Problem 28
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. \(\mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{i}+\left(\sin ^{3} t\right) \mathbf{j}, \quad 0 \leq t \leq 2 \pi, \quad t_{0}=\pi / 4\)
Step-by-Step Solution
Verified Answer
Plot the curve and calculate \( \kappa \) and \( \mathbf{C} \) at \( t_0 = \pi/4 \); overlay the osculating circle centered at \( (a, b) \).
1Step 1: Plot the Plane Curve
Use a computer algebra system (CAS) to plot the parametric curve given by \( \mathbf{r}(t) = (\cos^3 t) \mathbf{i} + (\sin^3 t) \mathbf{j} \) for \( 0 \leq t \leq 2\pi \). This will give us a visual representation of the curve.
2Step 2: Calculate the Curvature \( \kappa \)
To calculate the curvature \( \kappa \) at \( t_0 = \frac{\pi}{4} \), use the formula for the curvature of a parametric curve: \[ \kappa = \frac{|x'(t)y''(t) - y'(t)x''(t)|}{(x'(t)^2 + y'(t)^2)^{3/2}}. \] For \( \mathbf{r}(t) \), we have \( x(t) = \cos^3 t \) and \( y(t) = \sin^3 t \). Find the first and second derivatives of both \( x(t) \) and \( y(t) \), substitute \( t_0 \), and solve for \( \kappa(t_0) \).
3Step 3: Find the Unit Normal Vector \( \mathbf{N} \)
Calculate the unit tangent vector \( \mathbf{T}(t) \) at \( t_0 \), given by \( \mathbf{T}(t) = \frac{(x'(t), y'(t))}{\sqrt{x'(t)^2 + y'(t)^2}} \). The unit normal vector is perpendicular to \( \mathbf{T}(t) \). For clockwise rotation, \( \mathbf{N}(t) = (y'(t), -x'(t))/\sqrt{x'(t)^2 + y'(t)^2} \). Find \( \mathbf{N}(t_0) \).
4Step 4: Find the Center \( \mathbf{C} \) of the Osculating Circle
The center \( \mathbf{C} \) of the osculating circle is found using \( \mathbf{C} = \mathbf{r}(t_0) + \frac{1}{\kappa(t_0)} \mathbf{N}(t_0) \). Calculate \( \mathbf{r}(t_0) \) and substitute the found \( \mathbf{N}(t_0) \) and \( \kappa(t_0) \). This provides the coordinates \( (a, b) \).
5Step 5: Plot the Osculating Circle
Using \( \mathbf{C} = (a, b) \), plot the equation of the osculating circle, \( (x-a)^2 + (y-b)^2 = 1/\kappa(t_0)^2 \), implicitly in the CAS. Overlay this plot with the original curve to see how the circle fits at \( t_0 \). Ensure the window is square to accurately compare the curve and circle.
Key Concepts
CurvatureParametric CurveUnit Normal VectorComputer Algebra System
Curvature
Curvature is a measure of how much a curve bends at a given point. In mathematical terms, it quantifies how quickly the direction of the curve changes. Curves with a larger curvature are bending more sharply, while those with smaller curvature are gentler.
For a parametric curve like \( \mathbf{r}(t) = (x(t), y(t)) \), the curvature \( \kappa \) can be calculated using the formula: \[\kappa = \frac{|x'(t)y''(t) - y'(t)x''(t)|}{(x'(t)^2 + y'(t)^2)^{3/2}}.\] This formula uses the first and second derivatives of \( x(t) \) and \( y(t) \) with respect to the parameter \( t \). These derivatives capture how the curve's position changes.
The specific point \( t_0 = \pi/4 \) is important because we want to know how the curve behaves at this precise place. By computing \( \kappa \) at \( t_0 \), we can accurately determine how tightly or loosely the curve bends there.
For a parametric curve like \( \mathbf{r}(t) = (x(t), y(t)) \), the curvature \( \kappa \) can be calculated using the formula: \[\kappa = \frac{|x'(t)y''(t) - y'(t)x''(t)|}{(x'(t)^2 + y'(t)^2)^{3/2}}.\] This formula uses the first and second derivatives of \( x(t) \) and \( y(t) \) with respect to the parameter \( t \). These derivatives capture how the curve's position changes.
The specific point \( t_0 = \pi/4 \) is important because we want to know how the curve behaves at this precise place. By computing \( \kappa \) at \( t_0 \), we can accurately determine how tightly or loosely the curve bends there.
Parametric Curve
Parametric curves are a way to represent curves by defining both the x and y coordinates in terms of a third variable, usually \( t \). This technique allows us to create more dynamic and varied curves than if we were to stick with the traditional y as a function of x.
In the exercise, the curve is given by the parametric equations\[\mathbf{r}(t) = (\cos^3 t) \mathbf{i} + (\sin^3 t) \mathbf{j}.\] Here, both the x and y components depend on \( t \), composing a curve as \( t \) ranges from 0 to \( 2\pi \). This particular form describes a shape known as the "deltoid".
Using parametric equations is beneficial for plotting complex shapes that cannot easily be described by a single function y = f(x). They provide flexibility to represent motion through both vertical and horizontal changes, being especially useful in computer graphics and physics simulations.
In the exercise, the curve is given by the parametric equations\[\mathbf{r}(t) = (\cos^3 t) \mathbf{i} + (\sin^3 t) \mathbf{j}.\] Here, both the x and y components depend on \( t \), composing a curve as \( t \) ranges from 0 to \( 2\pi \). This particular form describes a shape known as the "deltoid".
Using parametric equations is beneficial for plotting complex shapes that cannot easily be described by a single function y = f(x). They provide flexibility to represent motion through both vertical and horizontal changes, being especially useful in computer graphics and physics simulations.
Unit Normal Vector
The unit normal vector is an essential tool when determining the orientation of a curve relative to its direction of travel. It complements the unit tangent vector, which shows the direction of the curve at a point.
To find the unit normal vector \( \mathbf{N} \), we first derive the unit tangent vector \( \mathbf{T} \): \[ \mathbf{T}(t) = \frac{(x'(t), y'(t))}{\sqrt{x'(t)^2 + y'(t)^2}}.\] The unit normal vector is perpendicular to this and points towards the concave side of the curve. It can be computed using:\[\mathbf{N}(t) = \frac{(y'(t), -x'(t))}{\sqrt{x'(t)^2 + y'(t)^2}}.\]The exact direction (clockwise or counterclockwise) will impact the signs of the numerator. This calculation helps identify how the curve turns around a given point, crucial when constructing the osculating circle.
To find the unit normal vector \( \mathbf{N} \), we first derive the unit tangent vector \( \mathbf{T} \): \[ \mathbf{T}(t) = \frac{(x'(t), y'(t))}{\sqrt{x'(t)^2 + y'(t)^2}}.\] The unit normal vector is perpendicular to this and points towards the concave side of the curve. It can be computed using:\[\mathbf{N}(t) = \frac{(y'(t), -x'(t))}{\sqrt{x'(t)^2 + y'(t)^2}}.\]The exact direction (clockwise or counterclockwise) will impact the signs of the numerator. This calculation helps identify how the curve turns around a given point, crucial when constructing the osculating circle.
Computer Algebra System
A Computer Algebra System (CAS) is a software tool specifically designed to perform symbolic mathematics. This involves manipulating equations and expressions algebraically rather than solely numerically.
In the context of the exercise, a CAS is used to:
In the context of the exercise, a CAS is used to:
- Plot the parametric curve, providing a visual representation of complex shapes.
- Compute the curvature and find derivative functions, which are algebraically intensive tasks.
- Calculate the unit normal vector and others that may require intricate algebraic manipulation.
- Plot the osculating circle together with the curve for visual analysis.
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