Question

A position function \(\vec{r}(t)\) is given, where \(t=0\) corresponds to the initial position. Find the arc length parameter \(s,\) and rewrite \(\vec{r}(t)\) in terms of \(s ;\) that is, find \(\vec{r}(s)\). \(\vec{r}(t)=\langle 5 \cos t, 13 \sin t, 12 \cos t\rangle\)

Short answer

The arc length parameter is \(s = 13t\), and \(\vec{r}(s) = \langle 5 \cos(\frac{s}{13}), 13 \sin(\frac{s}{13}), 12 \cos(\frac{s}{13}) \rangle\).

Step-by-step solution

Understand the Problem

The problem asks us to find an arc length parameter \(s\) for the given vector function \(\vec{r}(t)\), and then express \(\vec{r}(t)\) in terms of \(s\), i.e., find \(\vec{r}(s)\). The arc length parameter \(s\) is calculated from the derivative of \(\vec{r}(t)\).

Calculate Derivative \(\vec{r}'(t)\)

Compute the derivative of \(\vec{r}(t)\), which is \(\vec{r}'(t) = \langle -5 \sin t, 13 \cos t, -12 \sin t \rangle\).

Compute the Magnitude of \(\vec{r}'(t)\)

The magnitude of \(\vec{r}'(t)\) is given by \(\|\vec{r}'(t)\| = \sqrt{(-5 \sin t)^2 + (13 \cos t)^2 + (-12 \sin t)^2}\). Simplifying inside the square root gives \(\sqrt{25 \sin^2 t + 169 \cos^2 t + 144 \sin^2 t}\).

Simplify the Magnitude Expression

Simplify \(\|\vec{r}'(t)\| = \sqrt{(25 + 144) \sin^2 t + 169 \cos^2 t} = \sqrt{169}\), which further simplifies to \(13\) because \(\sin^2 t + \cos^2 t = 1\). Hence, \(\|\vec{r}'(t)\| = 13\).

Find Arc Length Parameter \(s(t)\)

The arc length \(s\) from \(t = 0\) to \(t\) is \(s(t) = \int_{0}^{t} \|\vec{r}'(u)\| \, du = \int_{0}^{t} 13 \, du = 13t\).

Solve for \(t\) in Terms of \(s\)

From the equation \(s = 13t\), solve for \(t\): \(t = \frac{s}{13}\).

Substitute \(t(s)\) into \(\vec{r}(t)\)

Substitute \(t = \frac{s}{13}\) into \(\vec{r}(t)\): \(\vec{r}(s) = \langle 5 \cos \left(\frac{s}{13}\right), 13 \sin \left(\frac{s}{13}\right), 12 \cos \left(\frac{s}{13}\right) \rangle\).

Key concepts

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector-valued functions. We often work with functions that describe movement in space. These functions can represent motion along a path, such as the trajectory of a planet or an object moving along a spiral.
For example, a position function like \( \vec{r}(t) \) encodes information about the position of an object at any given time \( t \). This information is typically expressed in terms of dimensions such as \( x, y, \) and \( z \) coordinates.
When dealing with vectors in calculus, we focus on tasks like finding velocity, represented by derivatives, and calculating the length of paths, also known as arc length. Understanding how to differentiate and integrate these vector functions is crucial in applying vector calculus.

Position Function

A position function \( \vec{r}(t) \) describes the trajectory of a particle or object over time. Each component of the vector corresponds to a coordinate in space. For example, in the exercise, \( \vec{r}(t) = \langle 5 \cos t, 13 \sin t, 12 \cos t \rangle \) lets us track how the object moves in 3D space.
Generally, the position function uses parameters such as time \( t \) to express where the object is at any moment. This is vital for calculations involving motion analysis, such as velocity or acceleration.
To derive meaningful insights, transforming this function using the arc length parameter \( s \) can sometimes simplify the representation of the object's trajectory and offer a clearer understanding of the movement.

Derivative Calculation

Calculating the derivative of a vector function like \( \vec{r}(t) \) gives us the velocity vector \( \vec{r}'(t) \). This represents the rate at which the position changes with respect to time.
In our example, the calculation yields \( \vec{r}'(t) = \langle -5 \sin t, 13 \cos t, -12 \sin t \rangle \). Understanding this derivative helps us pinpoint instant velocity at any point \( t \) during the object's path.
This step is crucial because the magnitude of this velocity vector is needed to find the arc length parameter \( s \), which assists in re-parametrizing the function using \( s \) instead of \( t \).
It also connects to the calculation of total path length, allowing for applications in everything from physics to computer graphics.

Magnitude Simplification

The magnitude simplification step involves finding the length of the velocity vector obtained from the derivative. Calculating the magnitude involves using the formula \( \|\vec{r}'(t)\| = \sqrt{(-5 \sin t)^2 + (13 \cos t)^2 + (-12 \sin t)^2} \).
After simplifying this expression, we find that it reduces to \( 13 \), as shown by the expression \( \sqrt{25 \sin^2 t + 169 \cos^2 t + 144 \sin^2 t} = \sqrt{169} \).
This simplification occurs because \( \sin^2 t + \cos^2 t = 1 \), which is a Pythagorean identity.
This crucial step allows us to easily calculate the arc length parameter, a key element needed to re-express \( \vec{r}(t) \) in terms of \( s \). By understanding such simplifications, we can manage complex problem-solving tasks in vector calculus more effectively.