# Problem sketching the surface after figuring out minima

I have a firm understanding when it comes to 2-D graphs. However 3-D plots/graphs are confusing to me. I know there exists several software packages which neatly does the job. I need to sketch it by hand in order to understand it. I know 3-d graph is an extension to 2-D with an addition of $z$-axis, I am not sure where lies the $x$-axis, where lies the $y$-axis and the $z$-axis. Here's the situation.

Given the surface $f(x,y) = z = x^2 + y^2$. I have been told to determine the nature and sketch the surface after determining the partial derivatives with respect to $x$ and $y$ respectively. I obtained the critical point $(0,0)$. I also could determine that its minimum at that point by evaluating delta. But I'm not sure how to sketch this. Please help.

If you are curious to know from where I am solving this problem, it's from John Bird's higher engineering math, page 359.

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I fixed the TeX for you; hope that's ok. As a general suggestion, I feel a space after a full stop (before beginning a new sentence) enhances readibility. But I refrained from editing the post for this non-mathematical reason. Added: I guess @Willie Wong has done this for you anyway now. :) – Srivatsan Aug 31 '11 at 12:38

When you want to understand what a function like $f(x,y)$ does, it helps to think about special cases first: How does it look for $x=0$ ? for $y=0$? For $x=y$? In most cases, you can just interpolate from there and you will get a good impression of what's happening.

Here, I you can also introduce polar coordinates to get a nicer version of the function. It then writes like: $f(r,\phi)= r^2$. You now recognize that you have a symmetry when rotating along z-axis. Use the symmetry to draw your graph.

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@alok If you are wondering where the polar coordinates comes from (as you should be :-)), plotting the contour map of $z=x^2+y^2$ will give you one motivation. – Srivatsan Aug 31 '11 at 12:48

Note that another way to write $x^2 + y^2$ is $x^2 + y^2 = \left(\sqrt{x^2 + y^2}\right)^2 = ||(x,y)||^2$, where $||\cdot||$ is the Euclidean norm, while an equation of the form $||(x,y)|| = C$ describes a circle in $\mathbb{R}^2$ with radius $C$. So for fixed $z = z_0 \geq 0$, we have that $x^2 + y^2 = ||(x,y)||^2 = z_0$, or $||(x,y)|| = \sqrt{z_0}$. So at any fixed height $z = z_0$ on the $z$-axis, the points $(x,y)$ satisfying the equation are the points on a circle with radius $\sqrt{z_0}$. For increasing $z$, the radius of the circle increases as $\sqrt{z}$. So what you get is similar to a cone, except for the fact that the width of the cone now grows as $\sqrt{z}$ instead of linear in $z$.

Finally, you can always try verifying your plot with Wolfram|Alpha, which is a great tool. If we give it the following Mathematica-code

ContourPlot3D[x^2 + y^2 = z, {x, -10, 10}, {y, -10, 10}, {z, 0, 100}]

then it plots the following graph

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...and in fact the surface is what's termed as a "paraboloid of revolution". – J. M. Aug 31 '11 at 13:22