Problem 35 - Thomas Calculus

Question

In Exercises \(31-38\) 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: $$ \begin{array}{l}{\text { a. Plot the plane curve given in parametric or function form over }} \\ {\text { the specified interval to see what it looks like. }} \\ {\text { b. Calculate the curvature } \kappa \text { of the curve at the given value } t_{0}} \\ {\text { using the appropriate formula from Exercise } 5 \text { or } 6 . \text { Use the }} \\ {\text { parametrization } x=t \text { and } y=f(t) \text { if the curve is given as a }} \\ {\quad \text { function } y=f(x) \text { . }}\\\\{\text { c. Find the unit normal vector } N \text { at } t_{0} . \text { Notice that the signs of }} \\ {\text { the components of } N \text { depend on whether the unit tangent vector }} \\\ {\text { T is turning clockwise or counterclockwise at } t=t_{0} \text { . (See }} \\ {\text { Exercise } 7 . )}\\\\{\text { d. If } \mathbf{C}=a \mathbf{i}+b \mathbf{j} \text { is the vector from the origin to the center }(a, b)} \\ {\text { of the osculating circle, find the center } \mathbf{C} \text { from the vector }} \\ {\text { equation }} \\\ {\mathbf{C}=\mathbf{r}\left(t_{0}\right)+\frac{1}{\kappa\left(t_{0}\right)} \mathbf{N}\left(t_{0}\right)} \\ {\text { The point } P\left(x_{0}, y_{0}\right) \text { on the curve is given by the position }} \\ {\text { vector } \mathbf{r}\left(t_{0}\right) .}\\\\{\text { e. Plot implicitly the equation }(x-a)^{2}+(y-b)^{2}=1 / \kappa^{2}} \\ {\text { of the osculating circle. Then plot the curve and osculating }} \\ {\text { circle together. You may need to experiment with the size of }} \\ {\text { the viewing window, but be sure the axes are equally scaled. }}\end{array} $$ $$ \begin{array}{l}{\mathbf{r}(t)=(2 t-\sin t) \mathbf{i}+(2-2 \cos t) \mathbf{j}, \quad 0 \leq t \leq 3 \pi} \\ {t_{0}=3 \pi / 2}\end{array} $$

Step-by-Step Solution

Answer

The osculating circle's center and curvature are found; it closely matches the curve near \( t_0 = \frac{3\pi}{2} \).

Step 1: Plot the Plane Curve

First, we use a CAS to plot the parametric curve given by the position vector \( \mathbf{r}(t) = (2t - \sin t)\mathbf{i} + (2 - 2\cos t)\mathbf{j} \) for \( 0 \leq t \leq 3\pi \). This will allow us to visualize the shape and path of the curve.

Step 2: Calculate the Curvature at \( t_0 \)

We calculate the curvature \( \kappa \) at \( t_0 = \frac{3\pi}{2} \). Using the parametric form of curvature, \( \kappa(t) = \frac{|x'y'' - y'x''|}{(x'^2 + y'^2)^{3/2}} \), where \( x' = \frac{d}{dt}(2t - \sin t) = 2 - \cos t \) and \( y' = \frac{d}{dt}(2 - 2\cos t) = 2\sin t \). The second derivatives are \( x'' = \sin t \) and \( y'' = 2\cos t \). Evaluate these at \( t_0 \).

Step 3: Find the Unit Normal Vector \( N \)

Compute the unit tangent vector \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} \) at \( t_0 = \frac{3\pi}{2} \). Then find the unit normal vector \( \mathbf{N} \) by taking the derivative of \( \mathbf{T}(t) \) and normalizing it. The sign of \( \mathbf{N} \)'s components depends on whether the curve is turning clockwise or counterclockwise.

Step 4: Determine the Center of the Osculating Circle

Use the formula \( \mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)}\mathbf{N}(t_0) \) to calculate the center \( \mathbf{C} \) of the osculating circle. Evaluate \( \mathbf{r}(t_0) \) and use the previously calculated \( \kappa(t_0) \) and \( \mathbf{N}(t_0) \) to find \( a \) and \( b \), the components of \( \mathbf{C} \).

Step 5: Plot Osculating Circle

Plot the osculating circle using its implicit equation \( (x-a)^2 + (y-b)^2 = \frac{1}{\kappa^2} \). Superimpose this plot with the original parametric curve. Adjust the graph window for suitable viewing, ensuring the axes have equal scaling to correctly display the geometric relation of the curve and osculating circle.

Key Concepts

Curvature

Curvature is an essential concept in understanding how a curve bends at a particular point. To put it simply, curvature measures how fast a curve deviates from being a straight line. When looking at a plane curve, you can imagine tracing along it with a pencil. If the curve is very sharp or turns quickly, its curvature is high. In mathematical terms, curvature, denoted by \( \kappa \), can be calculated for parametric curves using this formula:

• \( \kappa(t) = \frac{|x'y'' - y'x''|}{(x'^2 + y'^2)^{3/2}} \)

Here, \( x' \) and \( y' \) are the first derivatives of the parametric equations, and \( x'' \) and \( y'' \) are the second derivatives.
Calculating at a specific point, like \( t_0 = \frac{3\pi}{2} \) in our exercise, means you'll find out exactly how "bendy" the curve is at that point.

Parametric Curve

A parametric curve is a fancy way of describing how points move in the plane. Instead of representing curves with a simple equation like \( y = f(x) \), parametric curves use two separate equations for \( x \) and \( y \) in terms of a different variable, often \( t \). These are called the parametric equations.
For example, in the provided exercise, the curve is described by:

• \( x(t) = 2t - \sin t \)
• \( y(t) = 2 - 2\cos t \)

As \( t \) varies over some interval, it traces out a path in the \( xy \)-plane, giving us the full shape of the curve. This approach is particularly useful for curves that loop back on themselves or have tricky turns that a simple \( y = f(x) \) can't represent well.

Unit Normal Vector

The unit normal vector helps us explore not just where a curve points but also how it turns. After finding the direction of the curve through the unit tangent vector \( \mathbf{T}(t) \), we use this to determine the unit normal vector, \( \mathbf{N}(t) \).
Why is it important? The unit normal vector gives us the direction "off" the curve at a typical point, showing where the curve turns toward.
Mathematically, to find \( \mathbf{N}(t) \) at any point:

• First, compute the unit tangent vector \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} \).
• Then, \( \mathbf{N}(t) \) is obtained by differentiating \( \mathbf{T}(t) \) and normalizing it.

Keep in mind that the direction of \( \mathbf{N}(t) \) can be influenced by the curve's turning direction - either clockwise or counterclockwise.

Center of a Circle

Finding the center of an osculating circle is like locating the heart of the circle that best hugs the curve at a specific point. The osculating circle is a circle that "kisses" the curve at that point and shares the same curvature.
To find this center \( \mathbf{C} \), use the formula:

• \( \mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)}\mathbf{N}(t_0) \)

Here, \( \mathbf{r}(t_0) \) is the position vector at the point of interest \( t_0 \), while \( \frac{1}{\kappa(t_0)} \) is the radius of curvature, translating the normal vector \( \mathbf{N}(t_0) \) into the correct position.
This point \( \mathbf{C} \) becomes the center of the osculating circle, effectively placing the circle snugly against the curve, reflecting its local geometry.

Tangent Vector

The tangent vector is the first step in understanding the direction a curve is moving at any given point. Think of it as the vector that points along the curve, showing the immediate path a moving point on the curve would follow. Calculating this is straightforward:

• The tangent vector \( \mathbf{T}(t) \) is derived from the derivative of your curve's position vector \( \mathbf{r}(t) \).
• Normalize it to obtain the unit tangent vector: \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} \)

This step allows us to capture the curve's local motion and direction precisely.
The unit tangent vector, being a vector of length 1, aligns perfectly along the path, indicating not just where the curve goes but how briskly or gently it must steer. Understanding this helps in building up the deeper geometry of the path, as it is linked to how sharply or gently the path continues to bend.