# Sketching graph in R3

• Dec 16th 2010, 12:30 AM
SyNtHeSiS
Sketching graph in R3
Let $\displaystyle f(x,y) = x^{1/3}y^{1/3}$, and let C be the curve of intersection of $\displaystyle z = f(x,y)$ with the plane $\displaystyle y = x$.

Sketch the graph of the curve C.

I know how to graph y = x in R3, but I am not sure how to graph z = f(x,y). Would you use the same steps as you would when sketching a graph in R2? (e.g. first, second derivatives, asymptotes etc).
• Dec 16th 2010, 01:08 AM
FernandoRevilla
Quote:

Originally Posted by SyNtHeSiS
Let $\displaystyle f(x,y) = x^{1/3}y^{1/3}$, and let C be the curve of intersection of $\displaystyle z = f(x,y)$ with the plane $\displaystyle y = x$. Sketch the graph of the curve C.

The intersection $\displaystyle S$ is:

$\displaystyle S=\{(x,x,\sqrt[3]{x^2}):\;x\in \mathbb{R}\}\subset \mathbb{R}^3$

You can determine the position of $\displaystyle z=\sqrt[3]{x^2}$ drawing the curve, as usual.

Fernando Revilla
• Dec 16th 2010, 04:08 AM
SyNtHeSiS
Thanks. I tried checking the graph on wolfram alpha and I dont get why when x < 0 the graph (real part) is negative. I mean if you let x = -2 in z = x^{2/3} you get a positive value.