91Ó°ÊÓ

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

Step by step solution

01

Plot the Plane Curve

Using a Computer Algebra System (CAS), input the parametric equations \( x(t) = t^2 \) and \( y(t) = t^3 - 3t \) to plot the curve within the interval \( -4 \leq t \leq 4 \). This visual representation will help us understand the behavior and shape of the curve.

02

Calculate the Curvature \( \kappa \)

To find the curvature at \( t_0 = \frac{3}{5} \), use the curvature formula for parametric curves: \[ \kappa(t) = \frac{|x'(t)y''(t) - y'(t)x''(t)|}{((x'(t))^2 + (y'(t))^2)^{3/2}}. \]Compute the derivatives: \( x'(t) = 2t, \quad y'(t) = 3t^2 - 3 \),\( x''(t) = 2, \quad y''(t) = 6t. \)Plug these into the curvature formula and evaluate at \( t_0 = \frac{3}{5} \).

03

Find the Unit Normal Vector \( \mathbf{N} \)

First, calculate the unit tangent vector \( \mathbf{T} = \frac{1}{\sqrt{(x'(t))^2 + (y'(t))^2}} (x'(t) \mathbf{i} + y'(t) \mathbf{j}) \) at \( t_0 = \frac{3}{5}\). Use the derivatives from Step 2: \( x'(t) = 2t \) and \( y'(t) = 3t^2 - 3 \). Then, find the normal vector \( \mathbf{N} \) by rotating \( \mathbf{T} \) 90 degrees (clockwise or counterclockwise) based on rotation direction of \( \mathbf{T} \).

04

Calculate the Center \( \mathbf{C} \) of the Osculating Circle

Use the vector equation given: \[ \mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)} \mathbf{N}(t_0). \]Substitute the position vector \( \mathbf{r}(t_0) = (\left(\frac{3}{5}\right)^2, \left(\frac{3}{5}\right)^3 - 3\left(\frac{3}{5}\right)) \) and curvature \( \kappa(t_0) \) from Step 2, along with the unit normal vector \( \mathbf{N}(t_0) \) from Step 3, to find \( \mathbf{C} \).

05

Plot the Osculating Circle and the Curve

Using the equation of the osculating circle center found in Step 4, \[ (x-a)^2 + (y-b)^2 = \frac{1}{\kappa(t_0)^2}. \]Plot this equation implicitly using the CAS. Overlay the curve from Step 1 and the osculating circle to ensure both plots are visible together. Adjust the view window if necessary to ensure clarity and that the scales of the axes are equal.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Curvature Calculation

Curvature is a measure of how sharply a curve bends at any given point. It is crucial to understand when studying the concepts of osculating circles, which fit the curve best at a particular point. The curvature at a point is calculated using derivatives of the parametric equations that define a curve.

For parametric curves, the formula for curvature \( \kappa \) is given by the equation: \[ \kappa(t) = \frac{|x'(t)y''(t) - y'(t)x''(t)|}{((x'(t))^2 + (y'(t))^2)^{3/2}}. \] Here, \( x'(t) \) and \( y'(t) \) are the first derivatives of the parametric functions \( x(t) \) and \( y(t) \), respectively, while \( x''(t) \) and \( y''(t) \) are the second derivatives.

To find the curvature at a specific point \( t_0 \), you need to evaluate this formula at \( t = t_0 \). Curvature gives insight into the nature of the curve at that point, with larger values indicating tighter bends. This concept is essential when determining the radius of the osculating circle, which is inversely related to the curvature.

Parametric Curves

Parametric curves provide a flexible way to represent curves through equations that express coordinates as functions of a variable, usually denoted as \( t \). In our exercise, the curve is given by the parametric equations \( x(t) = t^2 \) and \( y(t) = t^3 - 3t \).

These equations allow us to trace the curve more controllably by varying \( t \) over a specified interval. Parametric presentation is particularly beneficial when dealing with complex curves that might be difficult to represent or understand with simple functions like \( y=f(x) \).

Using parametric curves, you can plot the curve's path in the plane by calculating and connecting the points for various values of \( t \). They help visualize the trajectory explicitly, which is essential in understanding how the curve moves and bends, revealing its geometric properties.

Unit Normal Vector

The unit normal vector \( \mathbf{N} \) is perpendicular to the tangent vector \( \mathbf{T} \) at any point on the curve, pointing perpendicularly towards the circle's center at the point of tangency. To determine \( \mathbf{N} \), you first derive the unit tangent vector \( \mathbf{T} \), which is calculated as: \[ \mathbf{T} = \frac{1}{\sqrt{(x'(t))^2 + (y'(t))^2}} (x'(t) \mathbf{i} + y'(t) \mathbf{j}). \]
The unit normal vector then results from rotating \( \mathbf{T} \) 90 degrees. Depending on the direction of this rotation (clockwise or counterclockwise), the components of \( \mathbf{N} \) vary.

• For a clockwise rotation, the transformation is \( (-y'(t), x'(t)) \).
• For a counterclockwise rotation, it becomes \( (y'(t), -x'(t)) \).

This vector is crucial in finding the center of the osculating circle, which helps us understand the curve's local geometry. The application of \( \mathbf{N} \) allows one to determine how the curve diverges or converges from the osculating circle, providing a clearer geometric interpretation of the curve's behavior at that point.