Problem 38 - Thomas Calculus

Question

Use a CAS to perform the following steps. \begin{equation} \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} \end{equation} \begin{equation} \begin{array}{l}{\mathbf{r}(t)=\left(\ln \left(t^{2}+2\right) \mathbf{i}+\left(\tan ^{-1} 3 t\right) \mathbf{j}+\sqrt{t^{2}+1} \mathbf{k}\right.} \\ {-3 \leq t \leq 5, \quad t_{0}=3}\end{array} \end{equation}

Step-by-Step Solution

Answer

Plot the curve, compute the velocity, evaluate at \( t_0 = 3 \) for the tangent line equation, and plot both.

Step 1: Plot the Space Curve

Use a Computer Algebra System (CAS) to plot the curve defined by the position vector \( \mathbf{r}(t) = (\ln(t^2 + 2))\mathbf{i} + (\tan^{-1}(3t))\mathbf{j} + (\sqrt{t^2 + 1})\mathbf{k} \) over the interval \(-3 \leq t \leq 5\). This involves creating a parametric plot for \(t\) in this range.

Step 2: Compute the Velocity Vector

Find the components of the velocity vector by differentiating the position vector \( \mathbf{r}(t) \) with respect to \( t \). This gives \( \frac{d\mathbf{r}}{dt} = \left( \frac{d}{dt}\ln(t^2 + 2)\right)\mathbf{i} + \left(\frac{d}{dt}\tan^{-1}(3t)\right)\mathbf{j} + \left(\frac{d}{dt}\sqrt{t^2 + 1}\right)\mathbf{k} \).

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

Substitute \( t = 3 \) into the expression for \( \frac{d\mathbf{r}}{dt} \) (computed in Step 2) to find the velocity vector at \( t_0 = 3 \).

Step 4: Determine the Tangent Line Equation

Using the velocity vector at \( t_0 = 3 \), set up the equation of the tangent line. The tangent line at \( \mathbf{r}(t_0) \) has the form \( \mathbf{r}(t_0) + (t - t_0) \cdot \frac{d\mathbf{r}}{dt}\bigg|_{t_0} \). Calculate \( \mathbf{r}(3) \) and substitute into this formula.

Step 5: Plot the Tangent Line with the Curve

Use a CAS to plot the tangent line calculated in Step 4 over the interval along with the space curve plotted in Step 1. This visually shows the tangent relationship at \( t_0 = 3 \).

Key Concepts

space curves

Space curves are fascinating geometrical shapes in three-dimensional space. They are defined by parametric equations, where each component of the equation—generally denoted as \(x(t)\), \(y(t)\), and \(z(t)\)—is a function of a single parameter \(t\). This parameter represents time or another independent variable. Space curves offer a way to describe the motion of particles in space, as the vector changes over time.
In our exercise, the position vector \(\mathbf{r}(t)\) gives the space curve as \( (\ln(t^2 + 2)) \mathbf{i} + \left( \tan^{-1}(3t) \right) \mathbf{j} + \left( \sqrt{t^2 + 1} \right) \mathbf{k} \) over a specified interval from \(t = -3\) to \(t = 5\). Each component of \(\mathbf{r}(t)\) serves to determine the position in a three-dimensional space. This description creates a curve that threads through the space in a unique path, making space curves incredibly diverse and interesting to study.

velocity vector

The velocity vector is crucial when studying motion along a curve, as it describes both the speed and direction of an object moving over time. By differentiating the position vector \( \mathbf{r}(t) \) with respect to \( t \), we obtain the velocity vector \( \frac{d\mathbf{r}}{dt} \). This vector gives us a sense of how \( \mathbf{r}(t) \) changes at any given point, indicating how fast and in what direction the curve is being traced out.
The components of the velocity vector are derived by applying the derivative to each part of the position vector:

• \( \frac{d}{dt}\ln(t^2 + 2) \) for the \( \mathbf{i} \) component
• \( \frac{d}{dt}\tan^{-1}(3t) \) for the \( \mathbf{j} \) component
• \( \frac{d}{dt}\sqrt{t^2 + 1} \) for the \( \mathbf{k} \) component.

This calculation allows us to understand the precise dynamics of motion as the parameter \( t \) changes.

tangent line

The tangent line is a straight line that just "touches" the curve at a particular point. It gives a local linear approximation of the curve at that point, providing insight into the direction the curve takes locally. The mathematical expression of the tangent line involves the position vector at \( t_0 \) and the velocity vector evaluated at this point.
To find the tangent line at point \( t_0 = 3 \), we need to determine the value of the position vector \( \mathbf{r}(3) \) and the velocity vector \( \frac{d\mathbf{r}}{dt}\bigg|_{t=3} \). Then, the tangent line can be expressed in the form:
\[ \mathbf{r}(t_0) + (t - t_0) \cdot \frac{d\mathbf{r}}{dt}\bigg|_{t_0} \]
By substituting \( t_0 = 3 \), we obtain a linear equation representing the tangent line to the curve at that instant. This line illustrates how the curve behaves very closely at \( t_0 \), providing a simple way to deduce the direction of change.

Computer Algebra System

A Computer Algebra System (CAS) is a powerful tool used to perform symbolic mathematics operations quickly and accurately. It is particularly useful for dealing with complex calculations, such as finding derivatives, integrals, and plotting intricate graphs including space curves and tangent lines.
In our exercise, the CAS assists in plotting the space curve based on the position vector \( \mathbf{r}(t) \), computing the velocity vector, and plotting the tangent line along with the space curve. These tasks are often cumbersome to perform manually, especially when derivatives and intricate graphing are involved.
With a CAS, tasks such as differentiating \( \mathbf{r}(t) \), calculating \( \frac{d\mathbf{r}}{dt} \) at a specific \( t_0 = 3 \), and graphing the results become highly manageable and efficient. This makes CAS an invaluable resource for students and professionals working with complex mathematical problems.