Problem 36 - Thomas Calculus

Question

Use a CAS to perform the following steps. \(\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}$$ \mathbf{r}(t)=\sqrt{2} t \mathbf{i}+e^{t} \mathbf{j}+e^{-i} \mathbf{k}, \quad-2 \leq t \leq 3, \quad t_{0}=1\)

Step-by-Step Solution

Answer

Evaluate derivative at \( t_0 = 1 \) and find tangent line equation. Plot both curve and tangent line.

Step 1: Understanding the Position Vector

Given the position vector \( \mathbf{r}(t) = \sqrt{2}t \mathbf{i} + e^{t} \mathbf{j} + e^{-i} \mathbf{k} \), this is the equation of the curve we need to analyze. In this step, you would input this function into a CAS (Computer Algebra System) to generate a plot over the specified interval \(-2 \leq t \leq 3\).

Step 2: Calculating the Velocity Vector

The velocity vector \( \frac{d\mathbf{r}}{dt} \) is the derivative of the position vector with respect to \( t \). Using a CAS, take the derivative to find \( \frac{d\mathbf{r}}{dt} = \sqrt{2} \mathbf{i} + e^{t} \mathbf{j} + 0 \mathbf{k} \) since the derivative of \( e^{-i} \) with respect to \( t \) is zero.

Step 3: Evaluating the Velocity at \( t_0 = 1 \)

Substitute \( t_0 = 1 \) into the velocity vector to find the components of the velocity at this point: \( \frac{d\mathbf{r}}{dt} \Big|_{t=1} = \sqrt{2} \mathbf{i} + e \mathbf{j} \).

Step 4: Equations of the Tangent Line

The tangent line has a direction vector parallel to the velocity vector at \( t = 1 \). The equation for the tangent line is \( \mathbf{l}(t) = \mathbf{r}(t_0) + t \left( \frac{d\mathbf{r}}{dt} \Big|_{t=1} \right) \). Calculate \( \mathbf{r}(1) = \sqrt{2} \mathbf{i} + e \mathbf{j} + e^{-i} \mathbf{k} = \sqrt{2} \mathbf{i} + e \mathbf{j} + 1 \mathbf{k} \). Now, the tangent line equation is \( \mathbf{l}(t) = (\sqrt{2} + t\sqrt{2}) \mathbf{i} + (e + te) \mathbf{j} + 1 \mathbf{k} \).

Step 5: Plotting the Tangent Line and Space Curve

Using a CAS, plot the space curve \( \mathbf{r}(t) \) and the tangent line \( \mathbf{l}(t) \) over the interval \(-2 \leq t \leq 3\). The CAS will provide a visual representation to see how the tangent line relates to the curve at \( t_0 = 1 \).

Key Concepts

Space Curve

A space curve is a smooth, continuous path in three-dimensional space, traced by a moving point. Imagine it like a roller coaster track that extends in all directions. The curve provides information on how the point moves within the 3D field, determined by the path's shape and orientation. Every space curve can be represented by a position vector function \( \mathbf{r}(t) \), where \( t \) is the parameter that usually represents time.
This function consists of components along the x, y, and z axes, respectively represented by the unit vectors \( \mathbf{i}, \mathbf{j}, \text{ and } \mathbf{k} \). For example, in our case, the space curve \( \mathbf{r}(t) = \sqrt{2}t \mathbf{i} + e^{t} \mathbf{j} + e^{-i} \mathbf{k} \) is defined over the interval \(-2 \leq t \leq 3\).
By plotting this function in a Computer Algebra System (CAS), we can visualize how the curve traverses through three-dimensional space, providing insights into its structure and dynamics.

Position Vector

The position vector is a crucial tool in vector calculus, representing the location of a point in space at a specific time. It's typically expressed in the form \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \), where each component function \( x(t), y(t), \text{ and } z(t) \) defines the point's coordinates.
In the given example problem, the position vector \( \mathbf{r}(t) = \sqrt{2}t \mathbf{i} + e^{t} \mathbf{j} + e^{-i} \mathbf{k} \) describes the path of a point in three-dimensional space.

• The term \( \sqrt{2}t \mathbf{i} \) describes the linear movement in the x-direction.
• \( e^{t} \mathbf{j} \) contributes to the curve's exponential growth in the y-direction.
• \( e^{-i} \mathbf{k} \) provides an interesting visual element, but its derivative with respect to \( t \) is zero, indicating a constant or static movement in the z-direction.

Understanding the position vector is critical as it allows us to determine not only the point's location but also predict its future positions as the parameter \( t \) changes.

Tangent Line

The tangent line to a curve at a given point \( t_{0} \) is a straight line that just "touches" the curve at that point, matching its direction of travel and giving a linear approximation. It's the line of best fit at that exact moment, closely forecasting the immediate direction of motion.
For our space curve \( \mathbf{r}(t) \), finding the tangent line involves calculating the velocity vector, \( \frac{d\mathbf{r}}{dt} \), at a specific point \( t_{0} = 1 \). This vector points in the direction of the curve's path at the moment and is defined by \( \sqrt{2} \mathbf{i} + e \mathbf{j} \).
The equation of the tangent line is \( \mathbf{l}(t) =\mathbf{r}(t_0) + t \left( \frac{d\mathbf{r}}{dt} \Big|_{t=1} \right) \), using \( \mathbf{r}(1) = \sqrt{2} \mathbf{i} + e \mathbf{j} + 1 \mathbf{k} \). This gives us a clear understanding of how the curve behaves around \( t = 1 \). By plotting this line alongside the curve, we can visually grasp its behavior in that local "zoomed-in" area.

Velocity Vector

The velocity vector is the derivative of the position vector with respect to time \( t \), and it shows the speed and direction of the moving point at any given moment on the space curve. Calculating this vector involves differentiating the position vector component-wise.
For instance, the velocity vector derived from \( \mathbf{r}(t) = \sqrt{2}t \mathbf{i} + e^{t} \mathbf{j} + e^{-i} \mathbf{k} \) is \( \frac{d\mathbf{r}}{dt} = \sqrt{2} \mathbf{i} + e^{t} \mathbf{j} \). Notice here, since \( e^{-i} \) treats \( t \) as a constant, its derivative becomes zero in respect to \( t \), resulting in no alteration in the \( \mathbf{k} \) direction.

• The component \( \sqrt{2} \mathbf{i} \) signifies constant velocity along the x-axis.
• \( e^{t} \mathbf{j} \) indicates exponential velocity growth in the y-direction, growing faster as time progresses.

Performing calculations at \( t = 1 \) gives \( \frac{d\mathbf{r}}{dt} \Big|_{t=1} = \sqrt{2} \mathbf{i} + e \mathbf{j} \), precisely defining how fast and in what direction the curve's point moves at that instant. Understanding these concepts enables a deeper comprehension of the dynamics in vector calculus.