Quick Question about Converting Polar cordinates

[Quatros]

Homework Statement

I'm suppose to convert Sqrt[12x-2x^2] into a polar equation.

Homework Equations

The Attempt at a Solution

I went from that equation to r(sin(theta)^2 + 2cos(theta)^2)= 12cos(theta), I really don't know where to go from there.

 

[berkeman]

Homework Statement

I'm suppose to convert Sqrt[12x-2x^2] into a polar equation.

Homework Equations

The Attempt at a Solution

I went from that equation to r(sin(theta)^2 + 2cos(theta)^2)= 12cos(theta), I really don't know where to go from there.

Rectangular to polar conversion usually would involve a 2-dimensional quantity. Can you post an image of the full question along with any figure?

 

[Charles Link]

Could you please show a complete equation in part 1. The problem statement is incomplete.

 

[Quatros]

All I got was to convert y = sqrt[12x-2x^2] to polar form.

 

[Charles Link]

When you square both sides, you get $r^2$. You can let $sin^2(\theta)+cos^2(\theta)=1$, leaving one $r^2 cos^2(\theta)$. $\\$ Additional comment=the equation $y^2=12x-2x^2$ looks like an ellipse that is translated in the x-direction. In this case, it would only be taking the positive values of $y$. $\\$ Additional note: Except for simple graphs, polar coordinate expressions can often be somewhat clumsy to work with.

 

[Quatros]

When you square both sides, you get $r^2$. You can let $sin^2(\theta)+cos^2(\theta)=1$, leaving one $r^2 cos^2(\theta)$. $\\$ Additional comment=the equation $y^2=12x-2x^2$ looks like an ellipse that is translated in the x-direction. In this case, it would only be taking the positive values of $y$. $\\$ Additional note: Except for simple graphs, polar coordinate expressions can often be somewhat clumsy to work with.

I'm a bit confused,
I went from y^2+2x^2 = 12x

then I converted to

r^2cos(theta) ^2- 2r^2sin(theta ) = 12rsin(theta)

r^2(cos(theta)^2-2r^2sin(theta )) = r(12sin(theta)

 

[Charles Link]

$x=r \cos(\theta)$. You have it reversed.

 

[Mark44]

I'm a bit confused,
I went from y^2+2x^2 = 12x

then I converted to

r^2cos(theta) ^2- 2r^2sin(theta ) = 12rsin(theta)

This is incorrect. The relationship is $y = r\sin(\theta)$ and $x = r\cos(\theta)$, not the other way around, as you have it.
Use the advice that Charles Link is giving.

r^2(cos(theta)^2-2r^2sin(theta )) = r(12sin(theta)

 

[Quatros]

$x=r \cos(\theta)$. You have it reversed.

r^2(sin(theta)^2-2r^2cos(theta )^2) = r(12cos(theta))
12cos(theta) = (sin(theta)^2 +2cos(theta)^2)r ,

which is where i get stuck.

 

[Charles Link]

r^2(sin(theta)^2-2r^2cos(theta )^2) = r(12cos(theta))
12cos(theta) = (sin(theta)^2 +2cos(theta)^2)r ,

which is where i get stuck.

Can you see that $sin^2(\theta)+2 cos^2(\theta)=[sin^2(\theta)+cos^2(\theta)]+cos^2(\theta)$? That should be obvious.

 

[Quatros]

Ohhh, I see then i get 12cos(theta) = 1+ rcos^2(theta), then i just solve for r?

 

[Charles Link]

Ohhh, I see then i get 12cos(theta) = 1+ rcos^2(theta), then i just solve for r?

$12 \cos(\theta)=r(1+\cos^2(\theta) )$. You can then solve for $r$.