# Multivariable Calc Image-Graph Problem.

• Nov 14th 2010, 07:53 AM
r2d2
Multivariable Calc Image-Graph Problem.
1. The problem statement, all variables and given/known data

Given a function f: R-R2 , by f(t) = (t, t^2 - cos(t)), which represents a curve in the xy plane parametrically, give a function whose GRAPH represents this same curve.

2) Also, give a function h whose level set for height k=0 represents this same curve.

3. The attempt at a solution

From what I know, the graph of this function would be in R3. How would I make a function of a graph from that function? I'm probably missing an easy point but haven't picked up on it yet.

For number 2, I'm pretty confused on where to begin this problem.

Any thoughts?

Thanks a ton!
• Nov 14th 2010, 09:00 AM
roninpro
The question is asking you to deparameterize the curve. Since $x=t$ and $y=t^2-\cos(t)$, a substitution gives $f(x)=x^2-\cos(x)$.
• Nov 14th 2010, 09:11 AM
r2d2
I don't know if it's asking me to deparameterize it, but I think i found the answer to this qestion.

If I am finding the graph, then the equation for the graph of f(t) would be: (t,t, t^2-cos(t))
• Nov 14th 2010, 09:46 AM
roninpro
Quote:

Originally Posted by r2d2
I don't know if it's asking me to deparameterize it, but I think i found the answer to this qestion.

If I am finding the graph, then the equation for the graph of f(t) would be: (t,t, t^2-cos(t))

This actually makes no sense. Your function $f$ is a map from $\mathbb{R}$ to $\mathbb{R}^2$. In other words, $f$ gives some sort of curve on the plane. You all of a sudden gave a curve in space.

Try actually graphing the function I mentioned and compare it to your curve.
• Nov 14th 2010, 11:12 AM
r2d2
The graph of f is in R3, correct?

So wouldn't there be three points for f(t)?

If so, what would be those three points? That is what I believe I solved.
Is that correct
• Nov 14th 2010, 11:15 AM
HallsofIvy
Quote:

Originally Posted by roninpro
This actually makes no sense. Your function $f$ is a map from $\mathbb{R}$ to $\mathbb{R}^2$. In other words, $f$ gives some sort of curve on the plane. You all of a sudden gave a curve in space.

Try actually graphing the function I mentioned and compare it to your curve.

Yes, that is what is intended. The "graph" of f(x), a function on variable is a set of points in $R^2$, (x, f(x)). If f(t)= (x(t), y(t)) then its graph is the three dimensional set (t, x(t), y(t)).