Problem 36 - 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: $$ \begin{array}{l}{\text { a. Plot the plane curve given in parametric or function form over }} \\ {\text { the specified interval to see what it looks like. }} \\ {\text { b. Calculate the curvature } \kappa \text { of the curve at the given value } t_{0}} \\ {\text { using the appropriate formula from Exercise } 5 \text { or } 6 . \text { Use the }} \\ {\text { parametrization } x=t \text { and } y=f(t) \text { if the curve is given as a }} \\ {\quad \text { function } y=f(x) \text { . }}\\\\{\text { c. Find the unit normal vector } N \text { at } t_{0} . \text { Notice that the signs of }} \\ {\text { the components of } N \text { depend on whether the unit tangent vector }} \\\ {\text { T is turning clockwise or counterclockwise at } t=t_{0} \text { . (See }} \\ {\text { Exercise } 7 . )}\\\\{\text { d. If } \mathbf{C}=a \mathbf{i}+b \mathbf{j} \text { is the vector from the origin to the center }(a, b)} \\ {\text { of the osculating circle, find the center } \mathbf{C} \text { from the vector }} \\ {\text { equation }} \\\ {\mathbf{C}=\mathbf{r}\left(t_{0}\right)+\frac{1}{\kappa\left(t_{0}\right)} \mathbf{N}\left(t_{0}\right)} \\ {\text { The point } P\left(x_{0}, y_{0}\right) \text { on the curve is given by the position }} \\ {\text { vector } \mathbf{r}\left(t_{0}\right) .}\\\\{\text { e. Plot implicitly the equation }(x-a)^{2}+(y-b)^{2}=1 / \kappa^{2}} \\ {\text { of the osculating circle. Then plot the curve and osculating }} \\ {\text { circle together. You may need to experiment with the size of }} \\ {\text { the viewing window, but be sure the axes are equally scaled. }}\end{array} $$ $$ \mathbf{r}(t)=\left(e^{-t} \cos t\right) \mathbf{i}+\left(e^{-t} \sin t\right) \mathbf{j}, \quad 0 \leq t \leq 6 \pi, \quad t_{0}=\pi / 4 $$

Step-by-Step Solution

Answer

Plot the given curve, calculate \( \kappa(\pi/4) \), find \( \mathbf{N}(\pi/4) \), determine circle center, and plot both curve and circle.

Step 1: Plot the Plane Curve

To understand the shape of the curve, we need to plot the given plane curve in a CAS. The parametric equation is \( \mathbf{r}(t) = \left(e^{-t} \cos t\right) \mathbf{i} + \left(e^{-t} \sin t\right) \mathbf{j} \) for \( 0 \leq t \leq 6\pi \). Observe the spiral pattern of the curve as it descends toward the origin due to the exponential factor.

Step 2: Calculate Curvature κ

The curvature \( \kappa(t_0) \) can be calculated using the formula \( \kappa = \frac{|x'y'' - y'x''|}{(x'^2 + y'^2)^{3/2}} \), where \( x(t) = e^{-t} \cos t \) and \( y(t) = e^{-t} \sin t \). First, calculate the derivatives: \( x' = -e^{-t}(\sin t + \cos t) \), \( y' = e^{-t}(\cos t - \sin t) \), \( x'' = e^{-t}(\sin t - 2\cos t) \), and \( y'' = -e^{-t}(2\sin t + \cos t) \). Now, substitute \( t_0 = \pi/4 \) into these derivatives and use them to find \( \kappa(\pi/4) \).

Step 3: Find the Unit Normal Vector N

The unit tangent vector \( \mathbf{T} \) at any \( t \) is \( \mathbf{T}(t) = \frac{1}{\sqrt{x'^2 + y'^2}}(x' \mathbf{i} + y' \mathbf{j}) \). Evaluate \( \mathbf{T}(\pi/4) \). The unit normal vector \( \mathbf{N} \) is perpendicular to \( \mathbf{T} \) and can be obtained by rotating \( \mathbf{T} \) 90 degrees counterclockwise: \( \mathbf{N}(\pi/4) = \langle -T_y, T_x \rangle \). Use the calculated \( \mathbf{T}(\pi/4) \) to find \( \mathbf{N}(\pi/4) \).

Step 4: Find the Center of the Osculating Circle

The center \( \mathbf{C} \) of the osculating circle is given by the formula \( \mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)} \mathbf{N}(t_0) \). Calculate \( \mathbf{r}(\pi/4) \) to find \( P(x_0, y_0) \), then use the previously calculated \( \mathbf{N}(\pi/4) \) and \( \kappa(\pi/4) \) to find \( \mathbf{C} \).

Step 5: Plot the Osculating Circle

The equation of the osculating circle is \((x - a)^2 + (y - b)^2 = (1/\kappa)^2\) with center \((a, b)\) and radius \(1/\kappa\). Plot this circle along with the original curve within the viewing window, ensuring axes are equally scaled. Adjust the window size if necessary to adequately display both the curve and the circle.

Key Concepts

Curvature Calculation

Curvature is a way to measure how much a curve deviates from being a straight line at a specific point. It's a crucial element in understanding the geometry of a curve. To calculate curvature, we employ the formula: \( \kappa(t) = \frac{|x'y'' - y'x''|}{(x'^2 + y'^2)^{3/2}} \). This particular formula is chosen because it provides a straightforward calculation of curvature for plane curves.

Here's how it works:

• First, find the first derivatives \( x' \) and \( y' \) of the parametric equations \( x(t) \) and \( y(t) \).
• Next, calculate the second derivatives \( x'' \) and \( y'' \).
• Use these derivatives in the curvature formula. This will give the curvature \( \kappa \) at a specific parameter value \( t_0 \), providing insight into the curve's bending."

Curvature is more than a number; it tells us how tight or loose a curve is bending at any point, similar to how radius works for circles. This geometric insight is important for understanding how curves behave.

Unit Normal Vector

The unit normal vector to a curve provides essential information about the curve's orientation in space. It is a vector of length one, perpendicular to the curve at the given point. To compute it, we must first find the unit tangent vector, \( \mathbf{T} \), which points in the direction of the curve's immediate path.

The formula for \( \mathbf{T} \) is: \[ \mathbf{T}(t) = \frac{1}{\sqrt{x'^2 + y'^2}}(x' \mathbf{i} + y' \mathbf{j}) \]

• Compute \( \mathbf{T} \) using the derivatives \( x' \) and \( y' \).
• The unit normal vector \( \mathbf{N} \) is found by rotating \( \mathbf{T} \) by 90 degrees counterclockwise: \( \mathbf{N} = \langle -T_y, T_x \rangle \).

With the unit normal vector computed, it gives rise to the application in determining the center of the osculating circle. Understanding this orientation helps to know which direction the curve turns, contributing greatly to curve interviews and optimizations.

Parametric Equations

Parametric equations are used to express curves in terms of parameters, usually \( t \). They offer a flexible way to describe a curve's path in a plane or space. The benefit of parametric equations is their ability to represent complex curves that aren't simply functions \( y = f(x) \). Here, both \( x \) and \( y \) are described as independent functions of \( t \):

• \( x(t) = e^{-t} \cos t \)
• \( y(t) = e^{-t} \sin t \)

These equations allow us to trace the curve's path as \( t \) changes. In our exercise, they delineate a spiral forming as \( t \) increases from 0 to 6\( \pi \).

Parametric representation is especially useful when analyzing motion or other problems where the path and direction relate to time or another parameter. It provides clarity and precision where standard polynomial functions may fail.

Center of Osculating Circle

The osculating circle of a curve at a point is the best circle approximation or the "kissing circle," which shares the same tangent and curvature as the curve at that point. Finding the center of this osculating circle involves using the curvature \( \kappa \) and the unit normal vector \( \mathbf{N} \).

After calculating the position vector \( \mathbf{r}(t_0) \), the center \( \mathbf{C} \) of the osculating circle is determined by:\[ \mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)} \mathbf{N}(t_0) \]

• Calculate \( \mathbf{r}(t_0) \), which gives the point \( P(x_0, y_0) \) on the curve.
• Use \( \mathbf{N}(t_0) \) and \( \kappa(t_0) \) to find the precise vector \( \mathbf{C} \).

This center point \( \mathbf{C} \), along with the radius \( 1/\kappa \), help construct the osculating circle, offering a deep understanding of the curve's immediate geometry. This circle not only aids in visualizing curvature but also finds applications in understanding movement dynamics and design.