Problem 34 - 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. \(y=x(1-x)^{2 / 5}, \quad-1 \leq x \leq 2, \quad x_{0}=1 / 2\)

Step-by-Step Solution

Answer

Find the curvature \( \kappa \) at \( x=1/2 \), compute the unit normal vector at that point, locate the center of the osculating circle, and then plot the curve together with the circle.

Step 1: Plot the Plane Curve

First, let's plot the given curve to understand its shape. The function is given as \( y = x(1-x)^{2/5} \) where \( -1 \leq x \leq 2 \). Use a graphing tool or software to plot this curve over the specified interval.

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

The curvature \( \kappa \) of a curve given by \( y = f(x) \) can be calculated using the formula \( \kappa = \frac{|f''(x)|}{(1+(f'(x))^2)^{3/2}} \). Find the first and second derivatives, \( f'(x) \) and \( f''(x) \), for the function \( y = x(1-x)^{2/5} \). Evaluate \( \kappa \) at \( x_{0} = \frac{1}{2} \).

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

The unit tangent vector \( \mathbf{T} \) is \( \left(1, f'(x_{0})\right)/\sqrt{1+(f'(x_{0}))^2}\). The unit normal vector \( \mathbf{N} \) is perpendicular to \( \mathbf{T} \). If \( \mathbf{T} = (T_x, T_y) \), then \( \mathbf{N} = (-T_y, T_x) \).

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

Using \( \mathbf{r}(t_{0}) = \left(\frac{1}{2}, f\left(\frac{1}{2}\right)\right) \), which denotes the position vector of point \( P \), and the formula \( \mathbf{C} = \mathbf{r}(t_{0}) + \frac{1}{\kappa(t_{0})} \mathbf{N}(t_{0}) \), find the coordinates \( (a, b) \) of the center of the osculating circle.

Step 5: Plot the Curve and Osculating Circle

The equation of the osculating circle is \( (x-a)^{2}+(y-b)^{2}=1 / \kappa^{2} \). Solve this equation to find the circle's radius and plot it along with the original curve. Adjust the graphing window as required to visualize both the curve and the osculating circle clearly.

Key Concepts

Curvature

Curvature is a measure of how sharply a curve bends at a particular point. It determines the bendiness or the rate of turning of a curve. In mathematical terms, curvature can be defined for a curve given by parametric equations or by a function. When the curve is in the form of a function, such as \[ y = f(x) \], the formula for curvature is usually given by:\[\kappa = \frac{|f''(x)|}{(1+(f'(x))^2)^{3/2}}\]This formula involves the first and second derivatives of the function. Here's why these derivatives are important:

• First derivative \(f'(x)\): Indicates the slope or gradient of the tangent line at a particular point on the curve.
• Second derivative \(f''(x)\): Provides information about the rate of change of the first derivative, helping to understand how the slope of the tangent line itself changes.

Curvature is crucial for determining the osculating circle, which is the best possible circle that fits to a small arc of the curve at a given point.

Unit Normal Vector

After finding the curvature, you need to determine the Unit Normal Vector at a point on the curve. The Unit Normal Vector is a vector that is perpendicular to the tangent vector at a particular point, representing the direction towards which the curve is bending. It is important for understanding the orientation and curvature of a curve.To find the Unit Normal Vector:

• Calculate the tangent vector, \( \mathbf{T} \).
• If the curve is defined by a parametric function, say \( \mathbf{r}(t) = (x(t), y(t)) \), then \( \mathbf{T} \) can be expressed as:\[\mathbf{T} = \frac{(dx/dt, dy/dt)}{\sqrt{(dx/dt)^2 + (dy/dt)^2}}\]
• Once you have the tangent vector \( \mathbf{T} = (T_x, T_y) \), the Unit Normal Vector \( \mathbf{N} \) is given by:\[\mathbf{N} = (-T_y, T_x)\]

The signs in the Unit Normal Vector depend on the turning direction of the curve: clockwise or counterclockwise. Understanding the Unit Normal Vector helps in locating the center of the osculating circle.

Parametric Equations

Parametric equations are a powerful way to represent curves. Instead of expressing a curve as a single equation in terms of \(x\) and \(y\), you use an auxiliary parameter, often denoted \(t\), to describe both coordinates.Typical parametric representations take the form: - \(x(t)\), the \(x\)-coordinate as a function of \(t\).- \(y(t)\), the \(y\)-coordinate as a function of \(t\).For example, a circle of radius \(r\) can be described parametrically as:\[x(t) = r \cos(t),\quad y(t) = r \sin(t)\quad (0 \leq t \leq 2\pi)\]Parametric equations can make it easier to compute derivatives and integrals, which are often necessary in geometrical analyses, such as finding curvature or tangent vectors. They are especially useful in problems involving intersections or trajectory analysis, because they allow you to directly manipulate the path traced by points on the curve.

Calculus

Calculus is the mathematical study of continuous change, using derivatives and integrals to analyze functions. In the context of osculating circles and curve analysis, calculus plays a crucial role.Some key concepts include:

• Derivative (\( \frac{dy}{dx} \)): Measures the rate of change of a function's output value as its input changes. It provides information about the slope of the tangent line to the curve at any point.
• Second Derivative (\( \frac{d^2y}{dx^2} \)): Relates to the curvature of the curve, giving insights into how the slope itself changes. It helps in determining the concavity of curves at specific intervals.
• Integral: Involves the summing up of areas or, in another sense, the reversing of differentiation. While not central to curvature calculations, integrals are vital in many applications involving area and volume.

Calculus provides the tools to rigorously define and manipulate the mathematical constructs necessary for understanding curves, direction, and curvature, making it indispensable in the analysis of osculating circles and related phenomena.