**level curves**of a function \(f\) of two variables are the curves with equations \(f(x,y) = k\), where \(k\) is a constant (in the range of \(f\)).

**contour map**of a function \(f\) is a graph consisting of several level curves of \(f\).

Find the domain and range of \(g(x,y) = \sqrt{16-x^2-y^2}\)

Recall that \(\sqrt{u}\) has domain \(u \geq 0\). So that the domain of \(g(x,y)\) is

$$

\begin{aligned}

D & = \{ (x,y) \; | \; 16-x^2-y^2 \geq 0 \} \\

& = \{ (x,y) \; | \; 16 \geq x^2 + y^2 \} \\

\end{aligned}

$$

That is, the domain of \(g(x,y)\) are all ordered pairs that lie inside or on the circle of radius 4 centered at the origin.

The range of \(g(x,y)\) is

$$

\left \lbrace z \; | \; z = \sqrt{16-x^2-y^2}, \; (x,y) \in D \right \rbrace

$$

The maximum value of \(z\) is \(4\) and occurs at \((0,0)\). The minimum value of \(z\) is \(0\) and occurs at a number of places in the domain: \((4,0), (0,4), (2,2 \sqrt{3}), \ldots \). Actually, the minimum occurs on the circle of radius 4. \(z\) also takes on every value between \(0\) and \(16\) as the following argument demonstrates:

$$

\begin{aligned}

0 & \leq 16-x^2-y^2 \leq 16 \\

\Rightarrow 0 & \leq \sqrt{16-x^2-y^2} \leq 4 \\

\Rightarrow 0 & \leq z \leq 4

\end{aligned}

$$

So the range is

$$

\{ z \; | \; 0 \leq z \leq 4 \} = [0,4]

$$

**function \(f\) of two variables**is a rule that assigns to each ordered pair of real numbers \((x,y)\) in a set \(D\) a unique real number denoted by \(f(x,y)\). The set \(D\) is the

**domain**of \(f\) and its

**range**is the set of values that \(f\) takes on, that is,

$$

\{ f(x,y) \; | \; (x,y) \in D \}

$$

**graph**of \(f\) is the set of all points \((x,y,z)\) in \(\mathbb{R}^3\) such that \(z= f(x,y)\) and \((x,y)\) is in \(D\).

Graph

$$

z = x^2 + y^2

$$

- Traces in \(z = k\):

If \( z = k\) then \(k = x^2 + y^2\), which is the standard equation of a circle centered at the origin with radius \(\sqrt{k}\) (provided \(k \geq 0\)). If \(k = 0\) then \(x = 0\) and \(y = 0\) which is a point in the \(xy\)-plane. If we graphed \(k = x^2 + y^2\) in the \(xy\)-plane, for increasing values of \(k >0\) then we would see concentric circles centered at the origin going outward. - Traces in \(x =k\):

If \(x = k\) then \(z = y^2 + k^2\), which is the standard equation of a parabola opening upward with vertex at \((0,k^2)\). If \(k = 0\) then \(z = y^2\) which is an equation of a parabola in the \(yz\)-plane opening upward with vertex at \((0,0)\). If we graphed \(z = y^2 +k^2\) in the \(yz\)-plane for increasing values of \(k>0\) we would see parabolas opening upward moving up along the \(z\)-axis. We would not be able to distinguish between the traces \(x= k\) and \(x = -k\), as their graphs would coincide. - Traces in \(y = k\):

If \(y = k\) then \(z = x^2 + k^2\), which is the standard equation of a parabola opening upward with vertex at \((0,k^2)\). If \(k = 0\) then \(z = x^2\) which is an equation of a parabola in the \(xz\)-plane with vertex at \((0,0)\). If we graphed \(z=x^2+k^2\) in the \(xz\)-plane for increasing values of \(k>0\) then we would see parabolas opening upward moving up along the \(z\)-axis. We would not be able to distinguish between the traces \(y= k\) and \(y = -k\), as their graphs would coincide.

The traces in \(z=k, x =k, \) and \(y =k\) help determine the shape of the three-dimensional object defined by the equation \(z = x^2+y^2\). The \(k\) value determines the height at which the trace exists outside of its respective plane. For example, the trace \(z = 0\) is the point \((0,0)\) and since \(k = 0\) this point lies in the \(xy\)-plane whereas the trace \(z = 4\) is the circle centered at the origin with radius 2 and since \(k = 4\) this curve lies four units above, and parallel to, the \(xy\)-plane. Continuing in this way to extend the traces into three space, we conclude that the graph of \(z = x^2 + y^2\) is shaped like an oval cup opening upward. Consequently, \(z = x^2+y^2\) has an absolute minimum value of \(z = 0\) at \((0,0)\).

A three dimensional surface that is shaped like an oval cup is called an **elliptic paraboloid** and can be represented, under suitable choices of coordinate axes, by the equation

$$

\frac{z}{c} = \frac{x^2}{a^2} + \frac{y^2}{b^2}

$$

were \(a,b,\) and \(c\) are non-zero constants.

Therefore, the graph of \(z = x^2 + y^2\) is that of an elliptic paraboloid, where \(a = 1, b=1,\) and \(c = 1\).

Graph

$$

z = y^2-x^2

$$

- Traces in \(x = k\):

If \(x = k\) then \(z = y^2 – k^2\), which is the standard equation of a parabola opening upward with vertex at \((0,-k^2)\). If \(k = 0\) then \(z = y^2\) which is an equation of a parabola opening upward in the \(yz\)-plane with vertex at \((0,0)\). If we graphed \(z = y^2-k^2\) in the \(yz\)-plane for increasing values of \(k>0\) then we would see parabolas opening upward moving down along the \(z\)-axis. We would not be able to distinguish between the traces \(x = k\) and \(x = -k\), as their graphs would coincide. - Traces in \(y =k\):

If \(y = k\) then \(z = -x^2 + k^2\), which is the standard equation of a parabola opening downward with vertex at \((0,k^2)\). If \(k = 0\) then \(z = -x^2\) which is an equation of a parabola in the \(yz\)-plane with vertex at \((0,0)\) opening downward. If we graphed \(z = -x^2+k^2\) in the \(yz\)-plane for increasing values of \(k>0\) then we would see parabolas opening downward moving up along the \(z\)-axis. We would not be able to distinguish between the traces \(y = k\) and \(y = -k\), as their graphs would coincide. - Traces in \(z = k\):

If \(z = k\) then \(k = y^2 – x^2\), which is the standard equation of a hyperbola. If \(k = 0\) then \(y=x\) or \(y=-x\) which are equations of lines going through the origin, the former making an angle of 45 degrees with the positive \(x\)-axis and the latter making an angle of 135 degrees with the positive \(x\)-axis. If \(k>0\) then we have hyperbolas along the \(y\)-axis. If we graphed \(k = y^2-x^2\) in the \(xy\)-plane for increasing values of \(k>0\), then we would see hyperbolas moving outward. If \(k<0\) then we have hyperbolas along the \(x\)-axis. If we graphed \(k = y^2-x^2\) in the \(xy\)-plane for increasing values of \(k<0\), then we would see hyperbolas moving inward.

The traces in \(z=k, x =k, \) and \(y =k\) help determine the shape of the three-dimensional object defined by the equation \(z = y^2-x^2\). The \(k\) value determines the height at which the trace exists outside of its respective plane. For example, the trace \(z = 0\) gives rise to the lines \(y=x\) and \(y=-x\); since \(k = 0\) these lines lie in the \(xy\)-plane whereas the trace \(z = 1\) is a hyperbola opening along the \(y\)-axis; since \(k = 1\) this curve lies one unit above, and parallel to, the \(xy\)-plane. Continuing in this way to extend all traces into three space, we conclude that the graph of \(z = y^2-x^2\) is shaped a saddle. Consequently, \(z = y^2-x^2\) has no minimum and no maximum values.

A three dimensional surface that is doubly ruled and shaped like a saddle is called a **hyperbolic paraboloid** and can be represented, under suitable choices of coordinate axes, by the equation

$$

\frac{z}{c} = \frac{y^2}{b^2}-\frac{x^2}{a^2}

$$

were \(a,b,\) and \(c\) are non-zero constants.

Therefore, the graph of \(z=y^2-x^2\) is that of a hyperbolic paraboloid, where \(a = 1, b=1,\) and \(c = 1\).

Graph

$$

x^2 + \frac{y^2}{9} + \frac{z^2}{4} = 1

$$

- Traces in \(z = k\):

If \(x = k\) then \(x^2 + \frac{y^2}{9} = 1 – \frac{k^2}{4}\) is the standard equation of an ellipse. If \(k = 0\) then \(x^2 + \frac{y^2}{9} = 1\) which is an equation of an ellipse in the \(yz\)-plane centered at the origin with major and minor axes coinciding with the coordinate axes. If \(k = 2\) then \(x = 0\) and \(y = 0\) which is the point \((0,0)\). If \(0< k < 2\) then we have an ellipse centered at the origin with major and minor axes coinciding with the coordinate axes. If these traces are graphed in the \(yz\)-plane for increasing values of \(k>0\) then we would see ellipses with major and minor axes centered at the origin moving outward. We would not be able to distinguish between the traces \(x = k\) and \(x = -k\), as their graphs would coincide. - Traces in \(y =k\) and \(x = k\):

Similar arguments to that used for traces in \(z=k\), suggest that the traces in \(y =k\) and \(x = k\) are ellipses as well.

Thus the graph of \(x^2 + \frac{y^2}{9} + \frac{z^2}{4} = 1\) is a three dimensional analogue of an ellipse. This shape is called an **ellipsoid** and can be represented, under suitable choices of coordinate axes, by the equation

$$

\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1

$$

were \(a,b,\) and \(c\) are non-zero constants.

Therefore, the graph of \(x^2 + \frac{y^2}{9} + \frac{z^2}{4} = 1\) is that of an ellipsoid, where \(a = 1, b=3,\) and \(c = 2\).