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 \(\mathbf{N}\) 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. \(y=x^{2}-x, \quad-2 \leq x \leq 5, \quad x_{0}=1\)

Step-by-Step Solution

Answer

The center of the osculating circle is at \((0, \sqrt{2})\). Its curvature at \(x_0 = 1\) is \(\frac{1}{\sqrt{2}}\).

Step 1: Understand the Parametric Form

First, we need to express the function \(y = x^2 - x\) in parametric form. We set \(t = x\), hence \(x = t\) and \(y = t^2 - t\). The interval becomes \(-2 \leq t \leq 5\).

Step 2: Plot the Plane Curve

Plot the parametric equations \(x(t) = t\) and \(y(t) = t^2 - t\) for \(-2 \leq t \leq 5\). This helps us visualize the curve.

Step 3: Calculate the Curvature \(\kappa\)

The formula for curvature of a function \(y=f(x)\) is \(\kappa = \frac{y''}{(1 +(y')^2)^{3/2}}\). Compute the derivatives: \(y'(t) = 2t - 1\) and \(y''(t) = 2\). Substituting into the curvature formula at \(t_0 = 1\), we get \(\kappa(1) = \frac{2}{(1^2 + (2*1-1)^2)^{3/2}} = \frac{2}{(1+1)^{3/2}} = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}\).

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

The unit tangent vector \(\mathbf{T}(t)\) is given by \(\frac{1}{\sqrt{(\dot{x})^2 + (\dot{y})^2}}(\dot{x}, \dot{y})\), where \(\dot{x} = 1\) and \(\dot{y} = 2t-1\). At \(t=1\), \(\mathbf{T}(1) = \frac{(1, 1)}{\sqrt{2}} = \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\). The normal vector \(\mathbf{N}\) has components \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) since the curve turns counterclockwise.

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

Use the vector equation \(\mathbf{C} = \mathbf{r}(1) + \frac{1}{\kappa(1)} \mathbf{N}(1)\). Since \(\mathbf{r}(1) = (1, 0)\), \(\kappa(1) = \frac{1}{\sqrt{2}}\), and \(\mathbf{N}\) is \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\), we have \(\mathbf{C} = (1, 0) + \sqrt{2}(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) = (0, \sqrt{2})\).

Step 6: Plot the Osculating Circle

The equation for the osculating circle is \((x-0)^2 + (y-\sqrt{2})^2 = 2\). Plot this circle along with the curve \(y = x^2 - x\) on a graph, adjusting the window as necessary to show both on a square viewing window.

Key Concepts

Curvature

Curvature is a measure of how sharply a curve bends at a given point. It gives us an idea about the "roundness" or "straigthness" of the curve. The larger the curvature, the sharper the curve.To calculate the curvature, we use the second derivative of the curve, which tells us about the change in slope.
For a curve given by a function \( y = f(x) \), the curvature \( \kappa \) is calculated using the formula:\[ \kappa = \frac{y''}{(1 + (y')^2)^{3/2}} \]In our exercise, the parametric form of the curve is \( y = t^2 - t \) where \( t \) is a parameter from \(-2\) to \(5\), and at \( t = 1 \), we found \( \kappa(1) = \frac{1}{\sqrt{2}} \). This value of curvature tells us how the curve behaves at that specific point with respect to bending.

Parametric Equation

A parametric equation represents a curve by expressing the points on the curve as functions of a parameter, usually denoted as \( t \).
This approach is particularly useful when dealing with curves that cannot be expressed with a single function like \( y = f(x) \).In the problem, the function \( y = x^2 - x \) is reformulated into the parametric equations:

• \( x(t) = t \)
• \( y(t) = t^2 - t \)

By defining \( x \) as the parameter \( t \), we simplify calculations related to the curve's geometry, like finding the tangent or normal vectors, especially in the specified interval \(-2 \leq t \leq 5\). This method provides great flexibility and an easier way to compute various properties of more complex curves.

Unit Normal Vector

The unit normal vector, \( \mathbf{N} \), at a given point on a curve is a vector that is perpendicular to the tangent vector. This concept helps in understanding which direction the curve is bending at.
It's crucial because it defines the direction of the osculating circle that best approximates the curve near that point.We first determine the tangent vector \( \mathbf{T} \) and then find \( \mathbf{N} \). For the example curve:

• The tangent vector \( \mathbf{T}(t) \) is \( \frac{1}{\sqrt{1+(2t-1)^2}}(1, 2t-1) \)
• At \( t = 1 \), the tangent vector is \( \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \)

The unit normal vector \( \mathbf{N} \) is obtained by rotating the tangent vector by 90 degrees. At \( t = 1 \), \( \mathbf{N}(1) \) is \( \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \), indicating the leftward and upward direction of curvature at that point.

Plane Curve

A plane curve is a curve that lies entirely within a single plane. The given exercise uses a plane curve represented in both function and parametric forms. Understanding plane curves helps in studying their geometric properties, such as curvature, length, and torsion.Besides, these curves are crucial in fields like physics and engineering where analyzing motion trajectories or fluid paths is necessary.In the exercise, the curve described by the parametric equations \( x(t) = t \) and \( y(t) = t^2 - t \) is a 2D curve lying in the plane defined by the axes \( x \) and \( y \). In practice, representing a curve parametrically within a plane enables more convenient analytical and numerical manipulations.The exercise further explores this plane curve by plotting it along with its corresponding osculating circle. This visualization helps in understanding how closely a circle can "hug" the curve at a specific point (like \( t = 1 \)) and illustrates the effective use of the parametric form for visualization and computation.