Problem 23 - Advanced Engineering Mathematics

Question

Graph the curve \(C\) that is described by \(\mathbf{r}\) and graph \(\mathbf{r}^{\prime}\) at the indicated value of \(\boldsymbol{t}\). $$ \mathbf{r}(t)=2 \mathbf{i}+t \mathbf{j}+\frac{4}{1+t^{2}} \mathbf{k} ; t=1 $$

Step-by-Step Solution

Answer

Graph \(C\) in 3D and plot the tangent vector (0, 1, -2) at \((2, 1, 2)\).

Step 1: Understanding the Components of the Vector Function

The vector function given is \(\mathbf{r}(t) = 2\mathbf{i} + t\mathbf{j} + \frac{4}{1+t^2}\mathbf{k}\). Here, each unit vector represents a direction in 3D space: \(\mathbf{i}\) is the x-direction, \(\mathbf{j}\) is the y-direction, and \(\mathbf{k}\) is the z-direction. This means we need to plot the curve in 3-dimensional space defined by these components.

Step 2: Evaluating the Vector Function at \(t = 1\)

Substitute \(t = 1\) into \(\mathbf{r}(t)\). This gives: \(\mathbf{r}(1) = 2\mathbf{i} + 1\mathbf{j} + \frac{4}{1+1^2}\mathbf{k} = 2\mathbf{i} + 1\mathbf{j} + 2\mathbf{k}\). The point \((2, 1, 2)\) is on the curve at \(t = 1\).

Step 3: Finding the Derivative \(\mathbf{r}'(t)\)

Find the derivative of each component of \(\mathbf{r}(t)\):- \(\mathbf{i}\) component: Derivative of \(2\) is \(0\).- \(\mathbf{j}\) component: Derivative of \(t\) is \(1\).- \(\mathbf{k}\) component: Derivative of \(\frac{4}{1+t^2}\) is done using the chain rule, resulting in \(-\frac{8t}{(1+t^2)^2}\).So, \(\mathbf{r}'(t) = 0\mathbf{i} + 1\mathbf{j} - \frac{8t}{(1+t^2)^2}\mathbf{k}\).

Step 4: Evaluating \(\mathbf{r}'(t)\) at \(t = 1\)

Substitute \(t = 1\) into \(\mathbf{r}'(t)\):\[\mathbf{r}'(1) = 0\mathbf{i} + 1\mathbf{j} - \frac{8(1)}{(1+1^2)^2}\mathbf{k} = 0\mathbf{i} + 1\mathbf{j} - \frac{8}{4}\mathbf{k} = \mathbf{j} - 2\mathbf{k}\]This gives the tangent vector \((0, 1, -2)\) at the point \((2, 1, 2)\).

Step 5: Graph the Curve and Tangent Vector

Plot the point \((2, 1, 2)\) in 3D space. Then, draw the curve based on the function \(\mathbf{r}(t)\), which changes according to the parameters defined by \(\mathbf{i}, \mathbf{j}\), and \(\mathbf{k}\). At the point \((2, 1, 2)\), draw the tangent vector \(\mathbf{j} - 2\mathbf{k}\), which shows the direction of the curve at \(t = 1\). This requires moving \(1\) unit in the positive \(y\)-direction and \(2\) units in the negative \(z\)-direction from the point.

Key Concepts

3D Space

To understand vector calculus better, it is crucial to comprehend the concept of 3-dimensional, or 3D, space. Unlike two-dimensional planes, which have only length and width, 3D space includes height as well. This third dimension allows us to describe and work with objects in a more realistic way.

In mathematical terms, any position in 3D space can be represented using coordinates \(x, y, z\). These coordinates are aligned with the axes represented by the unit vectors \(\mathbf{i}, \mathbf{j},\text{ and } \mathbf{k}\).

Each vector component, then, corresponds to movement along one of these axes. For instance:

• \(\mathbf{i}\) moves along the x-axis.
• \(\mathbf{j}\) moves along the y-axis.
• \(\mathbf{k}\) moves along the z-axis.

Understanding 3D space is essential when graphing curves like \(\mathbf{r}(t) = 2\mathbf{i} + t\mathbf{j} + \frac{4}{1+t^2}\mathbf{k}\). Its components dictate how the curve navigates the space along the three axes.

Derivative of Vector Function

In vector calculus, finding the derivative of a vector function helps us understand how a curve changes or behaves as its parameter varies. The derivative of a function like \(\mathbf{r}(t) = 2\mathbf{i} + t\mathbf{j} + \frac{4}{1+t^2}\mathbf{k}\) is found by differentiating each component separately.

Let's break down the process:

• The \(\mathbf{i}\) component, being constant \(2\), has a derivative of \(0\).
• The \(\mathbf{j}\) component is linear ( extit{t}), so its derivative is \(1\).
• The \(\mathbf{k}\) component is more complex as it involves the function \(\frac{4}{1+t^2}\). Using the chain rule, we derive this to get \(-\frac{8t}{(1+t^2)^2}\).

The derivative \(\mathbf{r}'(t) = 0\mathbf{i} + 1\mathbf{j} - \frac{8t}{(1+t^2)^2}\mathbf{k}\) represents the directional change of the vector function at each point \(t\). Differentiating vector functions is fundamental in physics and engineering to analyze motion and force.

Tangent Vector

A tangent vector illustrates the direction and rate of change of a curve at a particular point. Imagine it as a short arrow pointing in the direction that the curve is heading at that point. It helps us understand the object's motion or the slope of the curve in the space.

In our problem, after differentiating the vector function, we found the tangent vector at \(t = 1\) by evaluating \(\mathbf{r}'(t)\) at this point. Specifically, \(\mathbf{r}'(1) = 0\mathbf{i} + 1\mathbf{j} - 2\mathbf{k}\), which means:

• It has no movement along the x-axis, represented by \(\mathbf{i}\).
• It moves positively along the y-axis by \(1\) unit, represented by \(\mathbf{j}\).
• It moves negatively along the z-axis by \(2\) units, represented by \(\mathbf{k}\).

This vector is crucial for understanding how the curve behaves at that exact moment, providing both direction and rate. In many practical scenarios, such as trajectory planning, the tangent vector is indispensable.