Problem 37 - 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)=(\sin 2 t) \mathbf{i}+(\ln (1+t)) \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 4 \pi} \\ {t_{0}=\pi / 4}\end{array} \end{equation}

Step-by-Step Solution

Answer

The tangent line equation is \( \mathbf{r}(t) = \left(0, \ln\left(1+\frac{\pi}{4}\right), \frac{\pi}{4}\right) + t \left(0, \frac{1}{1+\frac{\pi}{4}}, 1\right) \). Plot this with the curve.

Step 1: Plot the Space Curve

To plot the space curve represented by the position vector \( \mathbf{r}(t) = (\sin 2t) \mathbf{i} + (\ln(1+t)) \mathbf{j} + t \mathbf{k} \) over the interval \( 0 \leq t \leq 4\pi \), use a Computer Algebra System (CAS). This involves generating and visualizing the parametric equations of \( x = \sin 2t \), \( y = \ln(1+t) \), and \( z = t \) over the specified period.

Step 2: Find the Velocity Vector

Differentiate the position vector \( \mathbf{r}(t) \) with respect to \( t \) to find the velocity vector \( \frac{d \mathbf{r}}{dt} \). Compute:\[ \frac{d}{dt}[(\sin 2t) \mathbf{i} + (\ln(1+t)) \mathbf{j} + t \mathbf{k}] = (2\cos 2t) \mathbf{i} + \left(\frac{1}{1+t}\right) \mathbf{j} + \mathbf{k} \].This vector represents the velocity components at any time \( t \).

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

Substitute \( t_0 = \frac{\pi}{4} \) into the velocity vector to evaluate it at that point.\[ (2\cos(2 \times \frac{\pi}{4})) \mathbf{i} + \left(\frac{1}{1+\frac{\pi}{4}}\right) \mathbf{j} + \mathbf{k} = 0 \mathbf{i} + \left(\frac{1}{1+\frac{\pi}{4}}\right) \mathbf{j} + \mathbf{k} \].Thus, the velocity vector at \( t_0 \) is \( \left(0, \frac{1}{1+\frac{\pi}{4}}, 1\right) \).

Step 4: Determine Tangent Line Equation

Use the point \( \mathbf{r}(t_0) \) and the velocity vector \( \left(0, \frac{1}{1+\frac{\pi}{4}}, 1\right) \) to formulate the equation of the tangent line. The tangent line is given by:\[ \mathbf{r}(t) = \mathbf{r}(t_0) + t \cdot \frac{d \mathbf{r}}{dt}\bigg|_{t_0} \] \[ \mathbf{r}(t) = \left( \sin\left(\frac{\pi}{2}\right), \ln\left(1+\frac{\pi}{4}\right), \frac{\pi}{4} \right) + t \left(0, \frac{1}{1+\frac{\pi}{4}}, 1\right) \].

Step 5: Plot Tangent Line and Curve

Use CAS to plot both the space curve from Step 1 and the tangent line from Step 4 over the interval \( 0 \leq t \leq 4\pi \). The tangent line should be drawn at the point \( t_0 = \frac{\pi}{4} \), illustrating the tangency to the curve.

Key Concepts

Space Curves

Space curves are smooth paths traced out by a position vector as a function of a parameter, often time ( t ). They often exist in three-dimensional space and provide a graphical representation of the motion along that path. In this example, the position vector is given by \( \mathbf{r}(t) = (\sin 2t) \mathbf{i} + (\ln(1+t)) \mathbf{j} + t \mathbf{k} \). This means:

• \( x = \sin 2t \) - represents the curve's x-coordinate changing with time.
• \( y = \ln(1+t) \) - represents the curve's y-coordinate in terms of time.
• \( z = t \) - the curve's z-coordinate increases linearly with time.

To visualize this curve over the interval \( 0 \leq t \leq 4\pi \), a Computer Algebra System (CAS) helps transform these parametric equations into a graph that illustrates the path traced in three-dimensional space. This path helps in understanding how the object moves through space, defined by its x, y, and z coordinates over time.

Velocity Vector

The velocity vector represents how the position vector changes with respect to time. It provides information about both the speed and direction of an object moving along a space curve. Mathematically, it is found by differentiating the position vector with respect to time,\[ \frac{d \mathbf{r}}{dt} \].For example, differentiating \( \mathbf{r}(t) = (\sin 2t) \mathbf{i} + (\ln(1+t)) \mathbf{j} + t \mathbf{k} \) yields the velocity vector: \[ (2\cos 2t) \mathbf{i} + \left(\frac{1}{1+t}\right) \mathbf{j} + \mathbf{k} \].This vector shows:

• Changes in the x-direction are given by \(2\cos 2t\).
• Changes in the y-direction are \(\frac{1}{1+t}\).
• In the z-direction, the change is constant at 1.

By substituting a specific value of time like \( t_0 = \frac{\pi}{4} \), one finds the velocity at that instant. This gives insight into how fast the curve is being traced out and the direction of motion at any given moment.

Tangent Line

The tangent line to a space curve at a given point is a line that represents the direction of the curve at that point. This is akin to how a tangent to a circle touches the circle at exactly one point. For a space curve, the tangent line is determined using a point on the curve and the velocity vector at that point. It can be expressed by the formula:\[ \mathbf{r}(t) = \mathbf{r}(t_0) + t \cdot \frac{d \mathbf{r}}{dt}\bigg|_{t_0} \].In our example, at \( t_0 = \frac{\pi}{4} \), the velocity vector is \( \left(0, \frac{1}{1+\frac{\pi}{4}}, 1\right) \). Using the position vector at \( t_0 \), \( \mathbf{r}(t_0) = (\sin(\frac{\pi}{2}), \ln(1+\frac{\pi}{4}), \frac{\pi}{4}) \), allows calculation of the tangent line equation: \( \mathbf{r}(t) = (1, \ln(1+\frac{\pi}{4}), \frac{\pi}{4}) + t(0, \frac{1}{1+\frac{\pi}{4}}, 1) \).This results in a line that extends in the direction of the velocity vector and is tangent to the curve at position \( \mathbf{r}(t_0) \). Plotting this with the original curve demonstrates how they locally share the same slope at the contact point.