Today's topics include functions of two or three variables, determining domains, and sketching graphs. The next lesson will cover limits and continuity.
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Consider the function \(f(x) = \sqrt{9 - x^2}\).
Question 1: What is the domain of \(f\)?
The domain consists of all possible inputs. For this function, we need \(9 - x^2 \geq 0\), which gives us \(x^2 \leq 9\). Therefore, \(-3 \leq x \leq 3\).
Question 2: Sketch the graph of \(f(x) = \sqrt{9 - x^2}\) where \(y \geq 0\).
We can rewrite this as \(y = \sqrt{9 - x^2}\), which after squaring both sides (with \(y \geq 0\)) gives \(y^2 = 9 - x^2\), or \(x^2 + y^2 = 9\). This represents a semicircle with radius 3 centered at the origin, but only the upper half where \(y \geq 0\).
Key points on the graph:
Graph Description: The graph shows a coordinate system with the x-axis marked from -3 to 3. A semicircular curve is drawn connecting the point (-3, 0) through the maximum at (0, 3) to the point (3, 0). The curve represents the upper half of a circle with radius 3. Several points on the semicircle are marked with dots, including the endpoints and the apex.
We now extend our understanding from single-variable functions to multivariable functions. The key distinction is in the number of inputs and whether the output is a number or a vector.
Function Types:
1-Variable Function: \(x \rightarrow f(x)\)
Number → Number (real-valued function)
Vector-Valued Function: \(t \rightarrow \vec{r}(t)\)
Number → Vector
2-Variable Function: \(x, y \rightarrow f(x, y)\)
Two Numbers → Number
3-Variable Function: \(x, y, z \rightarrow f(x, y, z)\)
Three Numbers → Number
In each case, we're mapping from some input space (either single numbers, pairs, or triples of numbers) to an output (either a single number or a vector).
Let \(f(x, y) = x^2 + y^2\).
\(f(1, 3) = 1^2 + 3^2 = 1 + 9 = 10\)
\(f(-10, 11) = (-10)^2 + 11^2 = 100 + 121 = 221\)
Let \(f(x, y) = \dfrac{\sqrt{x + y + 1}}{x - 1}\).
\(f(-10, 13) = \dfrac{\sqrt{-10 + 13 + 1}}{-10 - 1} = \dfrac{\sqrt{4}}{-11} = \dfrac{2}{-11} = -\dfrac{2}{11}\)
\(f(1, 5)\) is undefined because the denominator becomes \(1 - 1 = 0\), resulting in division by zero.
Let \(f(x, y, z) = \dfrac{\sqrt{9 - x^2 - y^2}}{\sqrt{25 - z^2}}\).
\(f(0, 0, 0) = \dfrac{\sqrt{9}}{\sqrt{25}} = \dfrac{3}{5}\)
\(f(0, 1, 3) = \dfrac{\sqrt{9 - 0 - 1}}{\sqrt{25 - 9}} = \dfrac{\sqrt{8}}{\sqrt{16}} = \dfrac{2\sqrt{2}}{4} = \dfrac{\sqrt{2}}{2}\)
\(f(0, 1, 5)\) cannot be evaluated because \(z = 5\) makes the denominator \(\sqrt{25 - 25} = 0\).
Definition - Domain: The domain of a multivariable function consists of all possible inputs.
For \(f(x, y) = x^2 + y^2\), the domain is all of \(\mathbb{R}^2\) since we can square any real numbers and add them together.
Diagram Description: The page shows a coordinate plane with x and y axes. A large, irregularly-shaped cyan region fills most of the plane, representing the domain as a region in 2-dimensional space. The region extends in all directions, illustrating that the domain can be the entire plane or a specific subset of it.
Find the domain of \(f(x, y) = \dfrac{1}{x^2 + y^2}\).
The denominator cannot be zero, so we need \(x^2 + y^2 \neq 0\).
This means the point \((0, 0)\) is not in the domain.
Therefore, the domain is all of \(\mathbb{R}^2\) except the origin.
Diagram Description: The coordinate plane shows a large cyan-shaded region representing the domain. At the origin (0, 0), there is a small white circle with a dot in the center, indicating that this point is excluded from the domain. The notation shows an open circle (○) to represent a point not included, contrasted with a filled circle (●) which would represent a point that is included.
Find the domain of \(f(x, y) = \sqrt{x + y + 1}\).
For the square root to be defined, we need the argument to be non-negative:
\[x + y + 1 \geq 0\]This simplifies to:
\[x + y \geq -1\]Equivalently, \(y \geq -x - 1\).
The boundary line \(x + y + 1 = 0\) (or \(x + y = -1\)) is included in the domain and represented by a solid line. The domain consists of all points on or above this line.
Testing points:
Diagram Description: A coordinate plane shows a diagonal line passing through points approximately at (-1, 0) and (0, -1), representing the boundary \(x + y = -1\). The region above and to the right of this line is shaded in cyan, representing the domain. The boundary line is drawn as a solid blue line, indicating that points on the line are included in the domain. The line has a slope of -1. A vertical dashed line at \(x = -1\) is also shown, and the notation indicates that the boundary line is "included" and "represented by solid line."
Find the domain of \(f(x, y) = y\ln(y^2 - x)\).
For the natural logarithm to be defined, we need the argument to be strictly positive:
\[y^2 - x > 0\]This gives us:
\[y^2 > x\]or equivalently:
\[x < y^2\]The boundary curve \(y^2 = x\) (a parabola opening to the right) is not included in the domain, represented by a dashed line.
Testing points:
The domain consists of all points to the left of the parabola \(x = y^2\).
Diagram Description: A coordinate plane shows a parabola opening to the right with vertex at the origin, representing the curve \(x = y^2\). The region to the left of the parabola is shaded in cyan, representing the domain. The parabola itself is drawn as a dashed curve, indicating that points on the curve are not included in the domain. Two specific points are marked: (-2, 0) with a filled dot (in the domain) and (2, 0) with an open circle (not in the domain).
Take a moment to:
Find the domain of \(f(x, y, z) = \dfrac{\sqrt{9 - x^2 - y^2}}{\sqrt{25 - z^2}}\).
For this function to be defined, we need two conditions:
Numerator condition: \(9 - x^2 - y^2 \geq 0\)
This gives \(x^2 + y^2 \leq 9\), which describes a solid cylinder of radius 3 along the z-axis when extended in three dimensions. In the xy-plane, this is a disk of radius 3 centered at the origin.
Denominator condition: \(25 - z^2 > 0\)
This gives \(z^2 < 25\), so \(-5 < z < 5\).
The domain is the intersection of these two conditions:
Diagram Description: Two representations are shown:
Definition - Graph of \(f(x, y)\): The graph of \(f(x, y)\) consists of all points \((x, y, f(x, y))\) in \(\mathbb{R}^3\) for all \((x, y)\) in the domain.
The result is a surface \(z = f(x, y)\) in three-dimensional space.
To visualize and sketch the surface \(z = f(x, y)\), we can use several methods:
Methods for Graphing:
We examine the surface by taking cross sections, traces, or level curves for different possible values of \(z\).
Sketch the graph of \(f(x, y) = \sqrt{9 - x^2 - y^2}\).
We can rewrite this as \(z = \sqrt{9 - x^2 - y^2}\). Squaring both sides (with \(z \geq 0\)):
\[z^2 = 9 - x^2 - y^2\] \[x^2 + y^2 + z^2 = 9\]This is the equation of a sphere with radius 3 centered at the origin. Since \(z \geq 0\), we have the upper hemisphere.
Level curves (contours): Set \(z\) equal to various constants and solve for the relationship between \(x\) and \(y\).
Diagram Description: Two representations are shown:
Sketch the graph of \(f(x, y) = x^2 + y^2\).
We have \(z = x^2 + y^2\), which is a paraboloid opening upward.
Level curves: Set \(z\) equal to various constants.
As \(z\) increases, the level curves are circles with increasing radius. The surface is a bowl-shaped paraboloid.
Diagram Description: Two representations are shown:
Note: \(z < 0\) is marked with an X, indicating that there are no points on the surface where \(z\) is negative.
Sketch the graph of \(f(x, y) = 6 - 3x - 2y\).
We have \(z = 6 - 3x - 2y\), which can be rewritten as:
\[3x + 2y + z = 6\]This is the equation of a plane with normal vector \(\langle 3, 2, 1 \rangle\).
Level curves: Set \(z\) equal to various constants.
All level curves are parallel lines in the xy-plane, which is characteristic of a planar surface.
Diagram Description: A contour map (xy-plane view) shows several parallel lines representing level curves of the plane. The lines are colored differently to indicate different z-values: green line (\(z = -6\)), blue line (\(z = 0\)), purple line (\(z = 6\)), and red line (\(z = 12\)). All lines have the same slope (\(-3/2\)) and are evenly spaced, indicating a constant rate of change in z. The spacing between contour lines is uniform, confirming that the surface is a plane rather than a curved surface.
Key Observation: For linear functions \(f(x, y) = ax + by + c\), the graph is always a plane, and the level curves are parallel straight lines in the xy-plane.