Question

Rewrite the given equation using the substitutions x=rcos(theta) & y=rsin(theta). Simplify.
x^2 + y^2 + 3x = 0

I know exactly how to solve it but what do the r's mean? :/

Answer 1

So how would I solve? I thought I could simplify rcos(theta)^2 + rsin(theta)^2 to 1. Can I still?

Answer 2

nope, you just replace x, y into the original one. and solve it. that's it

Answer 3

Write it like this!
\[r^2 (\sin \theta)^2 + r^2 (\cos \theta)^2 + 3r \sin \theta = 0\]
Or
\[r^2 + 3r \sin \theta = 0\]
The r here is actually a way of representing a complex number in short!
Have you learnt imaginary numbers? :)

Answer 4

the last term is 3x, not 3y friend

Answer 5

My bad! :)
Write it like this then @aodfuj2
r2+3rcosθ=0
Got it? :)

Answer 6

so how did you get r2 from sin(theta)2 + cos(theta)2? Sorry I'm trying to understand for my next problems.

Answer 7

x = rcosθ
So
x2 = (rcosθ)^2 = r^2 . cos^2 θ
Got it?
And then take r^2 common and use sin^2 θ + cos^2 θ = 1
Understood? :)

Answer 8

Got it. Thank you!

Answer 9

:)