Problem 52 - Thomas Calculus

Question

Use a CAS to perform the following steps in Exercises \(49-52 .\) \(\begin{array}{l}{\text { a. Plot the space curve traced out by the position vector } \mathbf{r} \text { . }} \\ {\text { b. Find the components of the velocity vector } d \mathbf{r} / d t \text { . }} \\ {\text { c. Evaluate } d \mathbf{r} / d t \text { at the given point } t_{0} \text { and determine the equa- }} \\ {\text { tion of the tangent line to the curve at } \mathbf{r}\left(t_{0}\right) .} \\ {\text { d. Plot the tangent line together with the curve over the given }} \\ {\text { interval. }}\end{array}\) $$ \begin{array}{l}{\mathbf{r}(t)=\left(\ln \left(t^{2}+2\right) \mathbf{i}+\left(\tan ^{-1} 3 t\right) \mathbf{j}+\sqrt{t^{2}+1} \mathbf{k}\right.} \\ {-3 \leq t \leq 5, \quad t_{0}=3}\end{array} $$

Step-by-Step Solution

Answer

Plot the space curve and tangent line; velocity vector is \( \mathbf{v}(3) = (\frac{6}{11}, \frac{3}{82}, \frac{3}{\sqrt{10}}) \).

Step 1: Plot the Space Curve

First, we need to plot the space curve described by the position vector \( \mathbf{r}(t) = (\ln(t^2+2))\mathbf{i} + (\tan^{-1}(3t))\mathbf{j} + (\sqrt{t^2+1})\mathbf{k} \) for \( -3 \leq t \leq 5 \). Use a Computer Algebra System (CAS) like Mathematica or GeoGebra to input this vector and generate the 3D plot of the curve.

Step 2: Find the Velocity Vector

To find the components of the velocity vector \( \frac{d\mathbf{r}}{dt} \), differentiate each component of \( \mathbf{r}(t) \) with respect to \( t \). The derivatives are as follows:- \( \frac{d}{dt}\left(\ln(t^2+2)\right) = \frac{2t}{t^2+2} \)- \( \frac{d}{dt}\left(\tan^{-1}(3t)\right) = \frac{3}{1+9t^2} \)- \( \frac{d}{dt}\left(\sqrt{t^2+1}\right) = \frac{t}{\sqrt{t^2+1}} \)Thus, the velocity vector is \( \mathbf{v}(t) = \frac{2t}{t^2+2}\mathbf{i} + \frac{3}{1+9t^2}\mathbf{j} + \frac{t}{\sqrt{t^2+1}}\mathbf{k} \).

Step 3: Evaluate the Velocity at the Given Point and Determine the Tangent Line

Substitute \( t_0 = 3 \) into the velocity vector \( \mathbf{v}(t) \) to find \( \mathbf{v}(3) \).- \( \frac{2(3)}{3^2+2} = \frac{6}{11} \)- \( \frac{3}{1+9(3)^2} = \frac{3}{82} \)- \( \frac{3}{\sqrt{3^2+1}} = \frac{3}{\sqrt{10}} \)The initial position vector is \( \mathbf{r}(3) \) calculated as:- \( \ln(3^2+2) = \ln(11) \)- \( \tan^{-1}(9) \)- \( \sqrt{10} \)The equation of the tangent line is given by \( \mathbf{r}(t_0) + t \cdot \mathbf{v}(t_0) \), which is \( (\ln(11), \tan^{-1}(9), \sqrt{10}) + t \left(\frac{6}{11}, \frac{3}{82}, \frac{3}{\sqrt{10}} \right) \).

Step 4: Plot the Tangent Line with the Curve

Use a CAS to plot the tangent line on the same graph as the space curve. The tangent line equation is: \[ \mathbf{r}(t) = (\ln(11) + t \cdot \frac{6}{11})\mathbf{i} + (\tan^{-1}(9) + t \cdot \frac{3}{82})\mathbf{j} + (\sqrt{10} + t \cdot \frac{3}{\sqrt{10}})\mathbf{k} \]This line should be tangent at \( t_0 = 3 \). Compare how the tangent line aligns with the curve at this point.

Key Concepts

Velocity Vector

In vector calculus, the velocity vector is a fundamental concept used to describe the rate of change of position with respect to time. For a given position vector \( \mathbf{r}(t) \), the velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \). This derivative gives us important information:

• The direction in which the object is moving.
• The speed (or magnitude) of the movement.

For the exercise, we differentiate each component of \( \mathbf{r}(t) \) to obtain the velocity vector:

• \( \frac{d}{dt}\left(\ln(t^2+2)\right) = \frac{2t}{t^2+2} \mathbf{i} \)
• \( \frac{d}{dt}\left(\tan^{-1}(3t)\right) = \frac{3}{1+9t^2} \mathbf{j} \)
• \( \frac{d}{dt}\left(\sqrt{t^2+1}\right) = \frac{t}{\sqrt{t^2+1}} \mathbf{k} \)

The velocity vector provides a convenient way to analyze motion in a wide range of mathematical applications.

Tangent Line

The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. It represents the curve's direction or slope precisely at that location. To find the tangent line: 1. Determine the direction of the tangent line by evaluating the velocity vector at the point of interest, known as \( t_0 \). 2. Calculate the position vector \( \mathbf{r}(t_0) \) for this point. 3. The equation for the tangent line is \( \mathbf{r}(t_0) + t \cdot \mathbf{v}(t_0) \). In our exercise, evaluating at \( t_0=3 \) yields: - Initial position vector: \( (\ln(11), \tan^{-1}(9), \sqrt{10}) \). - Velocity at \( t = 3 \): \( \left(\frac{6}{11}, \frac{3}{82}, \frac{3}{\sqrt{10}}\right) \). Thus, the tangent line equation conveys how the curve behaves near the point \( \mathbf{r}(3) \). When plotting both the curve and tangent line, observe how the tangent line aligns with the curve precisely at \( t=3 \). This visual representation helps confirm the correct calculation and understanding of the curve's geometric properties.

Space Curve

A space curve is a curve that exists in three-dimensional space. Instead of lying on a flat plane, it extends into three dimensions, described by a position vector \( \mathbf{r}(t) \) with three components: \( x(t), y(t), \) and \( z(t) \). This vector provides a full description of the curve's position in space as \( t \) changes.

• Space curves are used to model trajectories of particles, paths in physical systems, and other 3D phenomena.
• It is defined parametrically, giving more flexibility and detail compared to traditional methods.

In our problem, the space curve is described by \( \mathbf{r}(t) = (\ln(t^2+2))\mathbf{i} + (\tan^{-1}(3t))\mathbf{j} + (\sqrt{t^2+1})\mathbf{k} \), plotted for the interval \( -3 \leq t \leq 5 \). Using this, we can visualize the path traced by the curve in a 3D coordinate system, revealing the relationship between its components.

Computer Algebra System

A Computer Algebra System (CAS) is an invaluable tool for solving complex mathematical problems. CAS tools like Mathematica and GeoGebra allow efficient plotting, differentiation, and algebraic manipulation of expressions. For exercises involving space curves, velocity vectors, and tangent lines:

• CAS assists in 3D visualization of space curves, helping you see the curve from different angles.
• It provides symbolic differentiation to find velocity vectors quickly and accurately.
• Automates solving and verification of equations, such as tangent line equations, reducing manual errors.

Using a CAS for this exercise, we can plot the space curve, calculate both the velocity vector and tangent line equation, and visualize these mathematical entities together. Its capabilities turn abstract mathematical concepts into tangible, visual experiences that deeply enhance understanding and learning.