Problem 35 - Thomas Calculus

Question

Use a CAS to perform the following steps in Exercises \(35-38\) \begin{equation} \begin{array}{l}{\text { a. Plot the space curve traced out by the position vector } \mathbf{r} \text { . }} \\ {\text { b. Find the components of the velocity vector } d \mathbf{r} / d t \text { . }} \\ {\text { c. Evaluate } d \mathbf{r} / d t \text { at the given point } t_{0} \text { and determine the equation of }} \\ {\text { the tangent line to the curve at } \mathbf{r}\left(t_{0}\right) .} \\ {\text { d. Plot the tangent line together with the curve over the given interval. }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\mathbf{r}(t)=(\sin t-t \cos t) \mathbf{i}+(\cos t+t \sin t) \mathbf{j}+t^{2} \mathbf{k}} \\ {0 \leq t \leq 6 \pi, \quad t_{0}=3 \pi / 2}\end{array} \end{equation}

Step-by-Step Solution

Answer

Plot the space curve, find velocity, evaluate it at \( t_0=3\pi/2 \), and plot the tangent line.

Step 1: Plot the Space Curve

To plot the space curve, input the position vector \( \mathbf{r}(t) = (\sin t - t \cos t) \mathbf{i} + (\cos t + t \sin t) \mathbf{j} + t^2 \mathbf{k} \) into a Computer Algebra System (CAS). Set the parameter \( t \) to range from \( 0 \) to \( 6\pi \). The CAS will generate a 3D plot of the curve described by the position vector.

Step 2: Find Velocity Vector Components

Differentiate the position vector \( \mathbf{r}(t) \) with respect to \( t \) to obtain the velocity vector \( \frac{d\mathbf{r}}{dt} \). Apply the derivative to each component of \( \mathbf{r}(t) \): \[ \frac{d\mathbf{r}}{dt} = \left( \cos t + t \sin t \right) \mathbf{i} + \left(-\sin t + t \cos t \right) \mathbf{j} + 2t \mathbf{k} \].

Step 3: Evaluate Velocity at Given Point

Substitute \( t_0 = 3\pi / 2 \) into the velocity vector \( \frac{d\mathbf{r}}{dt} \) to find the specific velocity vector at this point:- For \( \mathbf{i} \) component: \( \cos(3\pi/2) + (3\pi/2) \sin(3\pi/2) = 0 + (3\pi/2)(-1) = -3\pi/2 \).- For \( \mathbf{j} \) component: \( -\sin(3\pi/2) + (3\pi/2) \cos(3\pi/2) = 1 + 0 = 1 \).- For \( \mathbf{k} \) component: \( 2(3\pi/2) = 3\pi \).Thus, the velocity vector at \( t_0 \) is \( -3\pi/2 \mathbf{i} + 1 \mathbf{j} + 3\pi \mathbf{k} \).

Step 4: Determine Tangent Line Equation

The equation of the tangent line at \( t_0 = 3\pi/2 \) is given by:\[ \mathbf{r}(t_0) + \lambda \left( \frac{d\mathbf{r}}{dt} \right) \]At \( t_0 = 3\pi/2 \), substitute the \( \mathbf{r}(t_0) \) into the equation along with the velocity vector:- \( \mathbf{r}(3\pi/2) = (-3\pi/2 - 3\pi/2 \cdot 0) \mathbf{i} + (0 + 3\pi/2 \cdot -1) \mathbf{j} + (3\pi/2)^2 \mathbf{k} = -3\pi/2 \mathbf{i} - 3\pi/2 \mathbf{j} + 9\pi^2/4 \mathbf{k} \).Thus, the tangent line is:\[ (-3\pi/2 + \lambda (-3\pi/2)) \mathbf{i} + (-3\pi/2 + \lambda (1)) \mathbf{j} + (9\pi^2/4 + \lambda (3\pi)) \mathbf{k} \].

Step 5: Plot Tangent Line with Curve

Using the CAS, plot the tangent line equation derived in Step 4 along with the space curve from \( 0 \leq t \leq 6\pi \). Use the tangent line parameters to ensure it becomes visible at \( t_0 = 3\pi/2 \). The plot should show the tangent line intersecting the space curve at \( t_0 \).

Key Concepts

Space Curve

A space curve is a three-dimensional curve that is defined by a position vector as a function of a parameter, usually denoted by \( t \). In this exercise, the position vector \( \mathbf{r}(t) = (\sin t - t \cos t) \mathbf{i} + (\cos t + t \sin t) \mathbf{j} + t^2 \mathbf{k} \) describes the trajectory of a point in space as \( t \) changes.
Space curves can take many shapes and forms as they can bend and twist through 3D space. They are useful for describing paths of moving objects, like a particle moving in space.
To better visualize a space curve, you can use a Computer Algebra System (CAS) to generate a 3D plot, which helps you understand how the curve moves. This visualization aids in grasping how different components like \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) contribute to the curve's structure.

Velocity Vector

The velocity vector is a crucial concept in vector calculus. It represents the rate of change of the position of a point on a space curve. Mathematically, it is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \).
Here, the velocity vector is derived from \( \mathbf{r}(t) \), resulting in \( \frac{d\mathbf{r}}{dt} = (\cos t + t \sin t) \mathbf{i} + (-\sin t + t \cos t) \mathbf{j} + 2t \mathbf{k} \).
By calculating this derivative, we determine how the position vector’s components change as time progresses. This helps us understand the speed and direction of the moving point at any instant. Evaluating the velocity vector at specific points, such as \( t_0 = 3\pi/2 \), gives insight into the object's instantaneous velocity at that particular moment.

Tangent Line

In vector calculus, a tangent line represents the straight line that just "touches" a curve at a specific point. It shows the direction in which the curve is heading at that point.
The equation for the tangent line at a point \( t_0 \) on a space curve is given by \( \mathbf{r}(t_0) + \lambda \left( \frac{d\mathbf{r}}{dt} \right) \). Here, \( \lambda \) is a parameter that extends the tangent from the point of contact in both directions.
For example, at \( t_0 = 3\pi/2 \), we substitute \( \mathbf{r}(t_0) \) and the velocity vector to get the tangent line equation. This line not only provides the direction but also can be plotted alongside the curve to visually verify where the tangent touches the space curve.

Computer Algebra System

A Computer Algebra System (CAS) is a software program designed to perform symbolic mathematics. It provides an invaluable tool for students and professionals working with complex mathematical concepts like space curves and calculus.
In this exercise, a CAS helps to perform several tasks:

• Plot the space curve by inputting the position vector into the system.
• Calculate derivatives to find the velocity vector.
• Evaluate the velocity at a specific point to find the tangent line's equation.
• Visualize both the space curve and the tangent line in a 3D plot.

Using a CAS offers clarity, enabling you to see the intricate relationships and changes within mathematical equations. It simplifies calculations that would otherwise be complex, allowing you to focus more on concepts and less on computational mechanics.