Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

sketch the curve over the indicated domain for \(t\). Find \(\mathbf{v}, \mathbf{a}, \mathbf{T}\), and \(\kappa\) at the point where \(t=t_{1} .\) $$ \mathbf{r}(t)=\frac{t^{2}}{4} \mathbf{i}+2 \cos t \mathbf{j}+2 \sin t \mathbf{k} ; \quad 0 \leq t \leq 4 \pi ; t_{1}=\pi $$

Step-by-Step Solution

Answer

The velocity \( \mathbf{v}(\pi) = \frac{\pi}{2} \mathbf{i} - 2 \mathbf{k} \). The acceleration \( \mathbf{a}(\pi) = \frac{1}{2} \mathbf{i} + 2 \mathbf{j} \). The unit tangent \( \mathbf{T}(\pi) = \frac{\frac{\pi}{2} \mathbf{i} - 2 \mathbf{k}}{\sqrt{\frac{\pi^2}{4} + 4}} \). The curvature \( \kappa(\pi) = \frac{\sqrt{16 + \frac{\pi^2}{4} + \frac{\pi^2}{16}}}{\left( \sqrt{\frac{\pi^2}{4} + 4} \right)^3} \).

Step 1: Calculate the Velocity Vector

Given the position vector \( \mathbf{r}(t) = \frac{t^2}{4} \mathbf{i} + 2 \cos t \mathbf{j} + 2 \sin t \mathbf{k} \), differentiate it with respect to \( t \) to find the velocity vector \( \mathbf{v}(t) \).\[ \mathbf{v}(t) = \frac{d}{dt} \left( \frac{t^2}{4} \right) \mathbf{i} + \frac{d}{dt} (2 \cos t) \mathbf{j} + \frac{d}{dt} (2 \sin t) \mathbf{k} = \frac{t}{2} \mathbf{i} - 2 \sin t \mathbf{j} + 2 \cos t \mathbf{k} \]

Step 2: Calculate the Acceleration Vector

Differentiate the velocity vector \( \mathbf{v}(t) = \frac{t}{2} \mathbf{i} - 2 \sin t \mathbf{j} + 2 \cos t \mathbf{k} \) with respect to \( t \) to find the acceleration vector \( \mathbf{a}(t) \).\[ \mathbf{a}(t) = \frac{d}{dt} \left( \frac{t}{2} \right) \mathbf{i} - \frac{d}{dt} (2 \sin t) \mathbf{j} + \frac{d}{dt} (2 \cos t) \mathbf{k} = \frac{1}{2} \mathbf{i} - 2 \cos t \mathbf{j} - 2 \sin t \mathbf{k} \]

Step 3: Evaluate the Vectors at \( t_1 = \pi \)

Substitute \( t = \pi \) into the velocity and acceleration vectors.For \( \mathbf{v}(\pi) \):\[ \mathbf{v}(\pi) = \frac{\pi}{2} \mathbf{i} - 2 \sin(\pi) \mathbf{j} + 2 \cos(\pi) \mathbf{k} = \frac{\pi}{2} \mathbf{i} + 0 \mathbf{j} - 2 \mathbf{k} \]For \( \mathbf{a}(\pi) \):\[ \mathbf{a}(\pi) = \frac{1}{2} \mathbf{i} - 2 \cos(\pi) \mathbf{j} - 2 \sin(\pi) \mathbf{k} = \frac{1}{2} \mathbf{i} + 2 \mathbf{j} + 0 \mathbf{k} \]

Step 4: Calculate the Tangent Vector \( \mathbf{T}(t) \)

The unit tangent vector \( \mathbf{T}(t) \) is obtained by normalizing the velocity vector \( \mathbf{v}(t) \).\[ \mathbf{T}(t) = \frac{\mathbf{v}(t)}{||\mathbf{v}(t)||} \]At \( t = \pi \), find the magnitude:\[ ||\mathbf{v}(\pi)|| = \sqrt{\left( \frac{\pi}{2} \right)^2 + (-2)^2} = \sqrt{\frac{\pi^2}{4} + 4} \]The tangent vector at \( t = \pi \) is:\[ \mathbf{T}(\pi) = \frac{\frac{\pi}{2} \mathbf{i} - 2 \mathbf{k}}{\sqrt{\frac{\pi^2}{4} + 4}} \]

Step 5: Calculate the Curvature \( \kappa \)

Curvature \( \kappa(t) \) is given by:\[ \kappa(t) = \frac{||\mathbf{v}(t) \times \mathbf{a}(t)||}{||\mathbf{v}(t)||^3} \]Calculate the cross product \( \mathbf{v}(\pi) \times \mathbf{a}(\pi) \) using the vectors \( \mathbf{v}(\pi) = \frac{\pi}{2} \mathbf{i} - 2 \mathbf{k} \) and \( \mathbf{a}(\pi) = \frac{1}{2} \mathbf{i} + 2 \mathbf{j} \):\[ \mathbf{v}(\pi) \times \mathbf{a}(\pi) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\pi}{2} & 0 & -2 \ \frac{1}{2} & 2 & 0 \end{vmatrix} = (0 - (-4)) \mathbf{i} - \left( -\frac{\pi}{2} \right) \mathbf{j} + \left( \frac{\pi}{4} \right) \mathbf{k} = 4 \mathbf{i} + \frac{\pi}{2} \mathbf{j} + \frac{\pi}{4} \mathbf{k}\]Calculate the magnitude of the cross product:\[ ||\mathbf{v}(\pi) \times \mathbf{a}(\pi)|| = \sqrt{4^2 + \left( \frac{\pi}{2} \right)^2 + \left( \frac{\pi}{4} \right)^2} = \sqrt{16 + \frac{\pi^2}{4} + \frac{\pi^2}{16}} \]Finally, calculate the curvature:\[ \kappa(\pi) = \frac{\sqrt{16 + \frac{\pi^2}{4} + \frac{\pi^2}{16}}}{\left( \sqrt{\frac{\pi^2}{4} + 4} \right)^3} \]

Key Concepts

Velocity Vector

In vector calculus, the velocity vector is a fundamental concept when working with motion along a curve. Given a position vector \( \mathbf{r}(t) \), differentiating it with respect to time \( t \) yields the velocity vector, denoted as \( \mathbf{v}(t) \). This vector describes the speed and direction of a point moving along a path. It provides insight into how the position changes over time.

To compute the velocity vector for a curve like \( \mathbf{r}(t)=\frac{t^{2}}{4} \mathbf{i}+2 \cos t \mathbf{j}+2 \sin t \mathbf{k} \), perform the derivative operation on each of the component functions:

• For \( \mathbf{i} \): \( \frac{d}{dt} \left( \frac{t^2}{4} \right) = \frac{t}{2} \)
• For \( \mathbf{j} \): \( \frac{d}{dt} (2 \cos t) = -2 \sin t \)
• For \( \mathbf{k} \): \( \frac{d}{dt} (2 \sin t) = 2 \cos t \)

This results in the velocity vector \( \mathbf{v}(t) = \frac{t}{2} \mathbf{i} - 2 \sin t \mathbf{j} + 2 \cos t \mathbf{k} \). By understanding the velocity vector at a specific point, such as \( t = \pi \), we gain valuable information about the behavior of the moving point at that time.

Acceleration Vector

The acceleration vector is another crucial element in vector calculus, especially in the context of motion analysis. It describes how the velocity of a point changes as it moves along a path. To find the acceleration vector \( \mathbf{a}(t) \), differentiate the velocity vector \( \mathbf{v}(t) \) with respect to time \( t \).

For our example trajectory, the initial velocity vector is \( \mathbf{v}(t) = \frac{t}{2} \mathbf{i} - 2 \sin t \mathbf{j} + 2 \cos t \mathbf{k} \). Differentiating each component of this vector gives:

• \( \frac{d}{dt} \left( \frac{t}{2} \right) = \frac{1}{2} \)
• \( \frac{d}{dt} (-2 \sin t) = -2 \cos t \)
• \( \frac{d}{dt} (2 \cos t) = -2 \sin t \)

This results in the acceleration vector: \( \mathbf{a}(t) = \frac{1}{2} \mathbf{i} - 2 \cos t \mathbf{j} - 2 \sin t \mathbf{k} \).

Evaluating it at \( t = \pi \) provides essential insight into how the velocity of the path is evolving at that specific moment.

Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is pivotal in understanding the direction of a curve at any point. It is a normalized version of the velocity vector, meaning it has a magnitude of 1, purely giving direction.

To find the unit tangent vector, divide the velocity vector \( \mathbf{v}(t) \) by its magnitude \( ||\mathbf{v}(t)|| \). For \( \mathbf{v}(\pi) = \frac{\pi}{2} \mathbf{i} - 2 \mathbf{k} \), the magnitude is computed as:

• \( ||\mathbf{v}(\pi)|| = \sqrt{\left( \frac{\pi}{2} \right)^2 + (-2)^2} = \sqrt{\frac{\pi^2}{4} + 4} \)

The tangent vector at \( t = \pi \) becomes:

\[ \mathbf{T}(\pi) = \frac{\frac{\pi}{2} \mathbf{i} - 2 \mathbf{k}}{\sqrt{\frac{\pi^2}{4} + 4}} \]

This vector is indispensable when analyzing the curve's trajectory, considering it's only concerned with the direction without the influence of speed.

Curvature

Curvature is an aspect of vector calculus that reveals how sharply a curve turns. It provides a measure of the deviation of a curve from being a straight line. Mathematically, curvature \( \kappa(t) \) can be calculated using the cross-product formula:

\[ \kappa(t) = \frac{||\mathbf{v}(t) \times \mathbf{a}(t)||}{||\mathbf{v}(t)||^3} \]

The cross product of \( \mathbf{v}(\pi) \) and \( \mathbf{a}(\pi) \) results in a vector showing perpendicularity related to the change in the direction of the curve. To compute it:

• Use the determinant method for \( \mathbf{v}(\pi) = \frac{\pi}{2} \mathbf{i} - 2 \mathbf{k} \) and \( \mathbf{a}(\pi) = \frac{1}{2} \mathbf{i} + 2 \mathbf{j} \)
• The result is \( 4 \mathbf{i} + \frac{\pi}{2} \mathbf{j} + \frac{\pi}{4} \mathbf{k} \)

The magnitude is:
\[ ||\mathbf{v}(\pi) \times \mathbf{a}(\pi)|| = \sqrt{16 + \frac{\pi^2}{4} + \frac{\pi^2}{16}} \]
Finally, dividing this by \( ||\mathbf{v}(t)||^3 \) offers the curvature at \( t = \pi \). Understanding curvature is crucial for applications like aircraft navigation and roller coaster design, where path behavior needs to be predicted and controlled.