Implicit function theorem => continuously differentiable functions

Question

Consider the curve in $R^3$ consisting of the intersection of the paraboloid $z=x^2 + y^2$ and the cylinder $x^2 + y^2 = 1$. Near which points of this curve does the implicit function theorem say we can write the curves as the graph of continuously differentiable functions $y = y(x), z=z(x)$?

Answer 1

First, the intersection is just $x^2+y^2=1$. So, $z$ is the constant function $1$. Second, $y$ is a function of $x$ precisely where $\frac{dy}{dx}$ exists. Using implicit differentiation, we land at $$ \frac{dy}{dx}=-\frac{x}{y}. $$ So, $y$ is a function of $x$ for $y\neq 0$.

The way to think about this is to draw the circle $x^2+y^2=1$ and observe around which points on the curve have an area around them that passes the vertical line test.