How Do You Convert Cylindrical Equations to Rectangular Form?

[FabioTTT]

I have the following cylindrical equation:

z = r^2 cos(2theta)

I am suppose to convert it into a rectangular equation. I'm stumped.

 

[FabioTTT]

would i be correct to assume the trig property that cos(2theta) = cos^2 theta - sin^2 theta ?

then i'd end up with

r^2 cos^2 theta - r^2 sin^2 theta = z

then substitute x^2 and y^2 to get z = x^2 - y^2 ?

 

[FabioTTT]

no if someone could help me convert this rectangular equation to cylindirical i'd be eternally greatful:

z^2 (x^2 - y^2) = 4xy

 

[matt grime]

You seem to be doing well enough on your own:

you know what x^2-y^2 is from the previous example.

What other trig identities do you know? say, sin(2theta)?

 

[FabioTTT]

Originally posted by matt grime
You seem to be doing well enough on your own:

you know what x^2-y^2 is from the previous example.

What other trig identities do you know? say, sin(2theta)?

yeah i realized i could use cos(2theta) from the previous problem but I am still stuck since i still have the 4xy on the other side of the equation :(

 

[matt grime]

Hope this doesn't appear too RTFM, but you have formulae:
x=rcos(theta)
y=rsin(theta)

so use them as you did in the previous question.

 

[FabioTTT]

i know what youre hinting at that i could divide both sides of the equation by 2 and then sub x and y for r cos(theta) and r sin(theta) and then cancel the r^2 on both sides and have sin(2theta) left.. but that would leave me with

(z^2 cos(2theta)) / 2 = sin(2theta) is that correct though? doesn't seem right.

 

[FabioTTT]

(z^2) / 2 = tan(2theta)

if sin(2theta) / cos(2theta) = tan(2theta) is valid... is it?

 

[FabioTTT]

im an idiot.. of course its valid.. its just an angle. thanks guys. i guess this thread was pretty pointless ;)

 

[HallsofIvy]

Since I am not particularly bright and have great difficulty remembering trig identities, I would probably change
z^2 (x^2 - y^2) = 4xy into polar coordinates in the obvious way:

Since x= r sinθ and y= r cosθ, x^2= r^2 sin^2θ, y^2= r^2 cos^2θ so x^2- y^2= r^2(cos^2θ- sin^2θ), and 4xy= 4r^2 cosθ sinθ. Now the "r^2" terms cancel leaving z^2(cos^2θ-sin^2θ)= 4sinθcosθ or

z^2= (4sinθcosθ)/(cos^2θ- sin^2θ).

Now IF I were smart I might remember (or look up!) both
"cos(2θ)= cos^2θ- sin^2θ" and
"sin(2θ)= 2sinθcosθ to write that equation as

z^2= 2sin(2θ)/cos(2θ)

 

[FabioTTT]

Originally posted by HallsofIvy
Since I am not particularly bright and have great difficulty remembering trig identities, I would probably change
z^2 (x^2 - y^2) = 4xy into polar coordinates in the obvious way:

Since x= r sinθ and y= r cosθ, x^2= r^2 sin^2θ, y^2= r^2 cos^2θ so x^2- y^2= r^2(cos^2θ- sin^2θ), and 4xy= 4r^2 cosθ sinθ. Now the "r^2" terms cancel leaving z^2(cos^2θ-sin^2θ)= 4sinθcosθ or

z^2= (4sinθcosθ)/(cos^2θ- sin^2θ).

Now IF I were smart I might remember (or look up!) both
"cos(2θ)= cos^2θ- sin^2θ" and
"sin(2θ)= 2sinθcosθ to write that equation as

z^2= 2sin(2θ)/cos(2θ)

yep that's what i got. thanks