# Could someone help me with cylindrical and rectangular conversions?

#### 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

Homework Helper
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 im still stuck since i still have the 4xy on the other side of the equation :(

#### matt grime

Homework Helper
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? doesnt seem right.

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#### FabioTTT

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

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

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#### FabioTTT

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

#### HallsofIvy

Homework Helper
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 sin&theta; and y= r cos&theta;, x^2= r^2 sin^2&theta;, y^2= r^2 cos^2&theta; so x^2- y^2= r^2(cos^2&theta;- sin^2&theta;), and 4xy= 4r^2 cos&theta; sin&theta;. Now the "r^2" terms cancel leaving z^2(cos^2&theta;-sin^2&theta;)= 4sin&theta;cos&theta; or

z^2= (4sin&theta;cos&theta;)/(cos^2&theta;- sin^2&theta;).

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

z^2= 2sin(2&theta;)/cos(2&theta;)

#### 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 sin&theta; and y= r cos&theta;, x^2= r^2 sin^2&theta;, y^2= r^2 cos^2&theta; so x^2- y^2= r^2(cos^2&theta;- sin^2&theta;), and 4xy= 4r^2 cos&theta; sin&theta;. Now the "r^2" terms cancel leaving z^2(cos^2&theta;-sin^2&theta;)= 4sin&theta;cos&theta; or

z^2= (4sin&theta;cos&theta;)/(cos^2&theta;- sin^2&theta;).

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

z^2= 2sin(2&theta;)/cos(2&theta;)
yep thats what i got. thanks