Problem 49 - Thomas Calculus

Question

Use a CAS to perform the following steps in Exercises \(49-52 .\) \(\begin{array}{l}{\text { a. Plot the space curve traced out by the position vector } \mathbf{r} \text { . }} \\ {\text { b. Find the components of the velocity vector } d \mathbf{r} / d t \text { . }} \\ {\text { c. Evaluate } d \mathbf{r} / d t \text { at the given point } t_{0} \text { and determine the equa- }} \\ {\text { tion of the tangent line to the curve at } \mathbf{r}\left(t_{0}\right) .} \\ {\text { d. Plot the tangent line together with the curve over the given }} \\ {\text { interval. }}\end{array}\) $$ \begin{array}{l}{\mathbf{r}(t)=(\sin t-t \cos t) \mathbf{i}+(\cos t+t \sin t) \mathbf{j}+t^{2} \mathbf{k}} \\ {0 \leq t \leq 6 \pi, \quad t_{0}=3 \pi / 2}\end{array} $$

Step-by-Step Solution

Answer

The space curve and its tangent line are identified and plotted using analytical and computational methods.

Step 1: Plot the Space Curve

To begin, plot the space curve described by the position vector \( \mathbf{r}(t) = (\sin t - t \cos t) \mathbf{i} + (\cos t + t \sin t) \mathbf{j} + t^2 \mathbf{k} \) for \( 0 \leq t \leq 6\pi \). Use a computer algebra system (CAS) to generate this plot, which will represent the path traced out by \( \mathbf{r}(t) \) in three-dimensional space.

Step 2: Derive the Velocity Vector

Find the velocity vector \( \frac{d\mathbf{r}}{dt} \) by differentiating \( \mathbf{r}(t) \) with respect to \( t \). Calculate:\[ \frac{d}{dt}((\sin t - t \cos t) \mathbf{i} + (\cos t + t \sin t) \mathbf{j} + t^2 \mathbf{k}) = ((\cos t + t \sin t)\mathbf{i} + (-\sin t + t \cos t)\mathbf{j} + 2t\mathbf{k}) \]

Step 3: Evaluate the Velocity Vector at \( t_0 \)

Substitute \( t_0 = \frac{3\pi}{2} \) into the velocity vector:\[ \mathbf{v}\left(\frac{3\pi}{2}\right) = \left(\cos\left(\frac{3\pi}{2}\right) + \frac{3\pi}{2} \sin\left(\frac{3\pi}{2}\right)\right) \mathbf{i} + \left(-\sin\left(\frac{3\pi}{2}\right) + \frac{3\pi}{2} \cos\left(\frac{3\pi}{2}\right)\right) \mathbf{j} + 2 \cdot \frac{3\pi}{2} \mathbf{k} = \mathbf{i} \]

Step 4: Equation of the Tangent Line

Since the tangent to \( \mathbf{r}(t) \) at \( t_0 \) is given by the point \( \mathbf{r}\left(\frac{3\pi}{2}\right) \) and direction \( \mathbf{v}\left(\frac{3\pi}{2}\right) \), calculate \( \mathbf{r}\left(\frac{3\pi}{2}\right) \): \[ \mathbf{r}\left(\frac{3\pi}{2}\right) = \left(-\frac{3\pi}{2}, 0, \left(\frac{3\pi}{2}\right)^2\right) = \left(-\frac{3\pi}{2}\right) \mathbf{i} + 0 \mathbf{j} + \left(\frac{9\pi^2}{4}\right)\mathbf{k} \] Then, the tangent line is:\[ \mathbf{r}(t) = \left(-\frac{3\pi}{2}\right) \mathbf{i} + \left(\frac{9\pi^2}{4} + t\right) \mathbf{k} \]

Step 5: Plot the Tangent Line with the Curve

Finally, use the CAS to plot the tangent line alongside the original curve \( \mathbf{r}(t) \) for \( 0 \leq t \leq 6\pi \) to visually interpret the behavior of the tangent.

Key Concepts

Space Curves

When dealing with vector calculus, space curves represent paths traced by position vectors in a three-dimensional space. In this context, a position vector like \( \mathbf{r}(t) = (\sin t - t \cos t) \mathbf{i} + (\cos t + t \sin t) \mathbf{j} + t^2 \mathbf{k} \) describes the curve. Here, each component function corresponds to the respective direction in space: \( x \), \( y \), and \( z \). This allows us to see how the curve evolves as the parameter \( t \) changes.

Understanding space curves is fundamental, as it helps visualize the movement of an object or particle in three dimensions. By plotting the space curve using a Computer Algebra System (CAS), you can see the intricate path formed over a given interval, such as \( 0 \leq t \leq 6\pi \). It provides a visual representation that clarifies both the shape and orientation of the curve in 3D space.

Velocity Vector

The velocity vector is a crucial concept in vector calculus, representing the rate of change of the position vector with respect to time. For our given position vector \( \mathbf{r}(t) \), the velocity vector is obtained via differentiation, resulting in \( \frac{d\mathbf{r}}{dt} = (\cos t + t \sin t)\mathbf{i} + (-\sin t + t \cos t)\mathbf{j} + 2t\mathbf{k} \).

Each component of the velocity vector provides insight into how fast and in which direction an object moves along the space curve at any given time. For instance, the component \( 2t\mathbf{k} \) indicates a consistent increase in velocity along the z-axis. This concept extends to finding how the object speeds up or slows down over time, making it foundational for understanding motion within physical systems.

Tangent Line

The tangent line to a curve at a specific point provides a linear approximation of the curve near that point. To find this, you can use the evaluated velocity vector and a point on the curve. For our scenario: the point \( \mathbf{r}\left(\frac{3\pi}{2}\right) = \left(-\frac{3\pi}{2}, 0, \left(\frac{9\pi^2}{4}\right)\right) \) and direction vector \( \mathbf{i} \) (from velocity).

The equation of this tangent line is given by: \( \mathbf{r}(t) = \left(-\frac{3\pi}{2}\right) \mathbf{i} + \left(\frac{9\pi^2}{4} + t\right) \mathbf{k} \). Plotting this tangent offers insights into the curve's behavior, showing how it varies in a linear fashion around a specific point. This concept is vital for approximations in calculus and easing the analyses of complex curves.

Differentiation

Differentiation is the mathematical process of finding the derivative of a function, indicating its rate of change. In vector calculus, we apply differentiation to vector functions like \( \mathbf{r}(t) \) to find velocity or acceleration vectors. By differentiating each component of \( \mathbf{r}(t) \), we obtain the velocity vector.

In our example, the derivative \( \frac{d}{dt}((\sin t - t \cos t) \mathbf{i} + (\cos t + t \sin t) \mathbf{j} + t^2 \mathbf{k}) \) involves using rules like the product and chain rules. This results in breaking down complex movements into understandable parts. Mastery of differentiation techniques facilitates advanced analyses of motion and change within various fields of science and engineering.

3D Plotting

Plotting in three dimensions is a powerful way to visualize complex space curves and their tangent lines. By using a CAS, one can accurately display both the curve \( \mathbf{r}(t) \) and its tangent line. This visualization assists in understanding the spatial relationships and changes in the curve's geometry.

For example, by plotting both the curve and its tangent from \( t = 0 \) to \( t = 6\pi \), the interaction between these elements becomes clearer. This technique is particularly beneficial for verifying mathematical results and conveying concepts that are otherwise difficult to grasp through equations alone. Masterful use of 3D plotting turns theoretical calculations into comprehensible visual dynamics, enhancing both learning and interpretation.