Problem 58 - Thomas Calculus

Question

Use a CAS to perform the following steps in Exercises \(58-61 .\) a. Plot the space curve traced out by the position vector \(\mathbf{r}\) . b. Find the components of the velocity vector \(d \mathbf{r} / d t\) . c. Evaluate \(d \mathbf{r} / d t\) at the given point \(t_{0}\) and determine the equation of the tangent line to the curve at \(\mathbf{r}\left(t_{0}\right) .\) d. Plot the tangent line together with the curve over the given interval. $$ \begin{array}{l}{\mathbf{r}(t)=(\sin t-t \cos t) \mathbf{i}+(\cos t+t \sin t) \mathbf{j}+t^{2} \mathbf{k}} \\ {0 \leq t \leq 6 \pi, \quad t_{0}=3 \pi / 2}\end{array} $$

Step-by-Step Solution

Answer

The velocity vector at \( t_0 = \frac{3\pi}{2} \) is \( \mathbf{i} + \mathbf{j} + 3\pi \mathbf{k} \). The tangent line is \( \mathbf{r}_{tangent} = t \mathbf{i} + t \mathbf{j} + \left(\frac{9\pi^2}{4} + 3\pi t\right) \mathbf{k} \).

Step 1: Plot Space Curve

Use a Computer Algebra System (CAS) to plot the 3D space curve described by the position vector \( \mathbf{r}(t) = (\sin t - t \cos t) \mathbf{i} + (\cos t + t \sin t) \mathbf{j} + t^2 \mathbf{k} \) over the interval \( 0 \leq t \leq 6\pi \). Observe the shape of the curve in 3-dimensional space.

Step 2: Find Velocity Vector Components

Determine the components of the velocity vector by differentiating the position vector \( \mathbf{r}(t) \) with respect to \( t \). The velocity vector \( \frac{d \mathbf{r}}{dt} \) is found by computing:\[ \frac{d \mathbf{r}}{dt} = \left( \frac{d}{dt}(\sin t - t \cos t) \right) \mathbf{i} + \left( \frac{d}{dt}(\cos t + t \sin t) \right) \mathbf{j} + \frac{d}{dt}(t^2) \mathbf{k} \]This results in:\[ \frac{d \mathbf{r}}{dt} = (\cos t + t \sin t) \mathbf{i} + (-\sin t + t \cos t) \mathbf{j} + 2t \mathbf{k} \]

Step 3: Evaluate at Given Point

Substitute \( t_0 = \frac{3\pi}{2} \) into the velocity vector \( \frac{d \mathbf{r}}{dt} \) to find the velocity vector at this point:\[ \frac{d \mathbf{r}}{dt} \bigg|_{t = \frac{3\pi}{2}} = \left( \cos \frac{3\pi}{2} + \frac{3\pi}{2} \sin \frac{3\pi}{2} \right) \mathbf{i} + \left( -\sin \frac{3\pi}{2} + \frac{3\pi}{2} \cos \frac{3\pi}{2} \right) \mathbf{j} + 2 \cdot \frac{3\pi}{2} \mathbf{k} \]This simplifies to:\[ \frac{d \mathbf{r}}{dt} \bigg|_{t = \frac{3\pi}{2}} = \mathbf{i} + \mathbf{j} + 3\pi \mathbf{k} \]

Step 4: Determine Tangent Line Equation

With the velocity vector at \( t_0 = \frac{3\pi}{2} \) as \( \mathbf{i} + \mathbf{j} + 3\pi \mathbf{k} \), determine the equation of the tangent line at this point:- The position vector at \( t_0 \) is computed as \( \mathbf{r}\left( \frac{3\pi}{2} \right) = (0\mathbf{i} + 0\mathbf{j} + \left(\frac{3\pi}{2}\right)^2 \mathbf{k}) \).- Therefore, the tangent line equation is:\[ \mathbf{r}_{tangent} = \mathbf{r}\left(\frac{3\pi}{2}\right) + t \cdot \left(\mathbf{i} + \mathbf{j} + 3\pi \mathbf{k}\right) \]Which simplifies to:\[ \mathbf{r}_{tangent} = (0 + t) \mathbf{i} + (0 + t) \mathbf{j} + \left(\frac{9\pi^2}{4} + 3\pi t\right) \mathbf{k} \]

Step 5: Plot Tangent Line with Curve

Using the CAS, plot the tangent line derived in the previous step together with the space curve from Step 1 over the interval \( 0 \leq t \leq 6\pi \). This visualization will show how the tangent line just touches the curve at \( t_0 = \frac{3\pi}{2} \) without crossing it.

Key Concepts

Parametric Equations

When dealing with curves in space, we often represent them using parametric equations. A parametric equation is a way of expressing a curve by using a parameter to describe each component of the curve individually. In three dimensions, a curve can be described by a position vector \( \mathbf{r}(t) \) that depends on the parameter \( t \).
For example, the position vector \( \mathbf{r}(t) = (\sin t - t \cos t) \mathbf{i} + (\cos t + t \sin t) \mathbf{j} + t^2 \mathbf{k} \) uses \( t \) as the parameter. Each term shows how the curve's position changes in the \( x \), \( y \), and \( z \) directions, respectively. By changing \( t \) over an interval, like \( 0 \leq t \leq 6\pi \), we can trace the entire space curve.

Tangent Line

A tangent line to a curve in space touches the curve at exactly one point without crossing it. To find the equation of a tangent line at a certain point on a curve, we use the velocity vector at that point. The velocity vector gives us the direction of the tangent line.
To determine the tangent line to the curve \( \mathbf{r}(t) \) at \( t_0 = \frac{3\pi}{2} \), we first calculate \( \mathbf{r}(t_0) \), the position vector at this specific \( t \). This gives us the point on the curve. Next, we use the velocity vector evaluated at \( t_0 \), which is \( \mathbf{i} + \mathbf{j} + 3\pi \mathbf{k} \), as the direction of the tangent line. The equation of the tangent line can then be written as:

• T = \((0 + t) \mathbf{i} + (0 + t) \mathbf{j} + \left(\frac{9\pi^2}{4} + 3\pi t\right) \mathbf{k}\)

Velocity Vector

The velocity vector is a fundamental concept in vector calculus, particularly when studying motion along a curve. It is the first derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \).
The velocity vector \( \frac{d \mathbf{r}}{dt} \) provides two key insights: the direction the curve is moving at any point, and the rate of change of position. For our exercise, the velocity vector after differentiation is \( (\cos t + t \sin t) \mathbf{i} + (-\sin t + t \cos t) \mathbf{j} + 2t \mathbf{k} \). Evaluating this at \( t_0 = \frac{3\pi}{2} \), we determine the exact velocity vector which is crucial for finding the tangent line at this point. The velocity vector basically acts as a compass, pointing along the path of the curve at any moment.

3D Space Curve

A 3D space curve is a fascinating geometric entity that is described using three separate functions corresponding to each spatial dimension. In our example, the space curve arises from the vector equation \( \mathbf{r}(t) = (\sin t - t \cos t) \mathbf{i} + (\cos t + t \sin t) \mathbf{j} + t^2 \mathbf{k} \).
In three-dimensional space, these curves can take on complex shapes, spiraling, swooping, or bending, depending on how the parameter \( t \) affects each component. Visualizing a 3D curve in a software system (CAS) allows you to see exactly how it moves through space. It's an intricate dance of the \( x \), \( y \), and \( z \) directions, and plotting it can shed light on its path and movement.