Here's the three part question:
A) Find parametric equations for curve which is the intersection of the cylinder $x^2 + z^2 = 1$ and the plane y = -x.
B) Show that the curve lies on the surface $x^2 + y^2 + 2z^2 = 2$.
C) What is the shape of the curve?
What have I done so far?
For (A), should I set the two equations equal to find the curve's equation?
For (B), I believe once I have the curve equation, I can enter the value of x, y, and z into the given equation in (B) to see if it's equal to 2. If it is, it lies on the surface.
For (C), I assume graphing the curve equation would give me the general shape…
Best Answer
A) The relation $x^2+z^2=1$ calls for the substitution $x=\cos t, z=\sin t$. Then $y=-x$ turns to $y=-\cos t$.
B) Indeed, $\cos^2t+(-\cos t)^2+2\sin^2t=2$. But you can establish this independently: $x^2+y^2+2z^2=x^2+(-x)^2+2z^2=2(x^2+z^2)=2$. The surface in question is an ellipsoid of revolution.
C) When projected onto $xz$, the curve is the unit circle. When projected onto $yz$, the curve is the unit circle. And when projected onto $xy$, the curve is a line segment with direction $-45°$. As is well known, the intersection of a cylinder and a plane is an ellipse (axes $1$ and $\sqrt2$).
You can check all of this by rotating space around the axis $z$ by $45°$, with the transform $x=\frac1{\sqrt2}(u+v), y=\frac1{\sqrt2}(u-v), w=z$.
The cylinder and the plane become $\frac{(u+v)^2}2+w^2=1$ and $u=0$, so that $\frac{v^2}2+w^2=1$. This is the implicit equation of an ellipse centered at the origin, with axis $\sqrt2$ and $1$.
Also, $u=\frac{x+y}{\sqrt2}=0, v=\frac{x-y}{\sqrt2}=\sqrt2\cos t, w=\sin t$, which is the parametric equation of an ellipse.