Question

find a cartesian equation r=2cos(theta) + 3sin(theta ...please help me

Answer 1

\[\large r=2 \cos \theta + 3 \sin \theta\]use \[\large \cos \theta = \frac{ x }{ r },\quad \sin \theta = \frac{ y }{ r }\]

Answer 2

can you do it? for me :((((((((( please help me..

Answer 3

\[\large r=2\frac{ x }{ r }+3\frac{ y }{ r }\]multiply both sides of the equation by r\[\large r^2=2x+3y\]Finally, use \[\large r^2=x^2+y^2\]

Answer 4

is it correct¿? x2+y2 = (2costheta + 3sintheta)2 ------> x2+y2 = 4cos^2theta + 12 sinthetacostheta + 9 cos2theta

Answer 5

x2+y2 = 4x2/r2 + 12*x*y/r2+9y2/r2

Answer 6

(x2+y2)2 = 4x2 + 12xy + 9 y2

Answer 7

is it correct¿?

Answer 8

yes it is. BUT you still need to take the square root of both sides.

Answer 9

\[\large 4x^2+12xy+9y^2=(2x+3y)^2\]

Answer 10

I'm I'm confused with that

Answer 11

if you look at my post, \[\large r^2 = 2x+3y\\\large r^2=x^2+y^2\]you can get rid of the r^2

Answer 12

but the fisrt is correct?