Problem 34 - 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: $$ \begin{array}{l}{\text { a. Plot the plane curve given in parametric or function form over }} \\ {\text { the specified interval to see what it looks like. }} \\ {\text { b. Calculate the curvature } \kappa \text { of the curve at the given value } t_{0}} \\ {\text { using the appropriate formula from Exercise } 5 \text { or } 6 . \text { Use the }} \\ {\text { parametrization } x=t \text { and } y=f(t) \text { if the curve is given as a }} \\ {\quad \text { function } y=f(x) \text { . }}\\\\{\text { c. Find the unit normal vector } N \text { at } t_{0} . \text { Notice that the signs of }} \\ {\text { the components of } N \text { depend on whether the unit tangent vector }} \\\ {\text { T is turning clockwise or counterclockwise at } t=t_{0} \text { . (See }} \\ {\text { Exercise } 7 . )}\\\\{\text { d. If } \mathbf{C}=a \mathbf{i}+b \mathbf{j} \text { is the vector from the origin to the center }(a, b)} \\ {\text { of the osculating circle, find the center } \mathbf{C} \text { from the vector }} \\ {\text { equation }} \\\ {\mathbf{C}=\mathbf{r}\left(t_{0}\right)+\frac{1}{\kappa\left(t_{0}\right)} \mathbf{N}\left(t_{0}\right)} \\ {\text { The point } P\left(x_{0}, y_{0}\right) \text { on the curve is given by the position }} \\ {\text { vector } \mathbf{r}\left(t_{0}\right) .}\\\\{\text { e. Plot implicitly the equation }(x-a)^{2}+(y-b)^{2}=1 / \kappa^{2}} \\ {\text { of the osculating circle. Then plot the curve and osculating }} \\ {\text { circle together. You may need to experiment with the size of }} \\ {\text { the viewing window, but be sure the axes are equally scaled. }}\end{array} $$ $$ \mathbf{r}(t)=\left(t^{3}-2 t^{2}-t\right) \mathbf{i}+\frac{3 t}{\sqrt{1+t^{2}}} \mathbf{j}, \quad-2 \leq t \leq 5, \quad t_{0}=1 $$

Step-by-Step Solution

Answer

The osculating circle is centered at \(\mathbf{C}(a, b)\) and can be plotted alongside the curve with equal axis scaling for clarity.

Step 1: Plot the Plane Curve

Using the given parametric equations, plot the curve \[ \mathbf{r}(t) = (t^3 - 2t^2 - t)\mathbf{i} + \frac{3t}{\sqrt{1+t^2}}\mathbf{j} \]over the interval \(-2 \le t \le 5\).This will provide a visual representation of the curve, helping us understand its behavior.

Step 2: Calculate the Curvature

Compute the curvature **\(\kappa\)** of the curve at \(t_0 = 1\) using the curvature formula:\[\kappa = \frac{|x'y'' - y'x''|}{((x')^2 + (y')^2)^{3/2}}\]where \(x(t) = t^3 - 2t^2 - t\) and \(y(t) = \frac{3t}{\sqrt{1+t^2}}\). Calculate the derivatives \(x'(t), x''(t), y'(t),\) and \(y''(t)\), then evaluate \(\kappa\) at \(t_0 = 1\).

Step 3: Find the Unit Normal Vector

Calculate the unit normal vector \(\mathbf{N}\) at \(t_0 = 1\) using the derived unit tangent vector \(\mathbf{T}\):\[\mathbf{N}(t) = \frac{d}{dt}\left( \frac{-y'(t)}{\sqrt{(x')^2 + (y')^2}}, \frac{x'(t)}{\sqrt{(x')^2 + (y')^2}} \right)\]Determine the sign of \(\mathbf{N}\) based on whether \(\mathbf{T}\) is turning clockwise or counterclockwise at \(t_0 = 1\).

Step 4: Determine the Center of the Osculating Circle

Using the equation:\[\mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)} \mathbf{N}(t_0)\]calculate \(\mathbf{r}(t_0)\), and then determine the vector \(\mathbf{C} = a \mathbf{i} + b \mathbf{j}\), which points from the origin to the center \((a, b)\) of the osculating circle.

Step 5: Plot the Osculating Circle

Implicitly plot the equation of the osculating circle:\[(x-a)^2 + (y-b)^2 = \frac{1}{\kappa^2}\]alongside the original curve. Ensure the axes are equally scaled for a clear view of the curve and circle interaction.

Key Concepts

Curvature Calculation

Understanding curvature is essential when studying curves in mathematics. It measures how quickly a curve changes direction at a point. Calculating the curvature \(\kappa\) is crucial for finding the osculating circle—a circle that best approximates the curve near that point.
Let's break down the steps:
1. **Formula:** We use the formula \ \kappa = \frac{|x'y'' - y'x''|}{((x')^2 + (y')^2)^{3/2}} \ to calculate the curvature at a specific point. Here, \ x(t) \ and \ y(t) \ are the parametric equations describing the curve, while \ x'(t), x''(t), y'(t), \ and \ y''(t) \ are the first and second derivatives with respect to the parameter \ t \.
2. **Evaluate:** Once these derivatives are found, we substitute them into the formula and evaluate at \ t_0 = 1 \ to find the curvature at that particular point.
This step gives us a numeric value that represents how sharply the curve bends at \ t_0 = 1 \, helping to determine the size of the osculating circle.

Parametric Equations

Parametric equations are a way to express curves using parameters, typically with equations for \ x(t) \ and \ y(t) \. In this exercise, the curve \(\mathbf{r}(t) = (t^3 - 2t^2 - t)\mathbf{i} + \frac{3t}{\sqrt{1+t^2}}\mathbf{j}\) is defined in terms of the parameter \ t \.
This method:

• Allows more flexibility compared to traditional Cartesian equations.
• Enables us to describe curves that loop or spiral, which are difficult to describe with single \ y = f(x) \ functions.

To work with parametric equations, it's crucial to understand how to differentiate them to find their derivatives, which are used in calculating curvature and other characteristics of the curve.
These equations give insight into how the point moves through space as \ t \ varies, which is key when plotting and analyzing the curve's behavior.

Unit Normal Vector

Once we have the curvature, we can determine the unit normal vector \(\mathbf{N}\), which is perpendicular to the tangent vector at a given point on the curve.
Steps involved include:

• Calculate the unit tangent vector \(\mathbf{T}\), which shows the direction of the curve.
• Using derivatives, derive the normal vector from the tangent vector components.
• Normalize this vector (making its magnitude equal to 1) to find \(\mathbf{N}\): \mathbf{N}(t) = \left( \frac{-y'(t)}{\sqrt{(x')^2 + (y')^2}}, \frac{x'(t)}{\sqrt{(x')^2 + (y')^2}} \right)\

The unit normal vector helps in determining the orientation of the curve, showing if the curve is turning clockwise or counterclockwise. It also plays a pivotal role in determining the center of the osculating circle.

Center of Osculating Circle

The center of an osculating circle gives you the exact point from which the circle best fits the curve at a particular spot. This center is vital in visualizing how the circle tangentially kisses the curve.
To determine this center:

• Use the equation \mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)} \mathbf{N}(t_0)\ where \mathbf{r}(t_0)\ is the position vector of the point on the curve, and \mathbf{N}(t_0)\ is the unit normal vector at \ t_0 = 1\.
• This formula calculates the vector \mathbf{C} = a \mathbf{i} + b \mathbf{j}\, which points from the origin to the osculating circle's center \(a, b\).

Understanding this center helps us plot the osculating circle accurately, ensuring it mirrors the curve's path as closely as possible near \ t_0 = 1 \.