Graphing 2 variable functions

Question

I have a few questions where they ask to sketch 2 variable functions.

I am just wondering if there any specific tricks or calculations to do when solving problems like this instead of just pure logical thinking. It just doesn't feel like I am doing this correctly when there are no calculations involved.

$$ f(x,y) = 4-x^2-y^2 $$ This is how I figured it out the answer. It's an upside down paraboloid that starts at $z = 4$.

Steps of execution I did.

1. Think of $x^2$ and it's function.
2. Think of $x^2+y^2$ and how it is the same but in 3D.
3. Think of $-x^2$ and it turns the function upside down.
4. Think of $-x^2-y^2$. It again must turn the function upside down.
5. Think of $4 + x^2$ and how it is the same function but starts at four.
6. Putting it altogether. It is an upside down paraboloid starting at $z=4$.

Answer 1

Some alternative approach:

In a 3D plot you would plot all points $(x,y,z)$ which fulfill $$ z = f(x,y) = 4 - x^2 - y^2 $$ as points of the the graph of $f$.

One way to structure this graph is to determine the level curves $f(x,y) = \text{const}$, where $z$ has the same value along that curve.

One interesting candidate is the curve for $z = 0$. So we check and get $$ 0 = z = f(x,y) = 4 - x^2 - y^2 \iff \\ x^2 + y^2 = 2^2 $$ which is a circle with radius $2$ and center $(0,0,0)$ in the $x$-$y$-plane.

As it turns out for each constant $c$ we get $$ z = f(x,y) = 4 - x^2 - y^2 \iff \\ x^2 + y^2 = (\sqrt{4 - z})^2 = r^2 $$ so we end up with a circle of radius $r = \sqrt{4 - z}$ and center $(0,0,z)$ and this is only possible for the radikand $4 - z \ge 0 \iff 4 \ge z \iff z \le 4$.

This means $f(x,y) \le 4$ and the graph is rotational symmetric around the $z$-axis.

The image above shows the graph of $f$ (red) and the level curves for $z = 0$ (yellow) and $z = 2$ (green).

Answer 2

Let us work with the equation $z=4-x^2-y^2$.

Instead of treating separately $x$ and $y$, you can take advantage of the fact that, with $x^2+y^2=r^2$, the equation becomes $z=4-r^2$ and can be considered/drawn in a sectional plane as a "generator" parabola, with the meaning that by revolving this sectional plane around $z$-axis, one generates the paraboloid as a volume "of revolution" (this parabola is a "template form' for the paraboloid).

Answer 3

I always find it helpful to think of how the graph looks in certain planes.

So we have the equation $z=4-x^2-y^2$. When $z=0$, then in the $xy$ plane we have $x^2+y^2=4$, i.e. a circle centered at the origin with radius two. Notice for a value of $z$, say $z=c$, we have $x^2+y^2=4-c$ so as $c$ increases, our circle shrinks...when $c=4$ our circle vanishes into a point.

Consider what happens in the $xz$ plane when $y=0$. Then we have $z = 4-x^2$ which we know is an upside down parabola with vertex $(x,z) = (0,4)$. We get a similar parabola in the $yz$ plane when we consider $x=0$ and we get $z = 4-y^2$.

Plotting the circle and the two parabolas in the respective planes should give you an idea of what the plot looks like.

Answer 4

You can study functions of two variables by "freezing" one of them.

The curves

$$g(x)=4-Y^2-x^2$$ are upside down parabolas with the apex at $x=0,g(0)=4-Y^2$, and belong to the planes $y=Y$.

No by varying $Y$, you indeed obtain a paraboloid by moving the parabola on the apex curve, which is also a parabola.

Answer 5

These are my lecture notes on Classification of Quadratic surfaces

• If one variable is missing,then it is a cylinder.(In fact, a cylinder along the missing variable.)
  • $ax^2+by^2+cz^2=d$---(1)

If $a,b,c>0$,then (1) will represent an Ellipsoid.

• $ax^2+by^2+cz^2=d$---(2)

If $a,b>0$ & $c<0$,then the following cases will arise

If $d=0$,then (2) will represent an Elliptic Cone.

If $d>0$,then (2) will represent an Hyperboloid of one sheet.

If $d<0$,then (2) will represent an Hyperboloid of two sheets.

• $ax^2+by^2 \pm cz=d$---(3)

If $a,b>0$ & $c \in \mathbb R$,then (3) will represent an Ellptic Paraboloid.

• $ax^2+by^2 \pm cz=d$---(4)

If $a>0$,$b<0$ & $c \in \mathbb R$,then (3) will represent a Hyperbolic Paraboloid.

From (3),$z=4-x^2-y^2 \implies x^2+y^2+(z-4)=0$ represents an Elliptic paraboloid with its vertex on $z=4$.

Hope this will help you.