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15.1 Functions of Several Variables

If \( y = f(x) = \sqrt{9 - x^2} \)

input↑

output↑

The set of all acceptable input values form the domain.

For this function, the domain is:

\[ 9 - x^2 \ge 0 \]

\[ -3 \le x \le 3 \quad \text{or} \quad [-3, 3] \]

Figure: A horizontal number line with square brackets at -3 and 3, indicating the closed interval [-3, 3].

Note: this is a line or portions of a line.

The set of all possible output values is the range.

Here, the range is \( [0, 3] \).

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Now let's look at a function of two variables:

\[ z = f(x, y) = \sqrt{9 - x^2} - \sqrt{25 - y^2} \]

input↑

output↑

The input is now a set of ordered pairs \( (x, y) \).

The output remains the same as before.

The domain is now more complicated:

Here, we need:

\( 9 - x^2 \ge 0 \) AND \( 25 - y^2 \ge 0 \)

\( -3 \le x \le 3 \) AND \( -5 \le y \le 5 \)

The graph of the domain is now a region.

Figure: A rectangular region shaded on a Cartesian plane, bounded by x = -3, x = 3, y = -5, and y = 5.

All \( (x, y) \) inside and on the boundary of this box

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if boundary values are not included, for example \(-3 < x < 3\) , \(-5 < y < 5\)

Figure: A 2D coordinate system showing a rectangular region bounded by dashed lines at x = -3, 3 and y = -5, 5.

the range doesn't change from 1-variable function

\[z = f(x,y) = \sqrt{9-x^2} - \sqrt{25-y^2}\]

\(\sqrt{9-x^2}\)

largest 3

smallest 0

\(\sqrt{25-y^2}\)

largest 5

smallest 0

collectively, largest \(z\) is 3

smallest \(z\) is -5

range: \(-5 \le z \le 3\)

or \([-5, 3]\)

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we know \(z = f(x,y)\) produces a surface (e.g. plane, paraboloid, etc)

to aid visualization, we sometimes use level curves or contours

these are produced from \(z = z_0 = \text{constant}\) then graph \(f(x,y) = z_0\)

very similar to a trace

example

\[z = f(x,y) = 5 - x^2 - y^2 = 5 - (x^2 + y^2) \quad \text{paraboloid}\]

\(z = 5 - (x^2 + y^2)\)

if \(z = z_0\) , then we get \(x^2 + y^2 = 5 - z_0\) circles radius \(\sqrt{5 - z_0}\)

Figure: 3D coordinate graph showing a downward-opening paraboloid with its vertex at z=5.

domain:

\(-\infty < x < \infty\)

\(-\infty < y < \infty\)

\(\rightarrow \{ (x,y) : \mathbb{R}^2 \}\)

range: \((-\infty, 5]\)

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Level Curves and Contours

Function: \( z = f(x,y) = 5 - x^2 - y^2 \)

Level curve / contour: Set \( z = z_0 = \text{constant} \)

Figure: 3D plot of a downward-opening paraboloid starting at z=5 on the z-axis.

Case 1: \( z = z_0 = 5 \)

\[ 5 = 5 - x^2 - y^2 \implies x^2 + y^2 = 0 \]

This represents a point at the origin.

Figure: A point at the origin of a 2D xy-coordinate system.

Case 2: \( z = z_0 = 1 \)

\[ 1 = 5 - x^2 - y^2 \implies x^2 + y^2 = 4 \]

This represents a circle of radius 2.

Figure: A circle of radius 2 centered at the origin in the xy-plane.

Case 3: \( z = z_0 = 0 \)

\[ x^2 + y^2 = 5 \]

This represents a circle of radius \( \sqrt{5} \).

Figure: A circle of radius square root of 5 centered at the origin in the xy-plane.

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The collection of level curves provide another way to visualize / understand the function.

Figure: Contour plot showing concentric circles for z=0, z=1, and a point for z=5 in the xy-plane.

Each curve represents the surface at a particular height.

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Example: Multivariable Function Analysis

f(x, y) = \sin(xy)

domain: \(\{ (x, y) : \mathbb{R}^2 \}\) because sine can take any real number
range: \([-1, 1]\)
level curves: \(z = z_0\)

Deriving Level Curves

\[ z = f(x, y) = \sin(xy) \]

\[ z_0 = \sin(xy) \quad \text{(constant)} \]

\[ xy = \sin^{-1}(z_0) = K \quad \text{(constant)} \]

So level curves are:

\[ xy = K \]

\[ y = \frac{K}{x} \quad \text{hyperbolas} \]

0 in the first and third quadrants, and K < 0 in the second and fourth.">

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\[ y = \frac{K}{x} \quad \text{where } z_0 = \sin(K) \iff K = \sin^{-1}(z_0) \]

if \( K = \frac{\pi}{4} \) this corresponds to \( z_0 = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \)
if \( K = \frac{\pi}{2} \) this corresponds to \( z_0 = \sin(\frac{\pi}{2}) = 1 \)
if \( K = \pi \) this corresponds to \( z_0 = \sin(\pi) = 0 \)
if \( K = -\frac{\pi}{4} \) this corresponds to \( z_0 = \sin(-\frac{\pi}{4}) = -\frac{1}{\sqrt{2}} \)

Figure: Detailed plot of level curves y = K/x for various values of K, showing multiple hyperbolic branches.

Visual representation of level curves for different constant values.

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Visualizing Multivariable Functions: Heat Maps

The following figure illustrates a 2D representation of a function where the height is represented by a color gradient. This is often referred to as a heat map or a surface plot viewed from above.

Figure: A 2D heat map on an x-y plane with axes ranging from -3 to 3. Colors transition from blue (low) to yellow (high).

Note on Visualization:

different colors are different heights

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Visualizing Multivariable Functions: 3D Surface Plots

This visualization provides a three-dimensional perspective of the same function shown previously. The vertical axis represents the function's value, while the horizontal axes represent the domain.

Figure: A 3D surface plot showing peaks and valleys. The z-axis ranges from -1 to 1, and x and y axes range from -3 to 3.

The surface exhibits periodic behavior, with multiple local maxima (peaks) and minima (valleys) across the domain defined by \( x \in [-3, 3] \) and \( y \in [-3, 3] \).

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3D Surface Visualization and Projections

Figure: A 3D surface plot showing a wave-like structure with peaks and valleys, colored from blue to yellow.

"edge" on with \( x \) or \( y \) pointing at us \( \rightarrow \) \( x = \text{constant} = C \)

\[ z = \sin(cy) \]

Figure: A small 2D sketch of a sine wave on a zy-coordinate plane.