Warm-up
We begin by reviewing the concept of a function:
(1) What is a function? A function has one output for each input.
(2) What are some examples of functions?
Diagram Description: Two mapping diagrams are shown side by side. The left diagram shows a non-function where multiple inputs (cat, dog, cow, horse) map to the same outputs (-1, 2, Apple, Banana), with some inputs having multiple outputs, labeled "NOT function". The right diagram shows a valid function where each input (cat, dog, cow, horse) maps to exactly one output (-1, 2, Apple, Banana), labeled "Function".
Definition: The domain of a function consists of all possible inputs. The range consists of all outputs.
For example, consider \(f(x) = x^2\). The domain is \((-\infty, \infty)\) and the range is \([0, \infty)\).
Another example: \(f(x) = \ln x\) has domain \((0, \infty)\) and range \((-\infty, \infty)\).
Recall: Scalar Functions
Scalar functions have the property that the input is a number and the output is a number.
Example: Properties of Scalar Functions
Consider \(f(x) = x^3 + 1\) with domain \([0, \infty)\) and range \([1, \infty)\).
Graph Description: A cubic curve is shown in a standard xy-coordinate system. The curve starts at the origin and extends upward and to the right, representing \(f(x) = x^3 + 1\) for \(x \geq 0\). Multiple arrows point outward from the origin, indicating the behavior of the function near \(x = 0\).
Limit: We can compute
\[\lim_{x \to 0} f(x) = 1 = f(0)\]
which shows the function is continuous at \(x = 0\).
Derivative:
\[\frac{d}{dx}(x^3 + 1) = 3x^2\]
Indefinite Integral:
\[\int (x^3 + 1) \, dx = \frac{x^4}{4} + x + C\]
where \(C\) is a constant.
Definite Integral (Signed Area):
\[\int_{-2}^{1} f(x) \, dx = F(1) - F(-2)\]
where \(F\) is an antiderivative. The notation FTC refers to the Fundamental Theorem of Calculus.
Vector Valued Functions
A vector valued function has a number as input and a vector as output.
Definition (Notation): We write
\[\vec{r}(t) = \langle f(t), g(t), h(t) \rangle\]
where each of \(f(t)\), \(g(t)\), and \(h(t)\) are scalar functions.
The domain of \(\vec{r}(t)\) is the intersection of the domains of \(f(t)\), \(g(t)\), and \(h(t)\).
Example: Finding the Domain
Find the domain of
\[\vec{r}(t) = \langle t^3 + 5, \ln(3-t), \sqrt{t} \rangle\]
Solution:
We examine each component function separately:
First component: \(f(t) = t^3 + 5\) has domain \((-\infty, \infty)\)
Second component: \(g(t) = \ln(3-t)\) requires \(3 - t > 0\), which gives \(t < 3\). Therefore the domain is \((-\infty, 3)\).
Third component: \(h(t) = \sqrt{t}\) requires \(t \geq 0\). Therefore the domain is \([0, \infty)\).
The domain of \(\vec{r}(t)\) is the intersection:
\[(-\infty, \infty) \cap (-\infty, 3) \cap [0, \infty) = [0, 3)\]
Number Line Description: Three horizontal number lines are shown. The first (blue) extends infinitely in both directions. The second (green) shows an interval from negative infinity to 3 with an open circle at 3. The third (purple) shows an interval from 0 (closed circle) to positive infinity. Below these, a final number line shows the intersection as the interval [0, 3), with a closed circle at 0 and an open circle at 3.
Example: Line in 3D Space
Consider
\[\vec{r}(t) = \langle 1+t, 2t, 2+3t \rangle\]
with domain \((-\infty, \infty)\).
We can rewrite this as:
\[\vec{r}(t) = \langle 1, 0, 2 \rangle + t\langle 1, 2, 3 \rangle\]
This represents a line passing through the point \((1, 0, 2)\) in the direction of the vector \(\langle 1, 2, 3 \rangle\).
3D Graph Description: A three-dimensional coordinate system with x, y, and z axes is shown. A straight line passes through the point (1, 0, 2) and extends in both directions. The direction vector \(\langle 1, 2, 3 \rangle\) is indicated, showing the line rises as it moves away from the point in both positive and negative parameter directions.
Interpretation: The vector \(\langle 1, 0, 2 \rangle\) gives one point on the line, and \(t\langle 1, 2, 3 \rangle\) is the direction vector that determines how the line extends through space.
Orientation
Definition (Positive Orientation): Positive orientation means we go from lower \(t\) values to higher \(t\) values along the curve.
Diagram Description: A diagonal line in teal extends from lower-left to upper-right. Points are marked along the line at \(t = -1\), \(t = 0\), \(t = 1\), and \(t = 2\), with corresponding position vectors shown. Arrows point in the direction of increasing \(t\), demonstrating positive orientation. The point \((1, 0, 2)\) is marked at \(t = 0\), and \((2, 2, 5)\) is marked at \(t = 1\).
Definition (Negative Orientation): Negative orientation means we go from higher \(t\) values to lower \(t\) values along the curve.
Diagram Description: A diagonal line in purple extends from upper-right to lower-left. Points are marked at \(t = 2\), \(t = 1\), \(t = 0\), and \(t = -1\). Arrows point in the direction of decreasing \(t\), demonstrating negative orientation.
Example: Circle in the xy-plane
Consider
\[\vec{r}(t) = \langle \cos t, \sin t \rangle, \quad 0 \leq t \leq \frac{\pi}{2}\]
Graph Description: A unit circle is shown in the xy-plane centered at the origin. Only the first quadrant arc is highlighted, corresponding to \(0 \leq t \leq \pi/2\). The point \((0, 1)\) is marked at \(t = \pi/2\), and the point \((1, 0)\) is marked at \(t = 0\). The arc traces counterclockwise from \((1, 0)\) to \((0, 1)\).
A table of values shows the correspondence between \(t\), \(x = \cos t\), and \(y = \sin t\):
| \(t\) |
\(x\) |
\(y\) |
| 0 |
1 |
0 |
| \(\pi/6\) |
\(\sqrt{3}/2\) |
\(1/2\) |
| \(\pi/4\) |
\(\sqrt{2}/2\) |
\(\sqrt{2}/2\) |
| \(\pi/3\) |
\(1/2\) |
\(\sqrt{3}/2\) |
| \(\pi/2\) |
0 |
1 |
We can verify that all points lie on the unit circle since
\[x^2 + y^2 = \cos^2 t + \sin^2 t = 1\]
Example: Cylinder in 3D
Consider
\[\vec{r}(t) = \langle \cos t, \sin t, 0 \rangle, \quad 0 \leq t \leq \frac{\pi}{2}\]
The equation \(x^2 + y^2 = 1\) describes a cylinder in 3D. Since \(z = 0\), this curve lies in the xy-plane.
3D Graph Description: A three-dimensional coordinate system is shown with x, y, and z axes. A vertical cylinder of radius 1 is indicated by dashed lines, representing the surface \(x^2 + y^2 = 1\). Within the xy-plane (\(z = 0\)), a quarter-circle arc is drawn from the point \((1, 0, 0)\) at \(t = 0\) to the point \((0, 1, 0)\) at \(t = \pi/2\), with arrows showing positive orientation. The arc lies flat in the xy-plane, traced counterclockwise when viewed from above.
Example: Circle at Height z = 3
Consider
\[\vec{r}(t) = \langle \cos t, \sin t, 3 \rangle, \quad 0 \leq t \leq \frac{\pi}{2}\]
This describes a circle of radius 1 centered on the z-axis at height \(z = 3\).
3D Graph Description: A three-dimensional coordinate system shows a vertical cylinder of radius 1. At the horizontal plane \(z = 3\), a quarter-circle arc is drawn from the point \((1, 0, 3)\) at \(t = 0\) to the point \((0, 1, 3)\) at \(t = \pi/2\). The arc is traced counterclockwise when viewed from above, parallel to the xy-plane but elevated to \(z = 3\). Dashed vertical and horizontal lines show the cylindrical surface and the plane at \(z = 3\).
Example: Helix
Consider
\[\vec{r}(t) = \langle \cos t, \sin t, 2t \rangle, \quad -2\pi \leq t \leq 2\pi\]
This equation describes a helix that spirals around the z-axis. We note that \(x^2 + y^2 = 1\), so the helix stays on the cylinder of radius 1.
Graph Description (left): A unit circle in the xy-plane is shown with x and y axes. This represents the projection of the helix onto the xy-plane, which traces around the circle multiple times as t varies.
A table shows selected values of the helix:
| \(t\) |
\(x\) |
\(y\) |
\(z\) |
| \(-2\pi\) |
1 |
0 |
\(-4\pi\) |
| \(-3\pi/2\) |
0 |
1 |
\(-3\pi\) |
| \(-\pi\) |
-1 |
0 |
\(-2\pi\) |
| \(-\pi/2\) |
0 |
-1 |
\(-\pi\) |
| 0 |
1 |
0 |
0 |
| \(\pi/2\) |
0 |
1 |
\(\pi\) |
| \(\pi\) |
-1 |
0 |
\(2\pi\) |
| \(3\pi/2\) |
0 |
-1 |
\(3\pi\) |
| \(2\pi\) |
1 |
0 |
\(4\pi\) |
Helix Visualization (continued)
3D Graph Description: A three-dimensional coordinate system shows the complete helix structure. The curve spirals upward around a vertical cylinder of radius 1. Key points are marked: \((1, 0, 4\pi)\) at the top, \((0, 1, \pi)\) partway up, \((1, 0, 0)\) at the middle, \((0, 1, -2\pi)\) partway down, and \((1, 0, -4\pi)\) at the bottom. The helix makes two complete revolutions as it rises from \(z = -4\pi\) to \(z = 4\pi\). The word "Helix" is written with decorative styling to emphasize this three-dimensional curve.
Key Observation: As \(t\) increases from \(-2\pi\) to \(2\pi\), the helix makes exactly two complete loops around the z-axis while rising linearly in the z-direction at a rate of 2 units per radian of \(t\).
Intersection with a Surface
Example: Helix Intersecting a Sphere
Does
\[\vec{r}(t) = \langle \cos t, \sin t, 2t \rangle, \quad -\infty < t < \infty\]
intersect with the sphere \(x^2 + y^2 + z^2 = 25\) of radius 5?
3D Graph Description: A sphere of radius 5 is shown centered at the origin. A helix (purple curve) winds around the z-axis, passing through the interior of the sphere. The helix intersects the spherical surface at exactly two points, marked as \(\vec{r}(\sqrt{6})\) near the top of the sphere and \(\vec{r}(-\sqrt{6})\) near the bottom. A vertical dashed cylinder indicates the surface on which the helix lies.
Note: The helix intersects the sphere at two points!
Finding Points of Intersection
Solution:
To find where \(\vec{r}(t) = \langle \cos t, \sin t, 2t \rangle\) intersects the sphere \(x^2 + y^2 + z^2 = 25\), we plug in \(x\), \(y\), \(z\) from \(\vec{r}(t)\) into the sphere equation:
\[\cos^2 t + \sin^2 t + (2t)^2 = 25\]
Using the Pythagorean identity \(\cos^2 t + \sin^2 t = 1\), we have:
\[1 + 4t^2 = 25\]
\[4t^2 = 24\]
\[t^2 = 6\]
\[t = \pm\sqrt{6}\]
Two points of intersection:
Top point:
\[\vec{r}(\sqrt{6}) = \langle \cos(\sqrt{6}), \sin(\sqrt{6}), 2\sqrt{6} \rangle\]
Bottom point:
\[\vec{r}(-\sqrt{6}) = \langle \cos(-\sqrt{6}), \sin(-\sqrt{6}), -2\sqrt{6} \rangle\]
Interpretation: The helix enters the sphere from below at \(t = -\sqrt{6}\), passes through the interior, and exits at the top at \(t = \sqrt{6}\).
