14.1 Vector-Valued Functions

Scalar-valued functions: \(f(t) = t^2 + 3\)

Vector-valued functions: \(\vec{r}(t) = \langle \cos t, \sin t, 0 \rangle\)

You have seen an example already: equation of line \(\vec{r}(t) = \vec{r}_0 + t\vec{v}\)

Quick example: line through \(P(1, 2, 3)\) and \(Q(4, 5, 6)\)

\(\vec{r}_0 = \langle 1, 2, 3 \rangle\), \(\vec{v} = \vec{PQ} = \langle 3, 3, 3 \rangle\)

That is called the positive orientation.

Example

Consider the vector function:

What is the graph?

Easy thing to do: find points with x = cos t, y = sin t, z = 0 by choosing t, then connect them.

Positive orientation is counterclockwise when viewed from above.

Another way to visualize: make a connection to surface we know

Relationship between x, y, z: x = cos t, y = sin t, z = 0

So, \vec{r}(t) = \langle \cos t, \sin t, 0 \rangle is a curve on the cylinder x^2 + y^2 = 1 where z = 0.

Example

\(\vec{r}(t) = \langle 0, \cos t, 2\sin t \rangle\), \(0 \le t \le \pi/2\)

\(x=0, y=\cos t, z=2\sin t\)

\(\cos^2 t + \sin^2 t = 1\)

\(\cos t = y, \sin t = \frac{1}{2}z \rightarrow y^2 + (\frac{1}{2}z)^2 = 1\)

In \(\mathbb{R}^3\) this is an elliptic cylinder.

\(\vec{r}(t)\) is a portion of that with \(x=0\).

\(t=0: \vec{r}(0) = \langle 0, 1, 0 \rangle\)

\(t=\pi/2: \vec{r}(\pi/2) = \langle 0, 0, 2 \rangle\)

Example

\(\vec{r}(t) = \langle \cos t, \sin t, t \rangle\), \(0 \le t \le 2\pi\)

\(x=\cos t, y=\sin t, z=t\)

Overall shape is still \(x^2+y^2=1\) (\(z=t\) really doesn't directly fit in).

This curve is still on the cylinder \(x^2+y^2=1\) but with varying \(z\).

\(\vec{r}(0) = \langle 1, 0, 0 \rangle\)

\(\vec{r}(\pi) = \langle -1, 0, \pi \rangle\)

\(\vec{r}(2\pi) = \langle 1, 0, 2\pi \rangle\)

Like a spiraling staircase.

Example

Given the vector function:

We have the components:

• \( x = t \cos t \)
• \( y = t \)
• \( z = t \sin t \)

Remember the identity: \( \cos^2 t + \sin^2 t = 1 \).

Notice that:

Since \( y = t \), we have:

In \( \mathbb{R}^3 \), this is a cone. The curve lies on the cone, moving toward positive \( y \).

Intersection Analysis

Does \( \vec{r}(t) = \langle t \cos t, t, t \sin t \rangle \) for \( 0 \le t \le 2\pi \) intersect the plane \( x - z = 0 \)?

For an intersection, the coordinates must be the same. If they intersect, then at some \( t \), the components must satisfy \( x - z = 0 \).

This implies either \( t = 0 \) or \( \cos t = \sin t \).

Solving \( \cos t = \sin t \) gives \( t = \frac{\pi}{4} \) and \( t = \frac{5\pi}{4} \).

Therefore, they intersect at \( t = 0, \frac{\pi}{4}, \frac{5\pi}{4} \).

Domain of a Vector-Valued Function

The domain of a vector-valued function is the intersection of the domains of all components, which is where ALL components are defined.

Example

\[ \vec{r}(t) = \langle \sqrt{1-t^2}, \sqrt{t}, \frac{1}{\sqrt{5-t}} \rangle \]

• \( \sqrt{1-t^2} \) defined on \( [-1, 1] \)
• \( \sqrt{t} \) defined on \( [0, \infty) \)
• \( \frac{1}{\sqrt{5-t}} \) defined on \( (-\infty, 5) \)

Intersection is \( [0, 1] \)

This is the domain of \( \vec{r}(t) \)

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