# 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.

By the way, what exactly is contour map? Please help me sketch that one too.

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

-
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

## 2 Answers

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.

-
@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

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