Problem 97 - Calculus 3

Question

Find the unit tangent vector \(\mathbf{T}(t)\) 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

The unit tangent vector \( \mathbf{T}(t) \) is \( \left\langle \frac{1}{\sqrt{1 + \frac{1}{t^4}}}, \frac{-1}{t^2 \sqrt{1 + \frac{1}{t^4}}} \right\rangle \).

Step 1: Understand the Given Vector Function

The vector-valued function is given as \( \mathbf{r}(t)=\left\langle t, \frac{1}{t} \right\rangle \). This represents a curve in the plane where \( t \) determines the position \( (x, y) = (t, \frac{1}{t}) \) on the curve.

Step 2: Find the Derivative of the Vector Function

To find the tangent vector, we need to differentiate \( \mathbf{r}(t) \) with respect to \( t \). The derivative, \( \mathbf{r}'(t) \), is given by differentiating each component:- The derivative of \( t \) is 1.- The derivative of \( \frac{1}{t} \) is \( -\frac{1}{t^2} \).Thus, \( \mathbf{r}'(t) = \left\langle 1, -\frac{1}{t^2} \right\rangle \).

Step 3: Compute the Magnitude of the Derivative

The magnitude of the tangent vector, \( ||\mathbf{r}'(t)|| \), is needed to find the unit tangent vector. It is computed as:\[ ||\mathbf{r}'(t)|| = \sqrt{(1)^2 + \left(-\frac{1}{t^2}\right)^2} \]Simplify the expression:\[||\mathbf{r}'(t)|| = \sqrt{1 + \frac{1}{t^4}}\]

Step 4: Determine the Unit Tangent Vector \( \mathbf{T}(t) \)

The unit tangent vector \( \mathbf{T}(t) \) is the normalized form of the derivative \( \mathbf{r}'(t) \). It is given by dividing \( \mathbf{r}'(t) \) by its magnitude:\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} = \left\langle \frac{1}{\sqrt{1 + \frac{1}{t^4}}}, \frac{-\frac{1}{t^2}}{\sqrt{1 + \frac{1}{t^4}}} \right\rangle \]This simplifies to:\[ \mathbf{T}(t) = \left\langle \frac{1}{\sqrt{1 + \frac{1}{t^4}}}, \frac{-1}{t^2\sqrt{1 + \frac{1}{t^4}}} \right\rangle \]

Key Concepts

Vector-Valued Function

A vector-valued function is a mathematical entity where each input value results in a vector output rather than a scalar. It is typically represented in the form \( \mathbf{r}(t) = \langle x(t), y(t) \rangle \), where \( t \) is the parameter that varies over a certain interval. The function describes a curve or a path in a multi-dimensional space.

- In this exercise, the vector-valued function is \( \mathbf{r}(t) = \left\langle t, \frac{1}{t} \right\rangle \). This means that for every value of \( t \), the position on the curve is given by the point \( (t, \frac{1}{t}) \).

- As \( t \) changes, the function traces out a path in the coordinate plane, typically used to describe motion or a trajectory. Understanding vector-valued functions is crucial for solving problems in calculus related to motion, force fields, and other vector-related scenarios.

Differentiation

Differentiation is the process of finding the derivative of a function, which shows how the function changes as its input changes. In the context of vector-valued functions, differentiation is performed on each component separately.

- For instance, considering \( \mathbf{r}(t) = \left\langle t, \frac{1}{t} \right\rangle \), the derivative \( \mathbf{r}'(t) \) is found by differentiating each component individually:

• The derivative of \( t \) with respect to \( t \) is 1.
• The derivative of \( \frac{1}{t} \) is \( -\frac{1}{t^2} \), using the power rule.

- Therefore, the derivative of the entire vector function is \( \mathbf{r}'(t) = \left\langle 1, -\frac{1}{t^2} \right\rangle \).

This derivative gives the tangent vector, which indicates the direction of the path at any point \( t \). Differentiation is key to understanding rates of change and motion in vector calculus.

Magnitude of a Vector

The magnitude of a vector is a measure of its 'length' in a spatial sense. For a two-dimensional vector \( \mathbf{v} = \langle a, b \rangle \), its magnitude is calculated using the formula: \( ||\mathbf{v}|| = \sqrt{a^2 + b^2} \).

- In our example, we need the magnitude of the derivative \( \mathbf{r}'(t) = \left\langle 1, -\frac{1}{t^2} \right\rangle \). This is computed as: \[||\mathbf{r}'(t)|| = \sqrt{1^2 + \left(-\frac{1}{t^2}\right)^2} = \sqrt{1 + \frac{1}{t^4}}\]- The magnitude is crucial because it normalizes the tangent vector to create a unit tangent vector, which has a magnitude of 1.

Understanding how to measure the magnitude of a vector helps in determining not only distances and lengths but also in normalizing vectors for various applications in physics and engineering.