Is the Intersection of Two Surfaces a Cylinder or Paraboloid in 3D?

[0kelvin]

I'm given equations of surfaces and asked for the vector function that represents the intersection of the two surfaces.

For ex: $$x^2 + y^2 = 4$$ and $$z = xy$$

In the solutions manual the answer is given like this: a sum of terms of cos t and sin t (is this polar form?). The way I did wasn't using cos t. I isolated y, substituted in z = xy and assumed x = t. Which resulted in something like this r(t) = (... , ... , ...). This form should be the same thing as r(t) = ai + bj + ck.

Do I have to use polar coordinates in this exercise? I believe that I went cartesian coordinates in my solution. (not forgetting that $y = \pm \sqrt{4 - x^2}$)

 

[LCKurtz]

You apparently don't have to use polar coordinates, but they are the obvious best choice because you have a cylinder. You have $x = 2\cos t,~y=2\sin t$ and put $z$ in terms of them and you get a parametric curve.

 

[pasmith]

The usual parametrization of $x^2 + y^2 = a^2$ is $(x,y) = (a\cos t, a \sin t)$. This avoids ambiguities over the signs of square roots.

This parametrization is in cartesian coordinates; in polar coordinates it would be $(r,\theta) = (a, t)$.

 

[0kelvin]

$r(t) = (t, \pm \sqrt{4 - t^2}, \pm t \sqrt{4 - t^2})$ is correct?

 

[0kelvin]

uhm? Are you saying that r(t) means Radius as a function of t? I didn't use "r" as "radius".

Using polar coordinates as in the book, the answer is r(t) = (2 cos t, 2 sin t, 4 cos t sin t). Because radius = 2.

 

[Ray Vickson]

uhm? Are you saying that r(t) means Radius as a function of t? I didn't use "r" as "radius".

Using polar coordinates as in the book, the answer is r(t) = (2 cos t, 2 sin t, 4 cos t sin t). Because radius = 2.

Since $r$ is commonly used as a notation for a polar coordinate, if you mean your $r$ to be a position vector you should embelish the notation and use either $\vec{r}$ or the bold-face form $\bf r$.

 

[0kelvin]

I just realized what is wrong with my solution in cartesian coordinates. If I isolate like this $y = \sqrt{4 - x^2}$, I'm looking at the equation as a circle, not as a cylinder. For it to be a cylinder I have to do this: $x^2 + y^2 - 4 = z$, where z = f(x,y,z). Now it's a surface in 3D.

 

[LCKurtz]

I just realized what is wrong with my solution in cartesian coordinates. If I isolate like this $y = \sqrt{4 - x^2}$, I'm looking at the equation as a circle, not as a cylinder. For it to be a cylinder I have to do this: $x^2 + y^2 - 4 = z$, where z = f(x,y,z). Now it's a surface in 3D.

No, that is absolutely incorrect. That equation isn't a cylinder, it's a paraboloid. $x^2 + y^2 =4$ is a circle in 2D but a cylinder in 3D.