# Thread: Identifying a curve by finding a Cartesian equation for the curve.

1. ## Identifying a curve by finding a Cartesian equation for the curve.

This question is in the Parametric Equations and Polar Coordinates section of Multivariable Calculus, 6th Edition by James Stewart. I have to find a Cartesian equation for the curve:

rcos(theta)=1

and then identify it. I find this section a little confusing.

Thanks for the help.

2. Originally Posted by Undefdisfigure
This question is in the Parametric Equations and Polar Coordinates section of Multivariable Calculus, 6th Edition by James Stewart. I have to find a Cartesian equation for the curve:

rcos(theta)=1

and then identify it. I find this section a little confusing.

Thanks for the help.
confusing? this is one of the easiest problems you get get!

recall, in polar coordinates, $x = r \cos \theta$

so the curve $r \cos \theta = 1$ is the line $x = 1$

3. DOH! I was turning this question into some monstrous equations and it should have been obvious all along. Don't worry, there's more complicated stuff I'm going to have to do and already did which I did okay on.

4. Originally Posted by Undefdisfigure
DOH! I was turning this question into some monstrous equations and it should have been obvious all along. Don't worry, there's more complicated stuff I'm going to have to do and already did which I did okay on.
yeah, sometimes it's the easy ones that allude you, because you're expecting them to be hard

5. As another question, and this one I'm sure is a little trickier (not to mention I think I got the answer). When you have r = 2sin(theta) + 2cos(theta), you have to transform it into Cartesian coordinates and identify the curve. The first question I was thrown off because it asked for a curve so I didn't think a line would qualify.

Anyways in this second question I put cos(theta) = x/r and sin(theta) = y/r.
I then plugged these values in the equation and got r = 2x/r + 2y/r. I eliminated the r and got 2x + 2y = 1. This gave me the equation of a line y = -x + 1/2. And I think this is the answer. Tell me if I'm wrong.

6. Originally Posted by Undefdisfigure
As another question, and this one I'm sure is a little trickier (not to mention I think I got the answer). When you have r = 2sin(theta) + 2cos(theta), you have to transform it into Cartesian coordinates and identify the curve. The first question I was thrown off because it asked for a curve so I didn't think a line would qualify.

Anyways in this second question I put cos(theta) = x/r and sin(theta) = y/r.
I then plugged these values in the equation and got r = 2x/r + 2y/r. I eliminated the r and got 2x + 2y = 1. This gave me the equation of a line y = -x + 1/2. And I think this is the answer. Tell me if I'm wrong.
you're wrong. you cannot "eliminate" the r like that. you method is okay, but a simpler one would be:

$r = 2 \sin \theta + 2 \cos \theta$ ........multiply through by r

$\Rightarrow r^2 = 2 r \sin \theta + 2 r \cos \theta$ ........now switch to Cartesian

$\Rightarrow x^2 + y^2 = 2y + 2x$

now simplify. i believe this is the equation of a circle or an ellipse or something of that sort

7. Originally Posted by Jhevon
you're wrong. you cannot "eliminate" the r like that. you method is okay, but a simpler one would be:

$r = 2 \sin \theta + 2 \cos \theta$ ........multiply through by r

$\Rightarrow r^2 = 2 r \sin \theta + 2 r \cos \theta$ ........now switch to Cartesian

$\Rightarrow x^2 + y^2 = 2y + 2x$

now simplify. i believe this is the equation of a circle or an ellipse or something of that sort
Tsk, tsk, Jhevon. Of course it is a circle:
$x^2 + y^2 = 2y + 2x$

$(x^2 - 2x) + (y^2 - 2y) = 0$

$(x^2 - 2x + 1) + (y^2 - 2y + 1) = 1 + 1$

$(x - 1)^2 + (y - 1)^2 = 2$
which is the standard form for a circle centered on (1, 1) and with a radius of $\sqrt{2}$.

-Dan

8. Originally Posted by topsquark
Tsk, tsk, Jhevon. Of course it is a circle:
$x^2 + y^2 = 2y + 2x$

$(x^2 - 2x) + (y^2 - 2y) = 0$

$(x^2 - 2x + 1) + (y^2 - 2y + 1) = 1 + 1$

$(x - 1)^2 + (y - 1)^2 = 2$
which is the standard form for a circle centered on (1, 1) and with a radius of $\sqrt{2}$.

-Dan
well, excuse me for being lazy. but there have been times when i thought in simplifying stuff that i thought would be a circle, i ended up with an ellipse. i just wanted to be on the safe side.

...well, now that i think about it, i guess we're only in danger of that if the coefficients of x^2 and y^2 are not 1, but oh, well. let's just say i wanted the poster to verify it for me, you know, so he can practice his algebra (i'm so considerate). stop picking on me

i see what Krizalid is talking about now you're just a bully

9. Originally Posted by Jhevon
well, excuse me for being lazy. but there have been times when i thought in simplifying stuff that i thought would be a circle, i ended up with an ellipse. i just wanted to be on the safe side.

...well, now that i think about it, i guess we're only in danger of that if the coefficients of x^2 and y^2 are not 1, but oh, well. stop picking on me

-Dan

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# identify the curve by finding a cartesian equation for the curve r=2 sin theta 2 cos theta

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