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Math Help - Sketching graph in R3

  1. #1
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    Sketching graph in R3

    Let f(x,y) = x^{1/3}y^{1/3}, and let C be the curve of intersection of z = f(x,y) with the plane 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).
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by SyNtHeSiS View Post
    Let f(x,y) = x^{1/3}y^{1/3}, and let C be the curve of intersection of z = f(x,y) with the plane y = x. Sketch the graph of the curve C.
    The intersection S is:

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

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

    Fernando Revilla
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  3. #3
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    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.
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