Problem 59 - 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. $$ \mathbf{r}(t)=\sqrt{2 t \mathbf{i}}+e^{t} \mathbf{j}+e^{-t} \mathbf{k}, \quad-2 \leq t \leq 3, \quad t_{0}=1 $$

Step-by-Step Solution

Answer

Tangent line: \( \mathbf{r}(1) + t \left( \frac{1}{\sqrt{2}} \mathbf{i} + e \mathbf{j} - \frac{1}{e} \mathbf{k} \right) \).

Step 1: Plot the Space Curve

Use a computer algebra system (CAS) to plot the space curve represented by the position vector \( \mathbf{r}(t) = \sqrt{2t} \mathbf{i} + e^{t} \mathbf{j} + e^{-t} \mathbf{k} \) over the interval \(-2 \leq t \leq 3\). Input this vector function into the plotting tool to visualize the trajectory of the space curve.

Step 2: Calculate Velocity Vector

Differentiate the position vector \( \mathbf{r}(t) = \sqrt{2t} \mathbf{i} + e^{t} \mathbf{j} + e^{-t} \mathbf{k} \) with respect to \( t \) to find the velocity vector. This involves computing the derivatives for each component individually, yielding: \( \frac{d\mathbf{r}}{dt} = \frac{1}{\sqrt{2t}} \mathbf{i} + e^{t} \mathbf{j} - e^{-t} \mathbf{k} \).

Step 3: Evaluate Velocity at t_0

Substitute \( t_0 = 1 \) into the velocity vector \( \frac{d\mathbf{r}}{dt} = \frac{1}{\sqrt{2t}} \mathbf{i} + e^{t} \mathbf{j} - e^{-t} \mathbf{k} \). This yields \( \frac{d\mathbf{r}}{dt} \bigg|_{t=1} = \frac{1}{\sqrt{2}} \mathbf{i} + e \mathbf{j} - \frac{1}{e} \mathbf{k} \).

Step 4: Determine Tangent Line Equation

The equation of the tangent line at \( t_0 = 1 \) is \( \mathbf{r}(t_0) + t \frac{d\mathbf{r}}{dt}\bigg|_{t=1} \). First, calculate \( \mathbf{r}(1) = \sqrt{2} \mathbf{i} + e \mathbf{j} + \frac{1}{e} \mathbf{k} \). Then, the tangent line is \( \mathbf{r}(1) + t \left( \frac{1}{\sqrt{2}} \mathbf{i} + e \mathbf{j} - \frac{1}{e} \mathbf{k} \right) \).

Step 5: Plot Tangent Line with Curve

Use the CAS to plot both the space curve and the tangent line on the same graph over the interval \(-2 \leq t \leq 3\). This will show how the tangent line touches the curve precisely at \( t_0 = 1 \).

Key Concepts

Position Vector

In vector calculus, a position vector is a fundamental concept used to describe the location of a point in space. It is essentially a vector that originates from a fixed reference point, known as the origin, and points towards the specific location of the point of interest. In our exercise, the position vector is given by \[ \mathbf{r}(t) = \sqrt{2t} \mathbf{i} + e^{t} \mathbf{j} + e^{-t} \mathbf{k} \] This position vector defines a curve in space as the parameter \( t \) varies over the interval \(-2 \leq t \leq 3\).

The components \( \sqrt{2t} \mathbf{i} \), \( e^{t} \mathbf{j} \), and \( e^{-t} \mathbf{k} \) specify the coordinates of the point along the \( x \), \( y \), and \( z \) axes respectively.

• \( \sqrt{2t} \mathbf{i} \) controls the movement along the \( x \)-axis.
• \( e^{t} \mathbf{j} \) influences the \( y \)-axis.
• \( e^{-t} \mathbf{k} \) modifies the \( z \)-axis.

By plotting this vector over the given interval, we can visualize its path through space, often referred to as a space curve.

Velocity Vector

The velocity vector is an important tool in understanding how the position of an object changes over time, represented by the derivative of the position vector with respect to time \( t \).

For our given position vector \[ \mathbf{r}(t) = \sqrt{2t} \mathbf{i} + e^{t} \mathbf{j} + e^{-t} \mathbf{k}, \] we determine the velocity vector by differentiating each component of the position vector:\[ \frac{d\mathbf{r}}{dt} = \frac{1}{\sqrt{2t}} \mathbf{i} + e^{t} \mathbf{j} - e^{-t} \mathbf{k}. \]

• The term \( \frac{1}{\sqrt{2t}} \mathbf{i} \) represents the rate of change of the position along the \( x \)-axis.
• \( e^{t} \mathbf{j} \) gives the speed of movement along the \( y \)-axis.
• The expression \( -e^{-t} \mathbf{k} \) dictates how fast the point is moving along the \( z \)-axis in the negative direction.

Velocity vectors are essential in predicting future positions and understanding the nature of the object's trajectory along the space curve. By evaluating these components individually, we get a snapshot of the object's instant rate of change of position at any specified \( t \).

Tangent Line

A tangent line is a straight line that touches a curve at exactly one point at a given time \( t_0 \). At this spot, the tangent line has the same direction as the curve, acting as a best linear approximation to the curve at that point.

To find the tangent line to our space curve at \( t_0 = 1 \), we follow these steps:1. **Evaluate the Position Vector at \( t_0 \):** Compute \( \mathbf{r}(1) = \sqrt{2} \mathbf{i} + e \mathbf{j} + \frac{1}{e} \mathbf{k} \).2. **Determine the Velocity Vector at \( t_0 \):** Substitute \( t_0 = 1 \) into the velocity vector \( \frac{d\mathbf{r}}{dt} = \frac{1}{\sqrt{2}} \mathbf{i} + e \mathbf{j} - \frac{1}{e} \mathbf{k} \).3. **Construct the Equation of the Tangent Line:** The equation is given by \[ \mathbf{r}(1) + t \left( \frac{1}{\sqrt{2}} \mathbf{i} + e \mathbf{j} - \frac{1}{e} \mathbf{k} \right). \]This equation describes a line that accurately reflects the direction of the curve precisely at the point \( \mathbf{r}(1) \). By analyzing this, students can understand how curves behave around specific points.

Space Curve

A space curve is a path traced by a point moving through three-dimensional space, described by a vector function of one variable, usually \( t \). The concept of a space curve helps us visualize how an object moves and changes position as \( t \) varies.

In our exercise, the space curve is represented by the position vector \[ \mathbf{r}(t) = \sqrt{2t} \mathbf{i} + e^{t} \mathbf{j} + e^{-t} \mathbf{k}, \] defining how the object moves along the \( x \), \( y \), and \( z \) dimensions over time. By generating a plot of this curve over the interval \(-2 \leq t \leq 3\), we can see the continuous path that the point follows.

• Explore the trajectory that resembles three-dimensional motion.
• Observe changes that occur due to exponential functions \( e^{t} \) and \( e^{-t} \).
• Predict how the curve progresses based on different values of \( t \).

Space curves are crucial in many fields like physics, engineering, and computer graphics, providing insights into the dynamics and movement of objects in a real-world context.

Differentiation

Differentiation is a key mathematical concept used to determine how a function changes at any given point. In vector calculus, it plays a vital role, particularly when dealing with motion along curves in space.

For our challenge, differentiating the position vector \( \mathbf{r}(t) = \sqrt{2t} \mathbf{i} + e^{t} \mathbf{j} + e^{-t} \mathbf{k} \) involves determining the derivative for each component separately: \( \frac{d\mathbf{r}}{dt} = \frac{1}{\sqrt{2t}} \mathbf{i} + e^{t} \mathbf{j} - e^{-t} \mathbf{k}. \)

• Apply rules of differentiation to find individual derivatives.
• Understand the rate at which each component changes over time.
• Use derivatives to calculate velocity and other rates of change.

Differentiation gives us essential information about the object's dynamics, helping us model and analyze moving systems. By handling derivatives effectively, students gain deep insights into how positional changes influence movement, essential for mastering vector calculus.