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

Step-by-Step Solution

Answer

Plot the curve, calculate curvature, find unit normal vector, determine osculating circle center, and plot both together.

Step 1: Plot the Plane Curve

To plot the curve defined by the function \( y = x(1-x)^{2/5} \) in the interval \( -1 \leq x \leq 2 \), we would use a graphing tool such as a CAS (Computer Algebra System). We input the function and the interval to generate a visual representation that will help us understand the curve's shape.

Step 2: Calculate the Curvature \( \kappa \)

The curvature \( \kappa \) of a function \( y = f(x) \) can be found using the formula: \[ \kappa = \frac{|y''(x)|}{(1+(y'(x))^2)^{3/2}} \] where \( y'(x) \) is the first derivative and \( y''(x) \) is the second derivative of \( y \). Calculate the derivatives at \( x_0 = 1/2 \) and substitute them into the formula to get \( \kappa \left( \frac{1}{2} \right) \).

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

To find the unit normal vector \( N \) at \( t_0 \), we use the formula: \[ N = (-y'(x_0), 1) / \sqrt{1 + (y'(x_0))^2} \] Compute this using the first derivative \( y'(x) \) at \( x_0 = 1/2 \). Ensure you consider the direction of the tangent vector to apply the correct sign.

Step 4: Determine the Center of the Osculating Circle

The center of the osculating circle \( C \) can be found using the vector equation: \[ C = r(t_0) + \frac{1}{\kappa(t_0)} N(t_0) \] where \( r(t_0) \) is the position vector (\( \frac{1}{2}, f(\frac{1}{2}) \)). Substitute the calculated \( \kappa \) and vector \( N \) values to find the vector \( C = (a, b) \).

Step 5: Plot the Osculating Circle

Plot the equation \((x-a)^2+(y-b)^2= 1/\kappa^2 \) using implicit plotting capabilities of the CAS. Ensure that the plot includes both the original curve and the osculating circle with equally scaled axes to observe their relation clearly.

Key Concepts

Osculating Circle

The concept of an osculating circle is central in understanding the curve at a specific point. Imagine a curve, and picture the best-fit circle that just lightly kisses or "osculates" the curve at that point. This circle shares the same tangent and curvature as the curve at the specific point of interest.
The center of this osculating circle represents the point from which its radius extends to touch the curve at exactly one point.
To find the center of the osculating circle, you use the position vector of the given point on the curve alongside the unit normal vector. This helps you find where this circle should sit to approach the curve precisely. The radius of this circle is the reciprocal of the curvature, revealing how tightly the circle hugs the curve.

Curvature

Curvature is a measure of how much a curve deviates from being a straight line. It tells us how "bent" a curve is at a specific point.

• If the curvature is large, the curve is tightly bended.
• If the curvature is small, the curve is closer to a straight line.

The formula for calculating curvature for a function \( y = f(x) \) is given by:
\[ \kappa = \frac{|y''(x)|}{(1 + (y'(x))^2)^{3/2}} \]
Here, \( y'(x) \) and \( y''(x) \) are the first and second derivatives of \( f(x) \) respectively. Once you compute these derivatives for a specific point, you substitute them into the formula to find the curvature at that point. This calculated curvature is pivotal in determining the radius of the osculating circle, where the radius is \( 1/\kappa \).

Parametric Equations

Parametric equations express a curve using parameters, typically denoted as \( t \), which allows a more flexible description of curves that might be complex or impossible to describe using regular equations. In parametric equations, the x and y coordinates are both expressed as functions of \( t \). This allows us to describe motions and paths in a plane.

• For example, \( x = t \) and \( y = f(t) \) where \( f(t) \) describes \( y \) in terms of \( t \).
• The use of parameter \( t \) provides a thorough representation of curves, especially helpful in complex calculus problems.

This representation isn't just a tool for plotting but also crucial for calculating other properties like curvature and unit normal vectors by providing derivatives concerning parameter \( t \).

Unit Normal Vector

The unit normal vector is orthogonal to the tangent vector of a curve at a specific point. It points perpendicular to the curve, helping us understand the orientation of space around the point.

• To derive this, we first find the tangent vector, which is often calculated using derivatives of the parametric equations.
• The formula for the unit normal vector, given the first derivative, is:
  \[ N = \frac{(-y'(x), 1)}{\sqrt{1 + (y'(x))^2}} \]

Understanding the unit normal vector is vital because it's used to determine the position of the osculating circle's center.
The direction of this vector also helps in investigating properties like the direction of curve bending at any point on the curve. This assists in drawing accurate representations and plotting intricate geometrical properties, especially in multi-step calculus problems.