Problem 50 - Thomas Calculus

Question

Use a CAS to perform the following steps in Exercises \(49-52 .\) \(\begin{array}{l}{\text { a. Plot the space curve traced out by the position vector } \mathbf{r} \text { . }} \\ {\text { b. Find the components of the velocity vector } d \mathbf{r} / d t \text { . }} \\ {\text { c. Evaluate } d \mathbf{r} / d t \text { at the given point } t_{0} \text { and determine the equa- }} \\ {\text { tion of the tangent line to the curve at } \mathbf{r}\left(t_{0}\right) .} \\ {\text { d. Plot the tangent line together with the curve over the given }} \\ {\text { interval. }}\end{array}\) $$ \mathbf{r}(t)=\sqrt{2} t \mathbf{i}+e^{t} \mathbf{j}+e^{-1} \mathbf{k}, \quad-2 \leq t \leq 3, \quad t_{0}=1 $$

Step-by-Step Solution

Answer

The tangent line is \( \mathbf{r}(t) = (\sqrt{2}, e, \frac{1}{e}) + t(\sqrt{2}, e, 0) \).

Step 1: Plot the Space Curve

We begin by plotting the space curve traced out by the position vector \( \mathbf{r}(t) = \sqrt{2} t \mathbf{i} + e^{t} \mathbf{j} + e^{-1} \mathbf{k} \). Use a computer algebra system (CAS) to graph this vector function over the interval \(-2 \leq t \leq 3\). The plot should illustrate how the vector evolves in three-dimensional space as \(t\) varies.

Step 2: Find Velocity Vector Components

The velocity vector \( \frac{d\mathbf{r}}{dt} \) is the derivative of the position vector with respect to \(t\). Calculate each component: \[ \frac{d}{dt}(\sqrt{2} t) \mathbf{i} = \sqrt{2} \mathbf{i}, \quad \frac{d}{dt}(e^{t}) \mathbf{j} = e^{t} \mathbf{j}, \quad \frac{d}{dt}(e^{-1}) \mathbf{k} = 0 \mathbf{k}. \] Hence, \( \frac{d\mathbf{r}}{dt} = \sqrt{2} \mathbf{i} + e^{t} \mathbf{j}. \)

Step 3: Evaluate Velocity at \(t_0\)

Substitute \(t_0 = 1\) into the expression for \( \frac{d\mathbf{r}}{dt} \). Thus, \[ \left. \frac{d\mathbf{r}}{dt} \right|_{t=1} = \sqrt{2} \mathbf{i} + e^{1} \mathbf{j} = \sqrt{2} \mathbf{i} + e \mathbf{j}. \] This gives the components of the velocity vector at \(t = 1\).

Step 4: Determine Tangent Line Equation

The tangent line to the curve at \( \mathbf{r}(t_0) \) has the direction of the velocity vector at \(t_0\) and passes through the point \( \mathbf{r}(t_0)\). Calculate \( \mathbf{r}(1) = \sqrt{2} \cdot 1 \mathbf{i} + e^1 \mathbf{j} + e^{-1} \mathbf{k} = \sqrt{2} \mathbf{i} + e \mathbf{j} + \frac{1}{e} \mathbf{k} \). The equation is \[ \mathbf{r}(t) = (\sqrt{2} \mathbf{i} + e \mathbf{j} + \frac{1}{e} \mathbf{k}) + t(\sqrt{2} \mathbf{i} + e \mathbf{j}). \]

Step 5: Plot Tangent Line with Curve

Using the CAS, plot the tangent line derived in Step 4 together with the space curve from Step 1. Make sure the plot covers the interval \(-2 \leq t \leq 3\) to clearly show how the tangent line is aligned with the curve at \(t = 1\).

Key Concepts

Space Curve

A space curve is essentially a path traced by a moving point in three-dimensional space. It's described by a vector function, such as \( \mathbf{r}(t) = \sqrt{2} t \mathbf{i} + e^{t} \mathbf{j} + e^{-1} \mathbf{k} \), over an interval \(-2 \leq t \leq 3\). This vector function dictates the position of a point at any given time \(t\). To visualize this, you can plot the vector function using a Computational Algebra System (CAS), which helps illustrate how the curve moves and changes direction as \(t\) varies.

• The \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) components represent movements along the x, y, and z axes, respectively.
• The expressions \(\sqrt{2} t\), \(e^{t}\), and \(e^{-1}\) define how far the point moves along each axis as \(t\) changes.
• Plotting helps in V/visualizing the path and understanding how complex curves behave in space.

Use CAS tools, like GeoGebra or Desmos, to gain a clear visual insight, making it easier to grasp complex motion in 3D space by observing the path of the curve.

Velocity Vector

The velocity vector is crucial when analyzing motion along a space curve. It's derived from the derivative of the position vector, giving us the rate of change of position over time. For our vector function \(\mathbf{r}(t)\), we calculate: \[ \frac{d \mathbf{r}}{dt} = \sqrt{2} \mathbf{i} + e^{t} \mathbf{j} \]

• The \(\sqrt{2} \mathbf{i}\) component indicates a constant rate along the \(x\)-axis.
• The \(e^{t} \mathbf{j}\) component shows that the speed changes exponentially with \(t\) along the \(y\)-axis.
• \(0 \mathbf{k}\) represents no change in the \(z\)-direction over time.

Evaluating this at \(t = 1\), we find the velocity vector components are \(\sqrt{2} \mathbf{i} + e \mathbf{j}\). This calculation is insightful as it lets us understand how quickly the object is moving and in what direction at any point in time on the curve.

Tangent Line

Imagine a tangent line as a line that just *touches* the curve at one point and has the same direction as the curve at that point. To find this line, we need two things:- A point on the curve, found by substituting \(t_0 = 1\) into \(\mathbf{r}(t)\), which yields \(\sqrt{2} \mathbf{i} + e \mathbf{j} + \frac{1}{e} \mathbf{k}\).- The direction determined by the velocity vector at this point, which is \(\sqrt{2} \mathbf{i} + e \mathbf{j}\).The formula for the tangent line is \[ \mathbf{r}(t) = (\sqrt{2} \mathbf{i} + e \mathbf{j} + \frac{1}{e} \mathbf{k}) + t(\sqrt{2} \mathbf{i} + e \mathbf{j}) \]This equation represents a line that not only passes through a point on the curve but has its direction following the path's direction exactly at that point. Such calculations are essential for predicting paths and analyzing motion in physics.

Computational Tools in Calculus

In modern calculus, computational tools like Computer Algebra Systems (CAS) are essential. They not only simplify calculations but also allow for the visualization and manipulation of complex functions. Tools like GeoGebra, Wolfram Alpha, or MATLAB can plot vectors, compute derivatives, and even help visualize tangent lines and curves.

• Plotting tools can show a curve and its tangent line, emphasizing relationships between functions.
• They simplify derivative calculations, allowing for easy handling of complex expressions.
• Visualization aids in understanding abstract calculus concepts by providing a graphical representation.

In calculus education, these tools transform abstract concepts into visual stories, offering dynamic and interactive ways to explore mathematical phenomena.