Problem 51 - Thomas Calculus

Question

Use a CAS to perform the following steps in Exercises \(49-52 .\) \(\begin{array}{l}{\text { a. Plot the space curve traced out by the position vector } \mathbf{r} \text { . }} \\ {\text { b. Find the components of the velocity vector } d \mathbf{r} / d t \text { . }} \\ {\text { c. Evaluate } d \mathbf{r} / d t \text { at the given point } t_{0} \text { and determine the equa- }} \\ {\text { tion of the tangent line to the curve at } \mathbf{r}\left(t_{0}\right) .} \\ {\text { d. Plot the tangent line together with the curve over the given }} \\ {\text { interval. }}\end{array}\) $$ \begin{array}{l}{\mathbf{r}(t)=(\sin 2 t) \mathbf{i}+(\ln (1+t)) \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 4 \pi} \\ {t_{0}=\pi / 4}\end{array} $$

Step-by-Step Solution

Answer

The tangent line at \( t_0 = \frac{\pi}{4} \) is \( \mathbf{L}(t) = (1 \mathbf{i} + 0.569 \mathbf{j} + \frac{\pi}{4} \mathbf{k}) + t(\frac{4}{5} \mathbf{j} + \mathbf{k}) \).

Step 1: Plot the Space Curve

Start by plotting the space curve defined by the position vector \( \mathbf{r}(t) = (\sin 2t) \mathbf{i} + (\ln(1+t)) \mathbf{j} + t \mathbf{k} \) over the interval \( 0 \leq t \leq 4\pi \). Use a graphing tool or a CAS (Computer Algebra System) such as Desmos, GeoGebra, or Mathematica to visualize this 3-dimensional curve.

Step 2: Compute the Velocity Vector Components

Find the components of the velocity vector by differentiating the position vector \( \mathbf{r}(t) \) with respect to \( t \). Thus, calculate:\[\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \left( \frac{d}{dt}(\sin 2t), \frac{d}{dt}(\ln(1+t)), \frac{d}{dt}(t) \right)\]This results in \( \mathbf{v}(t) = (2 \cos 2t) \mathbf{i} + \frac{1}{1+t} \mathbf{j} + \mathbf{k} \).

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

Evaluate the velocity vector \( \mathbf{v}(t) \) at \( t_0 = \frac{\pi}{4} \). Substitute \( t_0 \) into the velocity vector:\[\mathbf{v}\left(\frac{\pi}{4}\right) = (2 \cos \left(2 \times \frac{\pi}{4}\right)) \mathbf{i} + \frac{1}{1 + \frac{\pi}{4}} \mathbf{j} + \mathbf{k}\]Computing gives \( \mathbf{v}\left(\frac{\pi}{4}\right) = (2 \times 0) \mathbf{i} + \frac{1}{1.785} \mathbf{j} + \mathbf{k} = \frac{4}{5} \mathbf{j} + \mathbf{k} \).

Step 4: Determine the Tangent Line Equation

The equation of the tangent line at \( \mathbf{r}(t_0) \) is given by:\[\mathbf{L}(t) = \mathbf{r}(t_0) + t \cdot \mathbf{v}(t_0)\]Evaluate \( \mathbf{r}(t_0) \) by substituting \( t_0 = \frac{\pi}{4} \) into \( \mathbf{r}(t) \):\[\mathbf{r}\left(\frac{\pi}{4}\right) = (\sin \frac{\pi}{2}) \mathbf{i} + \ln(1 + \frac{\pi}{4}) \mathbf{j} + \frac{\pi}{4} \mathbf{k} = \mathbf{i} + 0.569 \mathbf{j} + \frac{\pi}{4} \mathbf{k}\]The tangent line is:\[\mathbf{L}(t) = (1 \mathbf{i} + 0.569 \mathbf{j} + \frac{\pi}{4} \mathbf{k}) + t \left(0 \mathbf{i} + \frac{4}{5} \mathbf{j} + 1 \mathbf{k}\right)\]

Step 5: Plot the Tangent Line and Curve

Plot the tangent line along with the original space curve within the interval \( 0 \leq t \leq 4\pi \). Use a graphical tool to illustrate the space curve and superimpose the tangent line at \( t_0 = \frac{\pi}{4} \). This visualization helps in understanding the relationship between the curve and its tangent at the specified point.

Key Concepts

Space Curve

A space curve can be thought of as a path that a point traces as it moves through three-dimensional space. In this exercise, the space curve is represented by the position vector \( \mathbf{r}(t) = (\sin 2t) \mathbf{i} + (\ln(1+t)) \mathbf{j} + t \mathbf{k} \), which traces a unique path as \( t \) varies from \( 0 \) to \( 4\pi \).
You can imagine the space curve like a roller coaster track moving up, down, twisting, and turning in space. Using a CAS (Computer Algebra System) like Desmos or GeoGebra helps visualize this tracking of the curve. It shows how the curve behaves and the patterns it forms as \( t \) changes within the given interval.
A graphical representation of the space curve allows us to see intersections, loops, and even where the curve speeds up or slows down. This visualization is crucial in understanding how objects move in 3D space.

Velocity Vector

The velocity vector provides information about how fast and in what direction an object is moving at any given point along a space curve. From a calculus perspective, we find the velocity vector by taking the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \).
This differentiation process yields \( \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = (2 \cos 2t) \mathbf{i} + \frac{1}{1+t} \mathbf{j} + \mathbf{k} \).

• The component \( (2 \cos 2t) \mathbf{i} \) reflects the change in the x-direction.
• The component \( \frac{1}{1+t} \mathbf{j} \) reflects the change in the y-direction.
• The constant component \( \mathbf{k} \) reflects constant change in the z-direction.

At any point \( t \), this vector gives the speed and direction of the movement along the curve. When evaluated at a specific point, like \( t = \frac{\pi}{4} \), it helps us understand exactly how the object is moving at that instant.

Tangent Line

The tangent line to a space curve at a certain point \( t_0 \) is like the best straight-line approximation of the curve at that point. To find this line, we need two pieces of information: the position vector evaluated at \( t_0 \) and the velocity vector at the same point.
Using the exercise, the tangent line at \( t_0 = \frac{\pi}{4} \) is found by the equation \( \mathbf{L}(t) = \mathbf{r}(t_0) + t \cdot \mathbf{v}(t_0) \).
This means:

• Start from the initial point given by \( \mathbf{r}(t_0) = (1 \mathbf{i} + 0.569 \mathbf{j} + \frac{\pi}{4} \mathbf{k}) \).
• Add the direction and magnitude of movement from \( \mathbf{v}(t_0) = 0 \mathbf{i} + \frac{4}{5} \mathbf{j} + \mathbf{k} \).

By plotting this line with the curve, you can visually see how the tangent line just 'kisses' the curve at \( t_0 \), matching the direction of the curve at this precise point.

Position Vector

A position vector \( \mathbf{r}(t) \) is used to describe a given point's location in three-dimensional space relative to an origin. It depends on the parameter \( t \), which often represents time. In this case, the position vector \( \mathbf{r}(t) = (\sin 2t) \mathbf{i} + (\ln(1+t)) \mathbf{j} + t \mathbf{k} \) uniquely defines the position of the object on the space curve for any \( t \) within the interval \[ 0 \leq t \leq 4\pi \].

• The \( \sin 2t \) component affects how the curve weaves in and out along the x-axis.
• The \( \ln(1+t) \) component manages the rise and fall along the y-axis.
• The \( t \) component ensures constant elevation change along the z-axis.

At any specific \( t \), the position vector gives us a direct 'snapshot' of the point's location on the 3D path.
Evaluating at \( t = \frac{\pi}{4} \) provides the point \( \mathbf{r}(t_0) = (1 \mathbf{i} + 0.569 \mathbf{j} + \frac{\pi}{4} \mathbf{k}) \), providing a precise location in the space where the tangent line is calculated.