Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

1 Answer 1

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.