Problem 33 - Thomas Calculus

Question

In Exercises \(31-38\) you will use a CAS to explore the osculating circle at a point \(P\) on a plane curve where \(\kappa \neq 0 .\) Use a CAS to perform the following steps: $$ \begin{array}{l}{\text { a. Plot the plane curve given in parametric or function form over }} \\ {\text { the specified interval to see what it looks like. }} \\ {\text { b. Calculate the curvature } \kappa \text { of the curve at the given value } t_{0}} \\ {\text { using the appropriate formula from Exercise } 5 \text { or } 6 . \text { Use the }} \\ {\text { parametrization } x=t \text { and } y=f(t) \text { if the curve is given as a }} \\ {\quad \text { function } y=f(x) \text { . }}\\\\{\text { c. Find the unit normal vector } N \text { at } t_{0} . \text { Notice that the signs of }} \\ {\text { the components of } N \text { depend on whether the unit tangent vector }} \\\ {\text { T is turning clockwise or counterclockwise at } t=t_{0} \text { . (See }} \\ {\text { Exercise } 7 . )}\\\\{\text { d. If } \mathbf{C}=a \mathbf{i}+b \mathbf{j} \text { is the vector from the origin to the center }(a, b)} \\ {\text { of the osculating circle, find the center } \mathbf{C} \text { from the vector }} \\ {\text { equation }} \\\ {\mathbf{C}=\mathbf{r}\left(t_{0}\right)+\frac{1}{\kappa\left(t_{0}\right)} \mathbf{N}\left(t_{0}\right)} \\ {\text { The point } P\left(x_{0}, y_{0}\right) \text { on the curve is given by the position }} \\ {\text { vector } \mathbf{r}\left(t_{0}\right) .}\\\\{\text { e. Plot implicitly the equation }(x-a)^{2}+(y-b)^{2}=1 / \kappa^{2}} \\ {\text { of the osculating circle. Then plot the curve and osculating }} \\ {\text { circle together. You may need to experiment with the size of }} \\ {\text { the viewing window, but be sure the axes are equally scaled. }}\end{array} $$ $$ \mathbf{r}(t)=t^{2} \mathbf{i}+\left(t^{3}-3 t\right) \mathbf{j}, \quad-4 \leq t \leq 4, \quad t_{0}=3 / 5 $$

Step-by-Step Solution

Answer

The curvature and osculating circle at \( t_0 = \frac{3}{5} \) show how the circle closely hugs the curve at this point.

Step 1: Plot the Curve

Use the given parameterization \( \mathbf{r}(t) = t^2 \mathbf{i} + (t^3 - 3t) \mathbf{j} \) to plot the curve for the interval \(-4 \leq t \leq 4\). This will help you visualize the overall shape of the plane curve.

Step 2: Calculate the Curvature

To find the curvature \( \kappa(t) \) at \( t_0 = \frac{3}{5} \), use the formula \( \kappa(t) = \frac{|r'(t) \times r''(t)|}{|r'(t)|^3} \). Compute the derivatives \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \), then find \( \kappa(t) \) at \( t_0 \).

Step 3: Determine the Unit Normal Vector

Calculate the unit tangent vector \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \). Then, find the unit normal vector \( \mathbf{N}(t) \) by rotating \( \mathbf{T}(t) \) by 90 degrees, depending on the curvature direction at \( t_0 = \frac{3}{5} \).

Step 4: Find the Center of the Osculating Circle

Using the vector equation \( \mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)} \mathbf{N}(t_0) \), calculate the vector \( \mathbf{C} \) to find the center \((a, b)\) of the osculating circle at \( t_0 = \frac{3}{5} \).

Step 5: Plot the Osculating Circle

Implictly plot the osculating circle with the equation \((x-a)^2 + (y-b)^2 = \frac{1}{\kappa^2}\). Overlay this plot with the curve to see how the osculating circle fits around the curve at \( t_0 \). Adjust the plot view to ensure scaled axes for accurate visualization.

Key Concepts

Curvature Calculation

Calculating the curvature of a curve is essential in understanding its bending at a particular point. Curvature helps us quantify how sharply a curve is turning. To find the curvature \(\kappa\) at a specific point \( t_0 \), especially when given a parametric equation, we use the formula:- \[ \kappa(t) = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \] Here's what each part of the formula means:

• **\(\mathbf{r}'(t)\)** is the first derivative of the position vector, representing the rate of change of the curve.
• **\(\mathbf{r}''(t)\)** is the second derivative, illustrating how the rate of change itself is changing.
• The cross product \(\times\) between \(\mathbf{r}'(t)\) and \(\mathbf{r}''(t)\) gives us a vector perpendicular to the plane of the curve.
• Finally, the absolute value (magnitude) of this cross product, divided by the cube of the magnitude of \(\mathbf{r}'(t)\), normalizes the curvature to a unit measure.

Calculating exact curvature helps understand the precise geometry of the curve at \(t_0\).

Parametric Curves

A parametric curve is a type of curve described by a set of equations, where each coordinate is defined as a function of one or more parameters (usually \(t\)). This form gives us much more flexibility in defining complex curves in contrast to a simple function form.Consider a parametric curve given by: - \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \)Here:

• **\(x(t)\)** specifies the X-coordinate of the curve in terms of the parameter \(t\).
• **\(y(t)\)** details the Y-coordinate, providing the complete path the curve takes.

Working with parametric curves allows you to:- Easily plot complex trajectories- Calculate curvature, tangent vectors, and other properties effectively- Parameterize even those curves which are not functions (e.g., circles and ellipses)This approach is especially useful in physics and engineering, where motion along a path needs clear descriptions.

Unit Normal Vector

The unit normal vector \(\mathbf{N}(t)\) is vital in analyzing curves, as it tells us the direction in which the curve is bending at any specific point. This vector is always perpendicular to the tangent vector of the curve.First, find the tangent vector \(\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}\). The unit tangent vector is then rotated by 90 degrees in the curve's plane to obtain the unit normal vector.Key characteristics of the unit normal vector:

• It is a unit vector, meaning its length is always one.
• Positioned perpendicularly to the curve at a considered point, pointing in the direction the curve turns.
• Helps in locating the center of the osculating circle.

Understanding the unit normal vector is fundamental for tasks in differential geometry, helping us precisely describe how a curve deviates from its tangent.

Parametrization

Parametrization refers to expressing a curve in terms of parameters, providing a comprehensive way to examine and manipulate curves. The process involves defining a curve not solely by an x-y relationship but by setting each coordinate as a separate function of a shared parameter, typically denoted by \( t \).When we parametrize a curve, we write it as:- \( \mathbf{r}(t) = [ x(t), y(t) ] \)Advantages of using parametrization include:

• Describes paths in complex curves effectively.
• Allows smooth plotting of curves over any specified interval.
• Simplifies calculus operations, such as finding derivatives and integrals.
• Useful in various applications, including graphics, physics simulations, and computer-aided design.

Through parametrization, we can handle diverse curves varying from simple lines to intricate spirals and beyond.

Tangent Vector

The tangent vector \(\mathbf{T}(t)\) is a core concept when analyzing the behavior of curves. This vector illustrates the direction in which the curve is moving at a given point. It is derived from taking the derivative of the position vector of the curve with respect to the parameter \(t\).Formula for the tangent vector is:- \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \)This normalization ensures the tangent vector is a unit vector, making it easier to analyze and interpret.Significance of the tangent vector includes:

• Indicating the immediate direction of the curve at any point \(t\).
• Helping determine the unit normal vector, which is perpendicular to it.
• Crucial for understanding the velocity direction in motion along a curve.

By examining the tangent vector, students can gain deeper insights into the curve's dynamic behavior along its path.