Problem 61 - 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. $$ \begin{array}{l}{\mathbf{r}(t)=\left(\ln \left(t^{2}+2\right)\right) \mathbf{i}+\left(\tan ^{-1} 3 t\right) \mathbf{j}+\sqrt{t^{2}+1} \mathbf{k}} \\\ {-3 \leq t \leq 5, \quad t_{0}=3}\end{array} $$

Step-by-Step Solution

Answer

The tangent line's equation at \( t_0 = 3 \) is \( \mathbf{l}(t) = (\ln(11) + \/t\frac{6}{11})\mathbf{i} + (\tan^{-1}(9) + t\frac{3}{82})\mathbf{j} + (\sqrt{10} + t\frac{3}{\sqrt{10}})\mathbf{k} \).

Step 1: Plot the Space Curve

Utilize a computer algebra system (CAS) to graph the space curve defined by the position vector \( \mathbf{r}(t) = (\ln(t^2 + 2))\mathbf{i} + (\tan^{-1}(3t))\mathbf{j} + \sqrt{t^2 + 1}\mathbf{k} \) for \( -3 \leq t \leq 5 \). Check the curve's behavior over the specified interval and ensure it is well-visualized.

Step 2: Find Velocity Vector Components

To find the components of the velocity vector \( \frac{d \mathbf{r}}{dt} \), differentiate each component of \( \mathbf{r}(t) \):- For the \( \mathbf{i} \) component: find \( \frac{d}{dt} \left( \ln(t^2 + 2) \right) \).- For the \( \mathbf{j} \) component: compute \( \frac{d}{dt} \left( \tan^{-1}(3t) \right) \).- For the \( \mathbf{k} \) component: determine \( \frac{d}{dt} \left( \sqrt{t^2 + 1} \right) \).

Step 3: Differentiate Each Component

The derivatives are:- \( \mathbf{i} \) component: \( \frac{d}{dt} (\ln(t^2+2)) = \frac{2t}{t^2+2} \).- \( \mathbf{j} \) component: \( \frac{d}{dt} (\tan^{-1}(3t)) = \frac{3}{1+9t^2} \).- \( \mathbf{k} \) component: \( \frac{d}{dt} (\sqrt{t^2+1}) = \frac{t}{\sqrt{t^2+1}} \).Thus, the velocity vector is \( \mathbf{v}(t) = \left(\frac{2t}{t^2+2}\right) \mathbf{i} + \left(\frac{3}{1+9t^2}\right) \mathbf{j} + \left(\frac{t}{\sqrt{t^2+1}}\right) \mathbf{k} \).

Step 4: Evaluate Velocity at \( t_0 = 3 \)

Substitute \( t = 3 \) into each component of the velocity vector \( \mathbf{v}(t) \):- \( \frac{2(3)}{3^2+2} = \frac{6}{11} \) for \( \mathbf{i} \).- \( \frac{3}{1+9(3)^2} = \frac{3}{82} \) for \( \mathbf{j} \).- \( \frac{3}{\sqrt{3^2+1}} = \frac{3}{\sqrt{10}} \) for \( \mathbf{k} \).Thus, \( \mathbf{v}(3) = \frac{6}{11} \mathbf{i} + \frac{3}{82} \mathbf{j} + \frac{3}{\sqrt{10}} \mathbf{k} \).

Step 5: Determine Equation of Tangent Line

The tangent line at \( t_0 = 3 \) can be expressed as \( \mathbf{r}(3) + t\mathbf{v}(3) \), where \( \mathbf{r}(3) = (\ln(11))\mathbf{i} + (\tan^{-1}(9))\mathbf{j} + \sqrt{10}\mathbf{k} \).The equation becomes:\( \mathbf{l}(t) = (\ln(11) + t\frac{6}{11})\mathbf{i} + (\tan^{-1}(9) + t\frac{3}{82})\mathbf{j} + (\sqrt{10} + t\frac{3}{\sqrt{10}})\mathbf{k} \).

Step 6: Plot Tangent Line with Space Curve

Using the CAS, plot the tangent line \( \mathbf{l}(t) \) together with the space curve \( \mathbf{r}(t) \) over the interval \(-3 \leq t \leq 5\). Observe how the tangent line corresponds with the space curve at \( t_0 = 3 \).

Key Concepts

Position Vector

A position vector is a crucial concept when dealing with space curves. It determines the specific location of a point in the three-dimensional space relative to an origin. In this context, the position vector is given by

• \( \mathbf{r}(t) = (\ln(t^2 + 2))\mathbf{i} + (\tan^{-1}(3t))\mathbf{j} + \sqrt{t^2 + 1}\mathbf{k} \)

This combines three functions, each corresponding to one of the dimensions (\(x\), \(y\), and \(z\)), and the parameter \(t\) represents a variable like time, tracing the path of the curve as it varies. As \(t\) changes from \(-3\) to \(5\), the vector gives us different points along the curve. It's like drawing a line through space where each part of the curve represents this vector depending on the value of \(t\). Understanding the role of a position vector lays the foundation for interpreting and analyzing space curves in mathematics.

Velocity Vector

The velocity vector is the derivative of the position vector with respect to \(t\). It shows the rate of change of position over time, essentially providing a sense of direction and speed along the space curve. To compute it:

• Differentiate each component of the position vector.
• \[ \mathbf{v}(t) = \left(\frac{2t}{t^2 + 2}\right)\mathbf{i} + \left(\frac{3}{1 + 9t^2}\right)\mathbf{j} + \left(\frac{t}{\sqrt{t^2 + 1}}\right)\mathbf{k} \]

This new vector \(\mathbf{v}(t)\) describes how the space curve moves at any instant and in which direction it progresses as each \(t\) approaches another point. Evaluating this at specific points can help understand the motion's dynamics, especially in physics, where velocity is a key concept.

Tangent Line

A tangent line refers to a straight line that touches a curve at a specific point without crossing it nearby. It reveals the curve's instantaneous direction at that point. To find the equation of the tangent line:

• Evaluate the position and velocity vectors at \(t_0 = 3\).
• Formulate it as \( \mathbf{r}(3) + t \mathbf{v}(3) \) which translates to:\( \mathbf{l}(t) = (\ln(11) + t\frac{6}{11})\mathbf{i} + (\tan^{-1}(9) + t\frac{3}{82})\mathbf{j} + (\sqrt{10} + t\frac{3}{\sqrt{10}})\mathbf{k} \)

This equation helps in visualizing how the curve tends to evolve at the particular point \(t_0\). The tangent line is essentially the best straight-line approximation of the curve at that point.

Differentiation

Differentiation is a mathematical operation used to compute an object's rate of change. In the context of space curves, it’s essential for finding the velocity vector. Each component of a position vector function is differentiated separately with respect to \(t\).

• For \(\ln(t^2 + 2),\) the derivative is \(\frac{2t}{t^2+2}.\)
• For \(\tan^{-1}(3t),\), the derivative is \(\frac{3}{1+9t^2}.\)
• For \(\sqrt{t^2 + 1},\) the derivative is \(\frac{t}{\sqrt{t^2 + 1}}.\)

Grasping differentiation helps translate the position vector into velocity, enabling more advanced understanding and analysis of the curve and motion, especially in three-dimensional space.

3D Plotting

3D plotting allows visual representation of equations and functions in a three-dimensional space. It is essential for graphing the position vector \(\mathbf{r}(t)\) and observing its path as a space curve between given intervals. With the help of a CAS, you can:

• Plot the space curve described by the equations for three different axes.
• Include additional elements, like tangent lines, to visualize their interactions.

3D plotting helps in understanding how functions behave over intervals and how different components come together in geometric space, providing invaluable insights into complex mathematical relationships.

Computer Algebra System (CAS)

A Computer Algebra System (CAS) is a sophisticated software that facilitates symbolic mathematics, including differentiation, integration, algebra, and more. It is indispensable for tackling complex calculations and visualizations in exercises involving space curves. Using a CAS provides:

• Efficient graphing and plotting of high-dimensional functions.
• Accurate computation of derivatives and evaluations.
• Visualization tools for better conceptual understanding.

When working on assignments that require plotting or solving intricate algebraic expressions, a CAS serves as a powerful tool, simplifying the process and providing detailed insights into mathematical phenomena.