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. $\mathbf{r}(t)=(3 \cos t) \mathbf{i}+(5 \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi, \quad t_{0}=\pi / 4$

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.

$\mathbf{r}(t)=(3 \cos t) \mathbf{i}+(5 \sin t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi, \quad t_{0}=\pi / 4$

Key Concepts

Osculating Circle

The osculating circle is the circle that best approximates a curve near a given point. Its center, known as the center of curvature, is found by moving from the point on the curve in the direction of the unit normal vector by a distance equal to the reciprocal of the curvature. This concept encapsulates the local geometric behavior of the curve.

Unit Tangent and Normal Vectors

The unit tangent vector to a curve indicates the direction in which the curve is heading at a given point, while the unit normal vector, which is perpendicular to the tangent, points in the direction in which the curve is turning. Together, these vectors are fundamental in the study of the curve’s geometry and are used to compute curvature and to construct the osculating circle.

Use of a Computer Algebra System (CAS)

A CAS is a powerful tool that can perform symbolic and numerical calculations, as well as generate plots. In the context of studying curves, a CAS can be used to compute curvature, unit tangent and normal vectors, and to generate plots that visualize both the curve and its osculating circle. This integration of computation and visualization greatly aids in understanding the geometry of plane curves.

Parametric Representation of Curves

This concept involves expressing a plane curve through a pair of functions that define its x and y coordinates in terms of a common parameter. This approach is particularly useful in calculus and analytical geometry because it allows for a clear description and manipulation of curves, especially when they do not represent functions in the usual sense.

Curvature

Curvature measures how sharply a curve bends at a given point. It is defined as the rate of change of the unit tangent vector with respect to arc length. A nonzero curvature indicates that the curve is not straight, and the reciprocal of the curvature gives the radius of the osculating circle, which is the best circular approximation of the curve at that point.

Transcript

00:01 We're asked to use a computer algebra system to explore the oscillating circle.

00:07 Sometimes that's called the circle of curvature.

00:11 At a point p on a plane, plane curve where kappa is not equal to zero.

00:17 So we went into, they ask us to do it to do things.

00:20 Plot the curve given in parametric form over the...

00:26 So in this case, we're given this curve here, which is just a circle.

00:31 No.

00:35 Did i write that wrong? indeed, did write that wrong.

00:40 I'm saying it's not a circle.

00:41 It's an ellipse.

00:43 The oscillating circle of a circle was just the circle itself.

00:47 So that would be a pretty, well, it would be a trivial problem.

00:53 So anyway, here we have this ellipse.

00:55 These are a curve, right this.

00:57 And we want the oscillating circle at this point here, at pi over four.

01:03 So what i did is because we want to solve a number of these problems the same way, is i just define all these functions.

01:11 And i use mathematica, so this is mathematica notation.

01:15 Again, you could translate it to whatever computer algebra package you want, but hopefully it's intuitive enough to see what's going on here.

01:25 And so i needed to find, let's see here, i needed to, we know where this point p is, right? because we can just plug this into here.

01:34 So this is where p is.

01:36 And then what we need is the normal vector, because the center of curvature, where the center of this oscillic circle is, is in the normal direction from there and a distance with a distance of the radius of curvature...

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