Problem 29 - 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. \(\mathbf{r}(t)=t^{2} \mathbf{i}+\left(t^{3}-3 t\right) \mathbf{j}, \quad-4 \leq t \leq 4, \quad t_{0}=3 / 5\)

Step-by-Step Solution

Answer

Plot the curve, calculate curvature and normal vector, find the circle's center, and plot both the curve and circle.

Step 1: Plot the Plane Curve

We begin with plotting the given curve \( \mathbf{r}(t) = t^2 \mathbf{i} + (t^3 - 3t) \mathbf{j} \) over the interval \(-4 \leq t \leq 4\). This will provide a visual representation of the curve. Use a CAS tool to input the parametric equations and view the plot.

Step 2: Calculate Curvature \( \kappa \)

Curvature \( \kappa \) at a point can be calculated using the formula \( \kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \). Compute the derivatives: \( \mathbf{r}'(t) = 2t\mathbf{i} + (3t^2-3)\mathbf{j} \) and \( \mathbf{r}''(t) = 2\mathbf{i} + 6t\mathbf{j} \). Evaluate at \( t = \frac{3}{5} \) to find \( \kappa\left(\frac{3}{5}\right) \).

Step 3: Evaluate Derivatives at \( t_0 \)

Find \( \mathbf{r}'\left(\frac{3}{5}\right) \) and \( \mathbf{r}''\left(\frac{3}{5}\right) \). This involves substituting \( t = \frac{3}{5} \) into the expressions for \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \). Calculate the cross product and magnitude to find \( \kappa \).

Step 4: Find Unit Normal Vector \( \mathbf{N} \)

First, calculate the unit tangent vector \( \mathbf{T}(t_0) = \frac{\mathbf{r}'(t_0)}{|\mathbf{r}'(t_0)|} \). Determine \( \mathbf{N} \) at \( t_0 = \frac{3}{5} \), ensuring you account for whether the tangent vector is turning clockwise or counterclockwise at that point.

Step 5: Determine Center \( \mathbf{C} \) of Osculating Circle

Use the formula \( \mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)} \mathbf{N}(t_0) \) to find the center of the osculating circle. First, calculate \( \mathbf{r}\left(\frac{3}{5}\right) \) and then find \( \mathbf{C} \) by substituting the values of \( \kappa(t_0) \) and \( \mathbf{N}(t_0) \).

Step 6: Plot the Osculating Circle

Plot the equation \((x-a)^2+(y-b)^2 = 1/\kappa^2\) to represent the osculating circle. Use the previously calculated center \((a, b)\) and the value of \(\kappa\) for your circle's radius. Ensure the viewing window is square to clearly visualize the circle and the curve together.

Key Concepts

Curvature Calculation

Curvature measures how sharply a curve bends at a given point. For a parametric curve defined by a function like \( \mathbf{r}(t) = t^2 \mathbf{i} + (t^3 - 3t) \mathbf{j} \), curvature \( \kappa \) can be calculated using the formula \( \kappa = \frac{|\mathbf{r}'(t) \times \mathbf{r}''(t)|}{|\mathbf{r}'(t)|^3} \). This involves:

• Finding the first derivative \( \mathbf{r}'(t) \), which represents the tangent vector.
• Deriving the second derivative \( \mathbf{r}''(t) \), reflecting the rate of change of the tangent vector.
• Computing the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \), which indicates the degree to which the velocities diverge.
• Evaluating these derivatives at the specific parameter \( t_0 \), which in this case is \( t_0 = \frac{3}{5} \).

This calculation tells us where the curve oscillates and helps in visualizing its shape.

Unit Normal Vector

Once we've found the curvature, the next step is to determine the unit normal vector \( \mathbf{N} \) at the point of interest. This vector is perpendicular to the tangent vector \( \mathbf{T} \) and points towards the center of curvature. The steps to find \( \mathbf{N} \) are:

• Calculate the unit tangent vector \( \mathbf{T}(t_0) = \frac{\mathbf{r}'(t_0)}{|\mathbf{r}'(t_0)|} \).
• Observe the direction of the curve's turn (clockwise or counterclockwise) to assign the right sign to the components of \( \mathbf{N} \).
• The normal vector can then be determined using the cross product and normalizing the resulting vector.

The unit normal vector is essential for defining the orientation of the osculating circle.

Parametric Equations

Parametric equations are expressions that define the coordinates of points on a curve. In this exercise, we use \( \mathbf{r}(t) = t^2 \mathbf{i} + (t^3 - 3t) \mathbf{j} \) to describe the curve. This formulation:

• Allows for complex curves to be represented without needing a single function \( y = f(x) \).
• Provides both \( x \) and \( y \) components as functions of a parameter \( t \), giving flexibility in defining curves.
• Makes it straightforward to calculate derivatives and subsequently the behavior of the curve, such as curvature and speed.

Parametric equations are a powerful tool for analyzing and visualizing curves in any dimension.

Curve Plotting

Plotting a curve using its parametric equations is the first step in visualizing its behavior. By plotting \( \mathbf{r}(t) = t^2 \mathbf{i} + (t^3 - 3t) \mathbf{j} \) over the interval \( -4 \leq t \leq 4 \), you can:

• Observe the overall shape and main features of the curve.
• Identify key points where the curve bends or has special characteristics.
• Use a Computer Algebra System (CAS) for precise graphing, which assists in subsequent exploratory tasks like finding curvature.

The graphical representation is essential for understanding how the mathematical equations translate to geometric shapes.

Derivatives Evaluation

Evaluating derivatives is crucial in studying parametric curves as it provides insight into their geometric properties. For the curve \( \mathbf{r}(t) = t^2 \mathbf{i} + (t^3 - 3t) \mathbf{j} \):

• The first derivative \( \mathbf{r}'(t) = 2t \mathbf{i} + (3t^2 - 3) \mathbf{j} \) gives the velocity vector, showing the direction and speed of the curve at any point.
• The second derivative \( \mathbf{r}''(t) = 2 \mathbf{i} + 6t \mathbf{j} \) highlights the acceleration or the rate of change in the curve's direction.
• Substitute \( t = \frac{3}{5} \) into these derivatives to pinpoint the curve's behavior precisely at the given parameter value.

Understanding these derivatives facilitates calculations of further properties like curvature and helps predict the curve's path.