Problem 11 - Thomas Calculus

Question

In Exercises \(11-14,\) find the arc length parameter along the curve from the point where \(t=0\) by evaluating the integral $$ s=\int_{0}^{t}|\mathbf{v}(\tau)| d \tau $$ from Equation \((3) .\) Then find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=(4 \cos t) \mathbf{i}+(4 \sin t) \mathbf{j}+3 t \mathbf{k}, \quad 0 \leq t \leq \pi / 2 $$

Step-by-Step Solution

Answer

The length of the curve is \( \frac{5\pi}{2} \).

Step 1: Find the derivative of the position vector

Given the position vector \( \mathbf{r}(t) = (4 \cos t) \mathbf{i} + (4 \sin t) \mathbf{j} + 3t \mathbf{k} \), we first find \( \mathbf{v}(t) \), the velocity vector, by differentiating each component of \( \mathbf{r}(t) \) with respect to \( t \).\[ \mathbf{v}(t) = \frac{d}{dt}[4 \cos t] \mathbf{i} + \frac{d}{dt}[4 \sin t] \mathbf{j} + \frac{d}{dt}[3t] \mathbf{k} \]\[ \mathbf{v}(t) = -4\sin t \mathbf{i} + 4\cos t \mathbf{j} + 3 \mathbf{k} \]

Step 2: Find the magnitude of the velocity vector

Compute the magnitude of \( \mathbf{v}(t) \), denoted as \( |\mathbf{v}(t)| \), which involves squaring each component, summing them, and taking the square root.\[ |\mathbf{v}(t)| = \sqrt{(-4\sin t)^2 + (4\cos t)^2 + 3^2} \]\[ = \sqrt{16\sin^2 t + 16\cos^2 t + 9} \]\[ = \sqrt{16(\sin^2 t + \cos^2 t) + 9} \]Since \( \sin^2 t + \cos^2 t = 1 \), the expression simplifies to:\[ = \sqrt{16 \cdot 1 + 9} \]\[ = \sqrt{25} \]\[ = 5 \]

Step 3: Set up the integral for arc length

Use the formula for arc length parameter \( s \) to set up the integral from \( 0 \) to \( t \) with the magnitude found.\[ s = \int_{0}^{t} |\mathbf{v}(\tau)| \, d\tau = \int_{0}^{t} 5 \, d\tau \]

Step 4: Evaluate the integral

Perform the integration.\[ s = \int_{0}^{t} 5 \, d\tau = [5\tau]_{0}^{t} \]\[ = 5t - 5 \times 0 \]\[ = 5t \]

Step 5: Find the length of the curve portion

To find the length of the curve from \( t = 0 \) to \( t = \pi/2 \), substitute \( t = \pi/2 \) into \( s = 5t \).\[ s = 5 \left( \frac{\pi}{2} \right) \]\[ = \frac{5\pi}{2} \]

Key Concepts

Velocity Vector

The velocity vector is crucial in understanding the motion of an object along a curve. It represents the rate of change of the position vector, effectively describing how fast and in which direction an object is moving. To determine the velocity vector, we differentiate the position vector \( \mathbf{r}(t) \) with respect to time \( t \).
In our exercise, the position vector is given by:

• \( \mathbf{r}(t) = (4 \cos t) \mathbf{i} + (4 \sin t) \mathbf{j} + 3t \mathbf{k} \)

Differentiating each component with respect to \( t \) gives us the velocity vector:

• \( \mathbf{v}(t) = -4\sin t \mathbf{i} + 4\cos t \mathbf{j} + 3 \mathbf{k} \)

This vector describes how the position changes as \( t \) varies, with each component representing the velocity in the direction of the corresponding coordinate axis. Understanding the velocity vector is foundational for calculating other properties, such as speed and arc length.

Integrals

Integrals are a fundamental tool in calculus, used to calculate areas, volumes, and lengths. In this arc length problem, we use integrals to sum an infinite number of infinitesimally small lengths, which gives us the total arc length of a curve.
The arc length parameter \( s \) is defined as:

• \( s = \int_{0}^{t} |\mathbf{v}(\tau)| \, d\tau \)

This integral computes the accumulation of small distances along the curve from the starting point \( t = 0 \) to a point \( t \) along the curve.
In the exercise, after determining the magnitude of the velocity vector to be constant at 5, the integral simplifies to:

• \( s = \int_{0}^{t} 5 \, d\tau = 5t \)

This tells us that as time progresses, the arc length grows linearly since the function is a simple integral of a constant.

Magnitude of a Vector

The magnitude of a vector, often referred to as the length or norm, quantifies the size or extent of the vector without considering its direction. To compute the magnitude of a vector \( \mathbf{v}(t) \), we sum the squares of its components and then take the square root.
For the velocity vector found in this exercise, \( \mathbf{v}(t) = -4\sin t \mathbf{i} + 4\cos t \mathbf{j} + 3\mathbf{k} \), the magnitude is:

• \( |\mathbf{v}(t)| = \sqrt{(-4\sin t)^2 + (4\cos t)^2 + 3^2} \)
• Simplifies using the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \): \( |\mathbf{v}(t)| = \sqrt{16 \cdot 1 + 9} = \sqrt{25} = 5 \)

This constant magnitude of 5 in the exercise signifies uniform motion along the curve, making the computation of arc length straightforward when integrating.