Problem 58 - Calculus III

Question

Find the unit tangent vector \(\mathrm{T}(\mathrm{t})\) for the following vector-valued functions.\(\mathbf{r}(t)=\left\langle t, \frac{1}{t}\right\rangle\). The graph is shown here:

Step-by-Step Solution

Answer

\( \mathbf{T}(t) = \left\langle \frac{1}{\sqrt{1 + \frac{1}{t^4}}}, \frac{-\frac{1}{t^2}}{\sqrt{1 + \frac{1}{t^4}}} \right\rangle \).

Step 1: Find the Derivative of r(t)

To find the unit tangent vector, we first need the derivative of the given vector function \( \mathbf{r}(t) = \left\langle t, \frac{1}{t} \right\rangle \). Calculate the derivative with respect to \( t \): \( \mathbf{r}'(t) = \left\langle \frac{d}{dt}(t), \frac{d}{dt}\left(\frac{1}{t}\right) \right\rangle = \left\langle 1, -\frac{1}{t^2} \right\rangle \).

Step 2: Find the Magnitude of the Derivative

Next, find the magnitude of the derivative \( \mathbf{r}'(t) = \left\langle 1, -\frac{1}{t^2} \right\rangle \). This is given by the formula: \( \|\mathbf{r}'(t)\| = \sqrt{1^2 + \left(-\frac{1}{t^2}\right)^2} = \sqrt{1 + \frac{1}{t^4}} \).

Step 3: Normalize the Derivative to Find T(t)

Divide the derivative \( \mathbf{r}'(t) \) by its magnitude to find the unit tangent vector \( \mathbf{T}(t) \): \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{\left\langle 1, -\frac{1}{t^2} \right\rangle}{\sqrt{1 + \frac{1}{t^4}}} = \left\langle \frac{1}{\sqrt{1 + \frac{1}{t^4}}}, \frac{-\frac{1}{t^2}}{\sqrt{1 + \frac{1}{t^4}}} \right\rangle \).

Key Concepts

Vector-Valued Functions

Think of a vector-valued function like a path that an object travels along in space.
For instance, if you see a function like:

• \(\mathbf{r}(t) = \left\langle t, \frac{1}{t} \right\rangle \)

This function gives you a set of points that depend on a parameter \(t\).
The first component tells us the \(x\)-coordinate, while the second gives the \(y\)-coordinate.
This implies when \(t\) changes, you'll see a different point on the graph.
These are incredibly useful in modeling real-world scenarios like the trajectory of a rocket or a curve of a roller coaster.
Once you get comfortable with this concept, you'll find it much easier to deal with.

Derivative of Vector Functions

The derivative of a vector-valued function helps us understand how the function behaves as the parameter changes.
It's much like taking the derivative in regular calculus but you do it for each component of the vector function separately.
For example, let's consider our vector function:

• \(\mathbf{r}(t) = \left\langle t, \frac{1}{t} \right\rangle \)

The derivative is:
• \(\mathbf{r}'(t) = \left\langle \frac{d}{dt}(t), \frac{d}{dt}\left(\frac{1}{t}\right) \right\rangle = \left\langle 1, -\frac{1}{t^2} \right\rangle\)
This derivative tells us the rate of change of each component with respect to \(t\).
The outcome is a new vector function that shows us where the path is going at any given moment.
Understanding this can help predict the future behavior of the function's curve.

Magnitude of a Vector

The magnitude of a vector is like measuring its size or length.
This is an important step because we often need to know how 'big' our vector is in the space it's occupying.
We calculate it using the formula:

• \(\|\mathbf{v}\| = \sqrt{x^2 + y^2}\)

For our derivative vector:

• \(\mathbf{r}'(t) = \left\langle 1, -\frac{1}{t^2} \right\rangle\)

The magnitude is:

• \(\|\mathbf{r}'(t)\| = \sqrt{1^2 + \left(-\frac{1}{t^2}\right)^2} = \sqrt{1 + \frac{1}{t^4}}\)

Getting the magnitude is crucial before we move on to normalization.
This gives us a scalar that shows how far the vector spans.

Normalizing Vectors

Normalizing a vector turns it into a unit vector, meaning it will have a magnitude of exactly 1.
This is handy because it allows you to maintain direction but standardizes the vector's length.
The process involves dividing the vector by its magnitude.
In our case, we want to normalize the derivative vector:

• \(\mathbf{r}'(t) = \left\langle 1, -\frac{1}{t^2} \right\rangle\)

And the magnitude we found is:

• \(\|\mathbf{r}'(t)\| = \sqrt{1 + \frac{1}{t^4}}\)

To normalize:

• \(\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \left\langle \frac{1}{\sqrt{1 + \frac{1}{t^4}}}, \frac{-\frac{1}{t^2}}{\sqrt{1 + \frac{1}{t^4}}} \right\rangle\)

This unit tangent vector \(\mathbf{T}(t)\) will follow the same direction as \(\mathbf{r}'(t)\) but be scaled down to a length of 1.
It simplifies scenarios where the direction is more important than size.
In fields like physics and engineering, this becomes particularly useful.