Looking for help in understanding a solution to a Calc III problem about surfaces

Question

Studying for a Calc III midterm and I'm trying to shore up my intuitions. The question I'm looking at asks:

Show that the curve with parametric equations $x = sin (t)$, $y = cos (t)$, $z = sin^2 (t)$ is the curve of intersection of the surfaces $z = x^2$ and $x^2 + y^2 = 1$.

Now I have the solution to the above problem but I'd like to understand why its the right answer.

The solution takes the equations of the two surfaces and sets the equal to each other, as both equations equal 0 with some algebraic manipulation. The resultant equation is:

$2x^2+y^2-z\:=\:1$

Straightforward enough. My first thought is that this equation represents a surface in 3d space, no? I believe its a parabolid. So this is currently the surface that represents the intersection of the two surfaces. But the question gives us a parametric equation of a curve (as an aside, are parametric equations typically used to describe surfaces, or only curves/lines?), so obviously we have more work to do. Is the curve basically the outline of the surface or am I missing something in my visualization of what this should look like?

The parametric equations can be subbed in for x, y, and z and put into the equation for the surface intersection given above (the $2x^2+y^2-z\:=\:1$ equation). After some more algebra, the resulting statement is found:

$sin^2\left(t\right)+cos^2\left(t\right)=1$

So this is a well know trig identity, right? I'm not sure how this shows that the given parametric equations are a curve of the intersection of the two equations above. Could someone unpack this for me a bit?

Answer 1

It's helpful to visualize what the surfaces you are to trying to intersect. In this case, you have two "cylinders". One is perpedicular to the xy plane, while the other is perpendicular to xz plane. So when you tried intersecting these surfaces, and obtain $$2x^2+y^2-z=1,$$ you have to keep in mind that this intersection is now a curve not a surface. And to set that in stone, you introduce a parametrization for the curve. Set $x=\sin(t) $ and $y=\cos(t)$ (as you said), then solve for $z=2x^2+y^2-1=2\sin^2(t)+\cos^2(t)-1=\sin^2(t)$.

Answer 2

Determining if an equation or equations defines a curve or surface can be tricky without visualizing them. To help with this I introduce the "dimension formula" in my calc. III course:

$$( \text{dimension of the object}) = (\text{dimension of the space}) - (\text{number of equations})$$

The two equations $z = x^2$ and $x^2 + y^2 = 1$ in $\mathbb{R}^3$ give:

$$1 = 3 - 2$$

dimension 1 objects are curves. You can also use the "dimension formula" to see why $x^2 + y^2 = 1$ is a circle (curve/dimension 1) in $\mathbb{R}^2$ but a surface (dimension 2) in $\mathbb{R}^3$. I hope this helps your intuition.

For parametric equations there's another version of the "dimension formula":

$$( \text{dimension of the object}) = (\text{number of parameters})$$