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Subsections


MA 1024 lab 1 - Curve Computations in the Plane

Purpose

The purpose of this lab is to introduce you to curve computations using Maple for parametric curves and vector-valued functions in the plane.

Getting Started

To assist you, there is a worksheet associated with this lab that contains examples and even solutions to some of the exercises. You can copy that worksheet to your home directory with the following command, which must be run in a terminal window, not in Maple.

cp ~bfarr/Vec2D_start.mws ~

You can copy the worksheet now, but you should read through the lab before you load it into Maple. Once you have read to the exercises, start up Maple, load the worksheet Vec2D_start.mws, and go through it carefully. Then you can start working on the exercises.

Background

By parametric curve in the plane, we mean a pair of equations $x=f(t)$ and $y=g(t)$ for $t$ in some interval $I$. A vector-valued function in the plane is a function $\mathbf{F}(t)$ that associates a vector in the plane with each value of $t$ in its domain. Such a vector valued function can always be written in component form as follows,

\begin{displaymath}\mathbf{F}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} \end{displaymath}

where $f$ and $g$ are functions defined on some interval $I$. From our definition of a parametric curve, it should be clear that you can always associate a parametric curve with a vector-valued function by just considering the curve traced out by the head of the vector. However, there are lots of situations where a vector-valued function is more appropriate. This happens most often when the quantity you want to describe with the function is natural to think of as a vector, for example, a force or a displacement. For the applications we will deal with, we only need to consider curvilinear motion of a point in the plane, That is, we have a vector-valued function $\mathbf{r}(t)$ that gives the position at time $t$ of a moving point $P$ in the plane. The velocity of this point is given by the derivative $\mathbf{r}'(t)$ and the acceleration is given by the second derivative, $\mathbf{r}''(t)$. If the velocity, $\mathbf{r}'(t)$, is not the zero vector, then it is clear from the way it is defined that $\mathbf{r}'(t)$ is a vector that is tangent to the curve at the point $\mathbf{r}(t)$. In many applications of curvilinear motion, we need to know the magnitude of the velocity, or the speed. This is easy to compute - just take the magnitude $\mid
\mathbf{r}'(t) \mid$. Using our idea of the parametric curve associated with $\mathbf{r}(t)$ and recalling the definition of arc length, we arrive at a different interpretation of speed as the rate of change of arc length, or

\begin{displaymath}\frac{ds}{dt} = \mid \mathbf{r}'(t) \mid \end{displaymath}

where $s$ is arc length. This should make sense, if you recall that speed is the rate of change of distance with time and arc length is distance measured along a curve.

If the speed is not zero for any value of $t$ in the interval $I$, then it is possible to define a unit vector, $\mathbf{T}(t)$ that is tangent to the curve as follows.

\begin{displaymath}\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\mid \mathbf{r}'(t) \mid} \end{displaymath}

Using this definition, you can write the velocity in the following form.

\begin{displaymath}\mathbf{r}'(t) = \frac{ds}{dt} \mathbf{T}(t) \end{displaymath}

This is not the most useful form for calculating the velocity, but it does lead to a useful way of thinking about the acceleration experience by a particle moving in a curvilinear path. If the path is a straight line, acceleration depends only on whether the particle is speeding up or slowing down. In a curve, however, there is an additional acceleration, called the centripetal acceleration, that is needed to keep the particle moving on the curve. The magnitude of this acceleration depends on the speed of the car and how much the path is curving. It turns out that you can quantify this with an intrinsic property of the curve called the curvature, usually denoted $\kappa$, defined by the following equation.

\begin{displaymath}\kappa = \left\vert \frac{d \mathbf{T}}{ds} \right\vert \end{displaymath}

That is, the curvature is the magnitude of the rate of change of the tangent vector $\mathbf{T}$ with respect to arc length. For example, the curvature of a straight line is zero and it can be shown that the curvature of a circle of radius $R$ is the same for every point on the circle and is given by $\kappa = 1/R$.

If the curvature is never zero for a particular curve, then we can define another intrinsic property of curve, the unit normal vector $\mathbf{N}$ by the following equation.

\begin{displaymath}\frac{d \mathbf{T}}{ds} = \kappa \mathbf{N} \end{displaymath}

It can be shown that at each point on the curve the vector $\mathbf{N}$ defined by this equation is a unit vector that is always perpendicular to the tangent vector $T$ at that point. Furthermore, the unit normal vector $\mathbf{N}$ always points in the direction of the centripetal acceleration required to keep a particle moving on the curve. One way to see this is to compute the acceleration by differentiating both sides of the equation

\begin{displaymath}\mathbf{r}'(t) = \frac{ds}{dt} \mathbf{T}(t) \end{displaymath}

Using the chain rule and the definition of the curvature and the normal vector one obtains the following important equation.

\begin{displaymath}\mathbf{r}''(t) = \frac{d^2s}{dt^2} \mathbf{T} + \kappa \left
( \frac{ds}{dt} \right)^2 \mathbf{N} \end{displaymath}

To see why this equation is useful, recall that $ds/dt$ is the speed, so $d^2s/dt^2$ is the rate of change of the speed. That is, this term measures whether the particle is speeding up or slowing down. Because this component of the acceleration is in the direction of the tangent vector it is often called the tangential acceleration, denoted by the symbol $a_T$. The component of the acceleration in the direction of the normal vector is called the normal acceleration, denoted $a_N$. In the case of motion on a circular path, the curvature is the reciprocal of the radius, so this term should be easily recognizable as the centripetal acceleration.

Computing these quantities is generally not an easy task. The Getting started worksheet for this lab describes commands from the CalcP package that simplify these calculations and provides examples for you to work from.

Exercises

  1. For each of the following parametrizations of (part) of the curve $y=x^3-x$, use ParamPlot to animate the curve and describe what you see.
    1. $\displaystyle \mathbf{r}(t) = t \mathbf{i} + (t^3-t)
\mathbf{j}$ for $-2 \leq t \leq 2$.
    2. $\displaystyle \mathbf{r}(t) = -t \mathbf{i} + (t^3-t)
\mathbf{j}$ for $-2 \leq t \leq 2$.
    3. $\displaystyle \mathbf{r}(t) = \sin(t) \mathbf{i} + (\sin^3(t)-\sin(t))
\mathbf{j}$ for $-\pi \leq t \leq \pi$.
  2. For each of the following vector functions, compute the unit tangent vector $\mathbf{T}(t)$, the unit normal vector $\mathbf{N}(t)$, the curvature $\kappa(t)$, and the normal and tangential accelerations. Include a plot of the curve in your worksheet.
    1. $\displaystyle \mathbf{r}(t) = t^2 \mathbf{i} + (t^3-t)
\mathbf{j}$ for $-2 \leq t \leq 2$.
    2. $\displaystyle \mathbf{r}(t) = t \cos(t) \mathbf{i} + t \sin(t)
\mathbf{j}$ for $0 \leq t \leq 2 \pi$.

  3. Consider ${\bf r}(t) = t{\bf i} + (t^3-t){\bf j}$. What is the normal vector at $t=0.01$? At $t=-0.01$? What about at $t=0$? What goes wrong?


next up previous
Next: About this document ... Up: lab_template Previous: lab_template
William W. Farr
2001-11-02