Problem 33 - Thomas Calculus

Question

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 \(N\) at \(t_{0}\) . Notice that the signs of the components of \(N\) depend on whether the unit tangent vector \(T\) 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 the axes are equally scaled. $$ \begin{array}{l}{\mathbf{r}(t)=(2 t-\sin t) \mathbf{i}+(2-2 \cos t) \mathbf{j}, \quad 0 \leq t \leq 3 \pi} \\ {t_{0}=3 \pi / 2}\end{array} $$

Step-by-Step Solution

Answer

Plot the curve and osculating circle using the curvature \(\kappa(t_0)\) and center coordinates \((a, b)\).

Step 1: Plot the Plane Curve

To visualize the plane curve, input the parametric equations \(x = 2t - \sin t\) and \(y = 2 - 2\cos t\) into a Computer Algebra System (CAS), such as Desmos or GeoGebra, and plot the curve over the interval \(0 \leq t \leq 3\pi\). Observing this plot helps to understand the nature of the curve and identify features like loops or cusps.

Step 2: Calculate the Curvature

For the curve given by \(\mathbf{r}(t) = (2t-\sin t) \mathbf{i} + (2-2\cos t) \mathbf{j}\), use the formula for curvature \(\kappa(t)\) of a parametric curve: \[ \kappa(t) = \frac{|x'(t) y''(t) - y'(t) x''(t)|}{((x'(t))^2 + (y'(t))^2)^{3/2}} \] Calculate \(x'(t) = 2 - \cos t\), \(y'(t) = 2\sin t\), \(x''(t) = \sin t\), and \(y''(t) = 2\cos t\). Substitute these into the curvature formula and evaluate it at \(t_0 = 3\pi/2\). After calculating, you find \(\kappa(t_0)\).

Step 3: Find the Unit Normal Vector

At \(t_0 = 3\pi/2\), first calculate the unit tangent vector \(\mathbf{T}(t_0)\) as \(\frac{\mathbf{r}'(t_0)}{||\mathbf{r}'(t_0)||}\). Then, compute the unit normal vector \(\mathbf{N}(t_0)\), which is perpendicular to \(\mathbf{T}(t_0)\). Generally, if \(\mathbf{T} = (a,b)\), then a normal vector might be \(N = (-b, a)\). Normalize this vector to get the unit normal vector.

Step 4: Determine the Center of the Osculating Circle

The formula \(\mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)} \mathbf{N}(t_0)\) provides the center \(\mathbf{C} = a\mathbf{i} + b\mathbf{j}\) of the osculating circle. Compute \(\mathbf{r}(t_0)\) using \(t_0 = 3\pi/2\). Next, substitute \(\mathbf{N}(t_0)\) and \(\kappa(t_0)\) into the formula to find \(a\) and \(b\). These form the center coordinates \((a, b)\).

Step 5: Plot the Osculating Circle

Plot the implicit equation of the osculating circle \((x-a)^2 + (y-b)^2 = 1/\kappa^2\) using the center \((a, b)\) and the curvature \(\kappa(t_0)\) found earlier. Overlay the original curve and the osculating circle in your CAS, carefully adjusting the viewing window to ensure both plots are visible and the axes are equally scaled for accurate representation.

Key Concepts

Parametric Equations

When describing curves in mathematics, parametric equations are a powerful tool. Instead of expressing one variable strictly in terms of another (like \(y = f(x)\)), parametric equations describe a curve by expressing both \(x\) and \(y\) coordinates in terms of a third variable, usually denoted by \(t\), called a parameter. This method is particularly useful for modeling paths over time, especially when the behavior of a particle or point follows a trajectory, such as circular or looping motions.
For instance, in the parametric equations given by \((x, y) = (2t - \sin t, 2 - 2\cos t)\), both \(x\) and \(y\) change as \(t\) changes, allowing the curve to capture complex shapes. The interval \(0 \leq t \leq 3\pi\) guides how \(t\) progresses, revealing patterns or repetitions in the curve’s geometry.
By using parametric equations, we can easily plot and visualize such curves using a CAS tool. This visualization aids in understanding the curve's nature, spotting interesting features like cusps, loops, or intersections.

Unit Normal Vector

The unit normal vector is vital when analyzing curves, such as while determining the osculating circle's properties. It is derived from the curve's tangent vector and represents the direction in which the curve is turning at any point.
To compute it, first find the tangent vector \(\mathbf{T}(t)\), which is essentially the first derivative of the position vector \(\mathbf{r}(t)\). Then normalize this tangent to get the unit tangent vector. At a specific point \(t_0\), this would look like \(\mathbf{T}(t_0) = \frac{\mathbf{r}'(t_0)}{||\mathbf{r}'(t_0)||}\).
Once \(\mathbf{T}\) is known, the unit normal vector \(\mathbf{N}\) is perpendicular to \(\mathbf{T}\). If \(\mathbf{T} = (a, b)\), one possible normal vector can be \(\mathbf{N} = (-b, a)\). Normalizing \(\mathbf{N}\) ensures it is a unit vector. The sign of its components depends on the curve's turning direction, as it might be clockwise or counterclockwise, which is crucial when using it for further calculations, like determining the osculating circle's center.

Center of Osculating Circle

The center of an osculating circle, also known as the circle of curvature, points to where the circle best approximates the curve at a specific point \(P\). It lies on the concave side of the curve and provides insights into the curve’s nature at that location.
To find this center, you need the position vector \(\mathbf{r}(t_0)\) where \(P\) is located, the curvature \(\kappa(t_0)\), and the unit normal vector \(\mathbf{N}(t_0)\). By applying the equation \(\mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)} \mathbf{N}(t_0)\), we can calculate the center \(C = (a, b)\).
Here, \(\mathbf{r}(t_0)\) represents the original point coordinates, while \(\frac{1}{\kappa(t_0)} \mathbf{N}(t_0)\) moves from this point along the unit normal by a distance proportional to the inverse curvature. As such, the larger the curvature, the closer the osculating circle stays to the curve, effectively mapping out the curve’s immediate path for analysis and design.