You will use a CAS to explore the osculating circle at a...

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 $t=t_{0} .$ (See Exercise $7 . )$ 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 it is square. $\mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{i}+\left(\sin ^{3} t\right) \mathbf{j}, \quad 0 \leq t \leq 2 \pi, \quad t_{0}=\pi / 4$

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 $t=t_{0} .$ (See Exercise $7 . )$
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 it is square.

$\mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{i}+\left(\sin ^{3} t\right) \mathbf{j}, \quad 0 \leq t \leq 2 \pi, \quad t_{0}=\pi / 4$

Key Concepts

Parametrization of Plane Curves

Parametrization provides a method to represent a plane curve using a parameter, typically t. This approach allows one to express the x and y coordinates as functions of t, which simplifies the process of differentiating and analyzing the geometric properties of the curve.

Curvature

Curvature quantifies how rapidly a curve changes direction at a given point. It is computed from the derivatives of the curve and captures the bending of the curve, with larger curvature indicating a sharper turn. The curvature is fundamental in understanding the local geometry of the curve.

Unit Tangent Vector

The unit tangent vector at a point on a curve is obtained by normalizing the derivative (velocity vector) of the position vector. It indicates the direction in which the curve is proceeding at that point and is essential for determining the rate of change in that direction.

Unit Normal Vector

The unit normal vector is orthogonal to the unit tangent vector and points in the direction in which the curve is turning. This vector plays a critical role in defining the osculating circle, as it provides the orientation for how and where the circle approximates the curve.

Osculating Circle

The osculating circle is the circle that best approximates a curve at a specific point. Its radius is the reciprocal of the curvature, and its center is located by moving along the unit normal vector by the distance equal to this radius. This concept is pivotal in differential geometry for understanding local approximations of curves.

Vector Operations in Differential Geometry

Vector operations, such as addition and scalar multiplication, are used to compute quantities like the center of the osculating circle. By combining the position vector with a scaled version of the normal vector, one can precisely locate the circle that closely matches the curvature of the original curve.

CAS Visualization

Utilizing a Computer Algebra System (CAS) for visualization allows for the plotting and analysis of curves and their osculating circles. Such tools facilitate a deeper understanding by providing graphical representations that illustrate complex concepts like curvature, tangent vectors, and local approximations.

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