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. $\mathbf{r}(t)=t^{2} \mathbf{i}+\left(t^{3}-3 t\right) \mathbf{j}, \quad-4 \leq t \leq 4, \quad t_{0}=3 / 5$

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.

$\mathbf{r}(t)=t^{2} \mathbf{i}+\left(t^{3}-3 t\right) \mathbf{j}, \quad-4 \leq t \leq 4, \quad t_{0}=3 / 5$

Key Concepts

Osculating Circle

The osculating circle at a point on a curve is the circle that best approximates the curve near that point. It shares the same tangent and curvature as the curve, and its center lies along the unit normal vector at the point, at a distance equal to the reciprocal of the curvature.

Curvature

Curvature quantifies how sharply a curve bends at a specific point. It is calculated from the derivatives of the position vector, and a nonzero curvature indicates that the curve is not a straight line. The reciprocal of the curvature gives the radius of the osculating circle.

Unit Tangent Vector

The unit tangent vector represents the direction of the curve at a given point and is obtained by normalizing the derivative of the position vector of the curve. It is fundamental in understanding the curve's direction and in the computation of curvature.

Unit Normal Vector

The unit normal vector is perpendicular to the unit tangent vector and indicates the direction in which the curve is turning. It is used to locate the center of the osculating circle by extending from the point along the normal direction by a distance equal to the reciprocal of the curvature.

Parametric Equations of Plane Curves

Parametric equations express the coordinates of points on a curve as functions of a parameter. This representation facilitates the computation of derivatives and the analysis of geometric properties such as curvature, tangent, and normal vectors.

Implicit Plotting

Implicit plotting refers to graphing curves defined by equations in which one variable is not explicitly solved in terms of another. This technique is used, for example, to plot the equation of the osculating circle when it is given in its standard implicit form.

Transcript

00:01 We asked to use, to look at this curve here with the computer algebra system to find the curve and the oscillating circle at this point at point on three -fifths.

00:19 I use mathematica.

00:21 Here's a bunch of functions i defined for dealing with space curves.

00:24 Well, you just pass some function of t into these expressions here.

00:33 You know, it was spit out, well, a lot of times, you know, these are going to be pretty ugly as a functions of t.

00:42 So i didn't actually write them out.

00:44 So you can just imagine that if you don't ask it to actually, you know, substitute in a value for t, you're going to get a very long output.

00:55 Again, i use simplifying here to try to at least simplify them some, but again, there's still going to be ugly expressions.

01:03 So here's our curve here, and it's x is t squared y is t -cube minus 3 t.

01:10 So what it looks like is kind of like this loop here.

01:16 Let's see here.

01:20 And when we started, it t equals zero.

01:22 We're here.

01:23 And then we go through zero again, y is zero at the root of this, which is square to 20 and square root of three and minus squared to three.

01:35 Now, we can figure out the point that we're on this curve where we want to draw the oscillating circle through.

01:44 And so just plug this into here and we get this kind of ugly.

01:48 We get x is 9 .25s and y is minus 198 over 125.

01:55 So, you know, over here.

02:00 Now, here's 1 -2.

02:02 Again, this kind of gets a little distorted.

02:05 I should have blown it up a little bit...

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