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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 $y = y(x), z=z(x)$?

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Probably near all the points, since the intersection is simply a circle moved up one unit in the z direction. – coffeemath Nov 14 '12 at 17:28
Hold on: probably not near $(1,0,1),(-1,0,1)$... – coffeemath Nov 14 '12 at 17:31
up vote 1 down vote accepted

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.

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Just a nitpick: the intersection is the circle $x^2+y^2=1$ lying in the plane $z=1$, not the entire cylinder $x^2+y^2=1$ (as a subset of 3-space). Otherwise this answer is on the mark. +1. – coffeemath Nov 14 '12 at 21:30

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