Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \mathbf{r}(t)=t \sin \pi t \mathbf{i}+t \cos \pi t \mathbf{j}+e^{-t} \mathbf{k}: t_{1}=2 $$

Step-by-Step Solution

Answer

The evaluated vector at \( t_1 = 2 \) is \( 0\mathbf{i} + 2\mathbf{j} + \frac{1}{e^2}\mathbf{k} \).

Step 1: Understand the Function

The vector function given is \( \mathbf{r}(t) = t \sin(\pi t) \mathbf{i} + t \cos(\pi t) \mathbf{j} + e^{-t} \mathbf{k} \). This is a parametric equation in terms of time \( t \) that describes a path or a trajectory in 3D space.

Step 2: Evaluate the Function at \( t_1 = 2 \)

Substitute \( t = 2 \) into the vector function: \[ \mathbf{r}(2) = 2 \sin(2\pi) \mathbf{i} + 2 \cos(2\pi) \mathbf{j} + e^{-2} \mathbf{k} \]. Since \( \sin(2\pi) = 0 \) and \( \cos(2\pi) = 1 \), the expression simplifies to: \[ \mathbf{r}(2) = 2 \times 0 \mathbf{i} + 2 \times 1 \mathbf{j} + e^{-2} \mathbf{k} \].

Step 3: Simplify the Expression

Simplify the result from the previous step: \[ \mathbf{r}(2) = 0 \mathbf{i} + 2 \mathbf{j} + \frac{1}{e^2} \mathbf{k} \], where \( e^{-2} = \frac{1}{e^2} \). This yields the evaluated vector.

Key Concepts

Parametric Equations

Parametric equations are quite fascinating as they provide a way to describe motion along a path using one or more parameters, typically expressed in terms of time.
In our exercise, we have a vector function \( \mathbf{r}(t) = t \sin(\pi t) \mathbf{i} + t \cos(\pi t) \mathbf{j} + e^{-t} \mathbf{k} \). This function presents a different approach from the standard \( y = f(x) \) equation. Here, each component of the vector is defined by a separate function of \( t \), which means:

• The \( x \)-component is \( t \sin(\pi t) \).
• The \( y \)-component is \( t \cos(\pi t) \).
• The \( z \)-component is \( e^{-t} \).

Each of these components changes as \( t \) changes, providing distinct values that together form a trajectory in three-dimensional space. This is particularly useful for modeling real-world scenarios like the path of a moving object.

3D Space

Understanding 3D space is crucial when analyzing vector functions. Unlike 2D space, where a point is defined by two coordinates \((x, y)\), 3D space requires a third coordinate \((x, y, z)\).
In the given exercise, the vector function \( \mathbf{r}(t) \) traces a path through this three-dimensional space.

• The \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) components are not arbitrary—they represent the unit vectors along the \( x \), \( y \), and \( z \) axes, respectively.
• This means that any position described by \( \mathbf{r}(t) \) can be decomposed into its respective influence along these three axes.

By evaluating \( \mathbf{r}(t) \) at a specific \( t \) value, like \( t = 2 \) in the exercise, we find a single point in this 3D space: \( 0 \mathbf{i} + 2 \mathbf{j} + \frac{1}{e^2} \mathbf{k} \). This point literally has no displacement along the \( x \)-axis, a full 2 units in the \( y \)-axis, and \( \frac{1}{e^2} \) units in the \( z \)-axis.

Exponential Function

The exponential function is an essential component of mathematics, frequently appearing in different areas like calculus, differential equations, and complex system modeling. In our exercise, we encounter it in the \( z \)-component: \( e^{-t} \).
This function is particularly interesting because it describes exponential decay:

• As \( t \) increases, \( e^{-t} \) rapidly approaches zero. This is because the negative exponent causes the value to decrease.
• The property of exponential decay makes it ideal for modeling processes where things diminish over time, like radioactive decay or cooling in thermodynamics.

For \( t = 2 \), \( e^{-2} \) results in \( \frac{1}{e^2} \), a fractional value showing this decline in the \( z \)-direction within our 3D vector space. Understanding how exponential functions behave is pivotal for interpreting dynamic systems and their transformations over time.