Problem 1 - Calculus 3

Question

Give the component functions \(x=f(t)\) and \(y=g(t)\) for the vector-valued function \(\mathbf{r}(t)=3 \sec t \mathbf{i}+2 \tan t \mathbf{j}\)

Step-by-Step Solution

Answer

The component functions are \( x = 3 \sec t \) and \( y = 2 \tan t \).

Step 1: Identify the Vector-Valued Function

The given vector-valued function is \( \mathbf{r}(t) = 3 \sec t \mathbf{i} + 2 \tan t \mathbf{j} \). It is expressed in terms of its component functions along the \( \mathbf{i} \) and \( \mathbf{j} \) unit vectors.

Step 2: Extract the Component Function for x

The component function for \( x \) along the \( \mathbf{i} \) direction is the coefficient of \( \mathbf{i} \) in \( \mathbf{r}(t) \). Here, it is given as \( x = 3 \sec t \).

Step 3: Extract the Component Function for y

The component function for \( y \) along the \( \mathbf{j} \) direction is the coefficient of \( \mathbf{j} \) in \( \mathbf{r}(t) \). Here, it is given as \( y = 2 \tan t \).

Key Concepts

Parametric Equations

Parametric equations are an essential concept in calculus and vector calculus. They allow us to express a set of related quantities as functions of an independent variable, often called a parameter. In the case of vector-valued functions, the parameter is typically denoted by \(t\), and each component of the vector is expressed as a function of \(t\). This approach provides a very flexible way to describe curves in the plane or space.

• These equations allow us to describe complex geometries and movements by focusing on the simplest components.
• Here, instead of directly linking \(x\) and \(y\), each is separately expressed using the same parameter \(t\).

For example, in the original exercise, the vector-valued function \(\mathbf{r}(t) = 3 \sec t \mathbf{i} + 2 \tan t \mathbf{j}\) is represented by parametric equations \(x = 3 \sec t\) and \(y = 2 \tan t\). These show how both \(x\) and \(y\) depend on \(t\). By adjusting \(t\), you can alter the position of a point on the curve represented by \(\mathbf{r}(t)\).

Trigonometric Functions

Trigonometric functions play a significant role in many areas of mathematics, including the study of vector-valued functions. They describe relationships among angles and sides within triangles and can generate periodic functions, which frequently model real-world phenomena like waves or circular motion.

• The functions \(\sec t\) and \(\tan t\) are derived from the basic sine and cosine functions.
• Secant (\(\sec t = \frac{1}{\cos t}\)) provides an inverse view of the cosine function, indicating reciprocal relationships.
• Similarly, tangent (\(\tan t = \frac{\sin t}{\cos t}\)) connects sine and cosine directly, offering insights into angles and slopes.

In the given problem, \(x = 3 \sec t\) and \(y = 2 \tan t\), trigonometric functions indicate how motion or change occurs with respect to time or some other parameter \(t\). This is particularly useful in physics and engineering, where modeling periodic behaviors is crucial.

Component Functions

Component functions are the building blocks of vector-valued functions. When working with vectors, each component function describes how the vector behaves in each directional axis. For a vector in two-dimensional space, like the one in the exercise, there are typically two component functions, one for the \(x\)-axis and one for the \(y\)-axis.

• Component functions simplify the understanding and analysis of complex vector behaviors by breaking them down into independent parts.
• In \(\mathbf{r}(t) = 3 \sec t \mathbf{i} + 2 \tan t \mathbf{j}\), \(x = 3 \sec t\) describes the movement or change in the \(x\) direction, while \(y = 2 \tan t\) represents similar behavior in the \(y\) direction.

This separation allows us to explore how each direction contributes to the overall vector function. Understanding these component functions assists in visualizing how the vector function \(\mathbf{r}(t)\) behaves over time or as the parameter \(t\) varies.