# 12.3 | Contour Diagrams

The 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 $$f$$).
A 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
$$\{ z \; | \; 0 \leq z \leq 4 \} = [0,4]$$

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

If $$f$$ is a function of two variables with domain $$D$$, then the 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 $$c = 2$$.