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. $\mathbf{r}(t)=\left(t^{3}-2 t^{2}-t\right) \mathbf{i}+\frac{3 t}{\sqrt{1+t^{2}}} \mathbf{j}, \quad-2 \leq t \leq 5, \quad t_{0}=1$

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.

$\mathbf{r}(t)=\left(t^{3}-2 t^{2}-t\right) \mathbf{i}+\frac{3 t}{\sqrt{1+t^{2}}} \mathbf{j}, \quad-2 \leq t \leq 5, \quad t_{0}=1$

Key Concepts

Osculating Circle

The osculating circle is the circle that best approximates a curve at a particular point. It shares the same tangent and curvature at that point as the curve. The center of the osculating circle, often referred to as the center of curvature, is found by moving from the point on the curve along the unit normal vector by a distance equal to the radius of curvature (which is the inverse of the curvature). This concept helps in visualizing and understanding the local geometric behavior of the curve.

Unit Normal Vector

Perpendicular to the unit tangent vector, the unit normal vector points in the direction in which the curve is turning. It is obtained by differentiating the unit tangent vector with respect to the parameter and then normalizing the result. The unit normal vector is instrumental when determining the center of curvature, as it indicates the direction in which the osculating circle must be constructed relative to the curve.

Use of a Computer Algebra System (CAS)

A CAS is a powerful tool used to perform symbolic and numerical computations with ease. In the context of exploring the osculating circle, a CAS can be used to plot curves, compute derivatives, calculate curvature, and visualize both the curve and its osculating circle. This allows for interactive exploration and a deeper understanding of the underlying geometric properties through visual and computational methods.

Curvature of a Curve

Curvature measures how quickly a curve deviates from being a straight line at a given point. Mathematically, it is defined by the rate of change of the unit tangent vector with respect to arc length. A higher curvature indicates a sharper turn in the curve. The concept is pivotal in understanding the geometric properties of curves and plays an essential role in the determination of the osculating circle, which best approximates the curve locally.

Parametric Representation of Curves

This concept involves describing a curve using a parameter, typically denoted by t. Instead of expressing one variable explicitly in terms of the other (like y as a function of x), both x and y coordinates are expressed as functions of a parameter. This representation is particularly useful for studying complex curves and for performing calculus-based analyses, such as finding derivatives and analyzing motion, as it easily accommodates curves that might not be functions in the traditional sense.

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 with respect to the parameter. It provides essential directional information about the curve and forms the basis for further analysis, such as computing curvature and constructing the osculating circle.

Transcript

00:07 We're given this curve here.

00:10 This is kind of a weird curve.

00:12 It basically is kind of obscured, but it kind of comes around and then quickly changes over to this.

00:17 So it's kind of this s -shaped thing.

00:25 So that's the curve.

00:26 And when one i find the oscillating circle at t equals 1.

00:30 So that point on the curve is actually at minus 2.

00:37 I plus three um over squared to two j so x is minus two y is three over squared or two they asked us to plot it out to a ridiculously long thing so i plotted the whole range that they asked us to plot but i should have just basically cut it off there and you could have seen better in what's going on here so this is you know minus two and then this is you know up a little bit um yeah so let's see here.

01:13 We can find, i used mathematica here.

01:16 I define all these functions.

01:18 I talk about them in the previous problems.

01:20 To find all these functions that i can pass this a vector to and get whatever i need.

01:26 So i passed it to this function here, which then obviously used this function to calculate the unit normal vector.

01:35 And you get a very ugly length of expression as a function of t, but then, if i plug in t equals one, it turns out to be minus 1 over square to 41 times 3i plus 4 times square to 2j.

01:49 So that normal, it's basically pointing into the third quadrant, which makes sense because if you look at the, you know, we're still, right before we kind of made this corner here...

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