Problem 91 - Calculus 3

Question

Given the vector-valued function \(\mathbf{r}(t)=\langle\tan t, \sec t, 0\rangle\) (graph is shown here), find the following: Speed

Step-by-Step Solution

Answer

The speed is \( \sqrt{2 \sec^4 t - \sec^2 t} \).

Step 1: Understanding Speed

The speed of a particle given by a vector-valued function \( \mathbf{r}(t) \) is the magnitude of its velocity vector \( \mathbf{v}(t) \). Hence, to find the speed, we first need to find the velocity vector which is the derivative of \( \mathbf{r}(t) \).

Step 2: Find the Velocity Vector

Differentiate \( \mathbf{r}(t) = \langle \tan t, \sec t, 0 \rangle \) with respect to \( t \) to get the velocity vector: \[ \mathbf{v}(t) = \frac{d}{dt} \langle \tan t, \sec t, 0 \rangle = \langle \sec^2 t, \sec t \tan t, 0 \rangle \]

Step 3: Calculate the Magnitude of the Velocity Vector

Calculate the magnitude of the velocity vector \( \mathbf{v}(t) = \langle \sec^2 t, \sec t \tan t, 0 \rangle \):\[ \text{Magnitude} = \sqrt{(\sec^2 t)^2 + (\sec t \tan t)^2 + 0^2} \]Simplify the expression inside the square root:\[ = \sqrt{\sec^4 t + \sec^2 t \tan^2 t} \]

Step 4: Simplify the Expression

Use the identity \( \sec^2 t = 1 + \tan^2 t \) to rewrite terms:\[ \sqrt{\sec^4 t + \sec^2 t \tan^2 t} = \sqrt{\sec^2 t (\sec^2 t + \tan^2 t)} = \sqrt{\sec^2 t (2 \sec^2 t - 1)} \]Simplifying further:\[ = \sqrt{2 \sec^4 t - \sec^2 t} \]

Step 5: Final Speed Expression

The speed of the function \( \mathbf{r}(t) = \langle \tan t, \sec t, 0 \rangle \) is given by: \[ \text{Speed} = \sqrt{2 \sec^4 t - \sec^2 t} \] This is the expression for speed in terms of \( t \).

Key Concepts

Vector-Valued Functions

Vector-valued functions are a way to represent a collection of functions as a single entity. Unlike a single value function which gives you one output, a vector-valued function gives you a vector as an output. Imagine you are tracking the position of a moving particle over time. A vector-valued function like \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \) can describe its position in a three-dimensional space. The components \( x(t) \), \( y(t) \), and \( z(t) \) are individual functions of the parameter \( t \), which often represents time.
Understanding these functions is crucial because they help in analyzing paths and movements, providing insight into dynamics and kinematics. They're essential in fields like physics and engineering where systems are ordered in multi-dimensional spaces, making it easier to study motions and forces.
For the given exercise, \( \mathbf{r}(t) = \langle \tan t, \sec t, 0 \rangle \), the vector describes a motion path in the 2D plane, since the third component is zero, meaning no movement in the z-direction.

Velocity Vector

The velocity vector is derived from the vector-valued function by taking its derivative. It represents the rate of change of the position vector, hence, showing how fast and in which direction a particle moves. So, if you're given a vector-valued function \( \mathbf{r}(t) \), its derivative with respect to \( t \), denoted as \( \mathbf{v}(t) \), gives the velocity vector.
Let's break this down with our original function, \( \mathbf{r}(t) = \langle \tan t, \sec t, 0 \rangle \). By differentiating each component individually with respect to \( t \), we get:

• The derivative of \( \tan t \) is \( \sec^2 t \).
• The derivative of \( \sec t \) is \( \sec t \tan t \).
• The derivative of 0 is still 0.

Thus, the velocity vector \( \mathbf{v}(t) = \langle \sec^2 t, \sec t \tan t, 0 \rangle \) completely describes the direction and speed of the particle at any time \( t \). It's a dynamic snapshot that tells us not just about where the particle is going, but also at what rate.

Derivative Calculation

Derivative calculation is a fundamental tool in calculus used to measure change. To find the derivative of a function with respect to a variable, like \( t \), means to determine how that function's output changes as \( t \) changes. It's straightforward when differentiating basic functions like powers of \( t \), but can get tricky with functions involving trigonometric terms.
In our exercise, we calculated the derivative of a vector-valued function \( \mathbf{r}(t) \). This means taking the derivative of each of its components. The approach involves:

• Applying the rule \( \frac{d}{dt} [\tan t] = \sec^2 t \).
• Using the derivative rule \( \frac{d}{dt} [\sec t] = \sec t \tan t \).
• Finding that \( \frac{d}{dt} [0] = 0 \).

These rules stem from understanding the basics of differentiating trigonometric functions. Such derivatives are key in both finding the velocity vector and further analyzing the behavior of the function.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are always true for any angle. They play a vital role in simplifying complex trigonometric expressions and are often used in derivative calculations as well as solving integrals.
In the exercise, while working to simplify the speed of the particle, a crucial trigonometric identity was used: \( \sec^2 t = 1 + \tan^2 t \). This identity helps in rewriting terms involving \( \sec t \) and \( \tan t \) to reach a simpler form.
Simplifications allow mathematicians to manage intricate expressions more easily, making calculations not only more elegant but often revealing deeper insights about the functions themselves. In our scenario, these identities enabled the reduction of the original expression for the magnitude of the velocity. This leads to the concise final expression for the speed: \( \sqrt{2 \sec^4 t - \sec^2 t} \). Knowing these identities deepens understanding and aids in solving similar problems with confidence.