Problem 31 - 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 \(\mathbf{N}\) 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. \(\mathbf{r}(t)=(2 t-\sin t) \mathbf{i}+(2-2 \cos t) \mathbf{j}, \quad 0 \leq t \leq 3 \pi\) \(t_{0}=3 \pi / 2\)

Step-by-Step Solution

Answer

Plot the curve, find \( \kappa \), and \( \mathbf{N}(t_0) \). Then, calculate \( \mathbf{C} \) and plot the osculating circle.

Step 1: Plot the Plane Curve

Use a CAS to plot the parametric curve defined by \( \mathbf{r}(t) = (2t - \sin t) \mathbf{i} + (2 - 2 \cos t) \mathbf{j} \) over the interval \( 0 \leq t \leq 3\pi \). This will help you visualize the shape of the curve and identify characteristics like inflection points or maximum curvature.

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

To find the curvature \( \kappa \) at \( t_0 = 3\pi/2 \), use the formula for the curvature of a parametric curve: \[ \kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \] First, compute the first derivative \( \mathbf{r}'(t) \) and second derivative \( \mathbf{r}''(t) \). Then evaluate these at \( t_0 \), and use them to find \( \kappa(t_0) \).

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

The unit normal vector \( \mathbf{N} \) is found by taking the derivative of the unit tangent vector \( \mathbf{T} \): \[ \mathbf{T} = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} \] Compute \( \mathbf{T}'(t) \), then normalize it to get \( \mathbf{N}(t) \). Evaluate at \( t_0 = 3\pi/2 \). Consider the direction of \( \mathbf{N} \) based on whether \( \mathbf{T} \) turns clockwise or counterclockwise.

Step 4: Calculate the Center \( \mathbf{C} \) of Osculating Circle

The center of the osculating circle \( \mathbf{C} \) is given by the equation: \[ \mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)} \mathbf{N}(t_0) \] Evaluate the position vector \( \mathbf{r}(t_0) \), then add the product of \( \frac{1}{\kappa(t_0)} \) and \( \mathbf{N}(t_0) \) to find \( \mathbf{C} \).

Step 5: Plot the Osculating Circle

Use the implicit equation \( (x-a)^2 + (y-b)^2 = 1/\kappa^2 \) to plot the osculating circle, where \((a, b)\) is the center calculated in the previous step. Plot the original curve and osculating circle together to visualize how the circle fits the curve at \( t_0 = 3\pi/2 \). Adjust the viewing window to see both plots clearly.

Key Concepts

Curvature Calculation

Curvature is a concept in mathematics that measures how much a curve deviates from being a straight line. In simpler terms, it quantifies the bendiness of a curve at a particular point. For a given parametric curve, the curvature \( \kappa \) can be calculated using the formula:

• \( \kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \)

Here’s a breakdown of this process:

• First Derivative \( \mathbf{r}'(t) \): This is the rate of change of the curve with respect to the parameter \( t \). It provides the tangent vector at any given point.


• Second Derivative \( \mathbf{r}''(t) \): This measures the rate of change of the tangent direction, effectively giving insight into the curve's concavity at a point.


• Cross Product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \): The magnitude of this product tells how quickly the curve is changing direction; a higher value indicates a sharp bend.

By understanding curvature, one can determine how 'tight' or 'loose' a curve is at a particular point, making it an essential concept when examining curves like circles, spirals, and even more complex geometries.

Parametric Curves

Parametric curves are a way to describe a curve using parameters, generally represented as \( t \). Unlike traditional functions, which use x and y coordinates, parametric forms use equations that define both variables in terms of a third parameter.

• Definition: A parametric curve is defined as \( \mathbf{r}(t) = (x(t), y(t), z(t)) \) in 3-dimensional space, or \( (x(t), y(t)) \) in 2-dimensional space.


• Advantages: Parametric forms can represent curves and shapes that are not easily described by standard functions, such as loops or waves.


• Flexibility: They allow for easier manipulation and representation of curves, as the relationship between x and y can be expressed without one being explicitly dependent on the other.

In exercise problems involving osculating circles and curvature, parametric equations make it convenient to calculate derivatives and curvature efficiently. One can easily derive tangent vectors, normal vectors, and other required components by manipulating these parametric equations directly.

Vector Calculus

Vector calculus is an area of mathematics that involves differentiation and integration of vector fields. It is very useful when dealing with multivariable functions and parametric curves.

• Key Components: Vector calculus often involves concepts like dot product, cross product, gradient, divergence, and curl, which are essential in computing curvature and evaluating directional properties of curves.


• Applications: In exercises requiring osculating circles, vector calculus helps find vectors like tangent and normal vectors. These vectors are foundational in locating the center of the osculating circle by determining how a curve turns.


• Tangent and Normal Vectors: The tangent vector \( \mathbf{T} \) is calculated from the first derivative of the parametric curve. For normal vectors \( \mathbf{N} \), one uses vector calculus to derive directions orthogonal to \( \mathbf{T} \).

By leveraging vector calculus, one can better understand complex geometrical shapes and how they behave under various conditions, making it indispensable in advanced mathematical problems.