Question

rewrite the equation x^2+y^2+3x=0 using the substitutions x=r cos thetha and y=r sin theta. simplify. please explain the steps!

Answer 1

centre -3/2, 0

radius 3

so

x+ 3/2 = 3cost
y=3sint

Answer 2

have you started out by plugging x=rcos(theta) and y=rsin(theta) into the equation?

what does that give you?

Answer 3

do you have to find the center and radius or is it done a different way?

Answer 4

dont' worry about center. just plug x=rcos(theta) and y=rsin(theat) into x^2+y^2+3x=0.

you'll get a cos^2 and a sin^2 that you can combine to get 1.

try it and see what you get

Answer 5

what do i do about the r?

Answer 6

im sorry i haven't done this in a very long time

Answer 7

\[(r \cos \theta)^{2} + (r \sin \theta)^{2} + 3r \cos \theta = 0\]\[r^{2}(\cos^{2}\theta + \sin^{2}\theta) + 3r \cos \theta = 0\]\[r^{2} + 3r \cos \theta = 0\]\[r(r + 3 \cos \theta) = 0\]

Answer 8

thank you so much

Answer 9

here...i'll get you started. :)

abtrehearn got you most of the way..

realistically, you wouldn't have r=0 since the radius can't be 0 for a circle.

Answer 10

so you could divide through by r.

Answer 11

to get r+3cos\[\theta\]=0?

Answer 12

sorry i messed up the format

Answer 13

True. r = 0 is just a point. It does, however, lie on the graph of the original equation, \[r = -3 \cos \theta\]is the complete answer.