_thompson5910
:)
hint: x = r*cos(theta) y = r*sin(theta) make sure you solve for r once you've made all the substitutions
I got as far as 5/r=2cos(theta)-3sin(theta), but I think that it is wrong...
2x - 3y = 5 2*r*cos(theta) - 3*r*sin(theta) = 5 r*(2*cos(theta) - 3*sin(theta)) = 5 r = 5/(2*cos(theta) - 3*sin(theta)) so you were (kinda) close
not sure how you got 5/r though
For this one, would it be r^2=sqrt(10)/2 ?
what do you mean
x2 + y2 = 10 My bad... That is the starting equation...
Oh, but it would be jut r=sqrt(10)/2
Right?
oh this is a circled centered at (0,0) with radius sqrt(10)
so it's just r = sqrt(10) since r is the radius
Oh! I see. I think that I am over-complicating this...
possibly, but that's what polar form is essentially: each point in the coordinate space is defined by the angle and how far it is from the center think of it like you're on a navy battleship and you are looking at the radar
and you can think of the angle as the direction you need to look in to see the point
I think I understand. So for x^2 - y^2 = 4 (You may leave your answer for this one as r2) It is r^2=sqrt(-2), and that is it, right?
hmm no that's a bit off
*sigh* I am bad at these...
x^2 - y^2 = 4 (r*cos(theta))^2 - (r*sin(theta))^2 = 4 r^2*cos^2(theta) - r^2*sin^2(theta) = 4 r^2(cos^2(theta) - sin^2(theta)) = 4 r^2*cos(2theta) = 4 r^2 = 4/cos(2theta) r = sqrt( 4/cos(2theta) )
its ok, just keep practicing
All right, well thanks bunches. You gave me a couple examples to go off of :)
you're welcome
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