Question

Need help transforming this polar coordinate:
r^2=4cos(theta)
into a parametric equation.

Answer 1

\[r=\sqrt{x^2+y^2}\]

Answer 2

\[x=rcos(\theta)\]

Answer 3

multiply both sides by r that might elp

Answer 4

\[r^3=4rcos(\theta)\]
\[r^3=4x\]

Answer 5

Oh I see you multiply each side by r...I guess I don't understand that part...

Answer 6

it has to do with getting an r on the right side so you can simplify

Answer 7

it or change it to cartesian

Answer 8

o wait parametric hold on

Answer 9

I just need to transform this equation so that I can graph it on a cartesian plane. I may have called it by the wrong name...

Answer 10

yeah that's polar to cartesian so this is what ou need to know

\[y=rsin(\theta)\]
\[x=rcos(\theta)\]
\[r=\sqrt{x^2+y^2}\]

Answer 11

since you have

\[r^2=4cos(\theta)\]

you have no way of changing the right side because it has no r in the equation itself so by multiplying both sides youget

\[r^3=4rcos(\theta)\]

\[(\sqrt{x^2+y^2})^3=4x\]

Answer 12

you write the left side as a exponential form

\[(x^2+y^2)^{3/2}=4x\]

Answer 13

now we want to get rid of the exponent by you can do that by multiply both sides by the reciprocal 3/2

\[x^2+y^2=(4x)^{3/2}\]

Answer 14

Okay, cool thanks! I understand now :)

Answer 15

nice! @Outkast3r09 but I have to point out the typo\[(x^2+y^2)^{3/2}=4x\implies x^2+y^2=(4x)^{2/3}\](exponent is reciprocal)

Answer 16

And then graph, I solve for y, right.
So, this would end up being...
y=\[\sqrt{(4x) ^{2/3}-x ^{2}}\]

Answer 17

I don't see why you need it solved for y, but if you did you would have to take \(\pm\) upon taking the sqrt

you can also factor out an x^2\[y=\pm2x\sqrt{4x^{1/3}-1}\]but that seems unnecessary

Answer 18

typo*\[y=\pm x\sqrt{(4x)^{1/3}-1}\]