Problem 2 - Essential Calculus Early Transcendentals

Question

Find the length of the curve. $$\mathbf{r}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+\ln \cos t \mathbf{k}, \quad 0 \leqslant t \leqslant \pi / 4$$

Step-by-Step Solution

Answer

The length of the curve is \( \ln 2 \).

Step 1: Find the Derivative

First, we need to find the derivative of the vector function \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + \ln \cos t \mathbf{k} \). The derivative \( \mathbf{r}'(t) \) is calculated by differentiating each component separately:- \( \frac{d}{dt} \cos t = -\sin t \)- \( \frac{d}{dt} \sin t = \cos t \)- \( \frac{d}{dt} \ln \cos t = \frac{-\sin t}{\cos t} = -\tan t \)Thus, the derivative is \( \mathbf{r}'(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} - \tan t \mathbf{k} \).

Step 2: Find the Magnitude of the Derivative

Next, we find the magnitude of \( \mathbf{r}'(t) \), which is given by \(\| \mathbf{r}'(t) \| = \sqrt{(-\sin t)^2 + (\cos t)^2 + (-\tan t)^2}\)This simplifies to:\(\| \mathbf{r}'(t) \| = \sqrt{\sin^2 t + \cos^2 t + \tan^2 t}\)Using the identity \( \sin^2 t + \cos^2 t = 1 \), we have:\(\| \mathbf{r}'(t) \| = \sqrt{1 + \tan^2 t}\)Given \( \tan^2 t = \frac{\sin^2 t}{\cos^2 t} \), the expression simplifies further to \( \| \mathbf{r}'(t) \| = \frac{1}{\cos t} \).

Step 3: Integrate the Magnitude over the Interval

We need to integrate the magnitude over the given interval \( [0, \frac{\pi}{4}] \). The length of the curve \( L \) is given by:\(L = \int_{0}^{\frac{\pi}{4}} \frac{1}{\cos t} \, dt = \int_{0}^{\frac{\pi}{4}} \sec t \, dt\)We know from calculus that \( \int \sec t \, dt = \ln |\sec t + \tan t| + C \). Thus, we evaluate:\(L = \left[ \ln |\sec t + \tan t| \right]_{0}^{\frac{\pi}{4}}\) { \(= \ln (\sec(\frac{\pi}{4}) + \tan(\frac{\pi}{4})) - \ln (\sec(0) + \tan(0))\)Both \( \sec(\frac{\pi}{4}) \) and \( \tan(\frac{\pi}{4}) \) are \( \sqrt{2} \). At \( t = 0 \), \( \sec(0) = 1 \) and \( \tan(0) = 0 \). Calculating, we find:\(L = \ln \sqrt{2} + \ln \sqrt{2} - \ln 1\)\(=2 \ln \sqrt{2}\)\(= \ln 2\) since \( 2 \ln \sqrt{2} = \ln 2 \).

Step 4: Final Answer

Thus, the length of the curve from \( t = 0 \) to \( t = \frac{\pi}{4} \) is \( \ln 2 \).

Key Concepts

Curve Length

The concept of curve length is essential in vector calculus, particularly when working with vector functions that describe paths in space. Curve length, often denoted as \( L \), represents the distance along the curve between two points. This is akin to measuring the distance along a path you might walk rather than the straight line from start to finish.
To find the curve length, one often computes the integral of the magnitude of the derivative of the vector function, \( \mathbf{r}'(t) \), over a specified interval. This derivative, \( \mathbf{r}'(t) \), reflects the curve's "direction and speed" at each point and helps us account for all axes’ contributions along the path.
Integrating the magnitude of this derivative over the path gives us the total length. This length isn't just the arc length in a two-dimensional plane but may involve three dimensions, as our example with a vector having \(i\), \(j\), and \(k\) components.

Derivative of Vector Functions

The derivative of a vector function is key when assessing a curve's properties, such as its length. For vector function \( \mathbf{r}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + \ln \cos t \mathbf{k} \), the derivative \( \mathbf{r}'(t) \) tells us how the curve changes with respect to the parameter \( t \).
Computing this derivative involves differentiating each component separately:

• The derivative of \( \cos t \) is \(-\sin t \)
• The derivative of \( \sin t \) is \( \cos t \)
• The derivative of \( \ln \cos t \) involves the chain rule: \( \frac{-\sin t}{\cos t} = -\tan t \)

Putting these together, the derivative becomes \( \mathbf{r}'(t) = -\sin t \mathbf{i} + \cos t \mathbf{j} - \tan t \mathbf{k} \).
This vector informs us of the rate of change in each spatial direction, which is essential for calculating further attributes of the curve, like its length.

Integration of Trigonometric Functions

Integration is a process of finding the accumulated value, like finding the total curve length over an interval. When dealing with trigonometric functions, it often involves solving integrals like \( \int \sec t \, dt \), which can appear challenging.
In our problem, the task was to integrate \( \sec t \) over the interval \([0, \frac{\pi}{4}]\). The integral of \( \sec t \) is known: \( \ln |\sec t + \tan t| + C \), where \( C \) is the constant of integration.
To find the definite integral, we evaluate this expression at the bounds of the interval:

• At \( t = \frac{\pi}{4} \), using trigonometric identities, both \( \sec \) and \( \tan \) equal \( \sqrt{2} \).
• At \( t = 0 \), \( \sec 0 = 1 \) and \( \tan 0 = 0 \).

This calculation yields a final result of \( \ln 2 \), representing the curve's total length. These evaluations simplify understanding and applying trigonometric integrals in calculus.