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MA 1024 Lab 2: Curve Computations in Three Dimensions

Purpose

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

Getting Started

To assist you, there is a worksheet associated with this lab that contains examples. 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/Curves3D_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 Curves3D_start.mws, and go through it carefully. Then you can start working on the exercises

Background

A parametric curve in three dimensions is a triple of functions $x=f(t)$, $y=g(t)$, $z=h(t)$ for $t$ in some interval $I$. A vector-valued function in three dimensions is a function $\mathbf{r}(t)$ 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{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \end{displaymath}

where $f$, $g$, and $h$ 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 our purposes, we will focus on the case of motion of a particle in three dimensions. That is, we have a vector-valued function $\mathbf{r}(t)$ that gives the position at time $t$ of a moving point $P$. 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 never zero, then we can define the unit tangent vector $\mathbf{T}(t)$ and the curvature $\kappa(t)$ the same way we did in two dimensions by

\begin{displaymath}\mathbf{T}(t) = \frac{1}{\mid \mathbf{r}'(t) \mid} \mathbf{r}'(t)
\end{displaymath}

and

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

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. Consider the circular helix $\mathbf{r}(t) = a \cos(t)
\mathbf{i} + a \sin(t) \mathbf{j} + ct \mathbf{k}$. Plot the graphs for the following sets of values of $a$ and $c$ using the VPlot command. You should also look at animations of the plots by using the ParamPlot3D command, though they won't appear in your printout.
    1. $a=2$, $c=1/2$ for $0 \leq t \leq 4 \pi$.
    2. $a=2$, $c=2$ for $0 \leq t \leq 4 \pi$.
    3. $a=4$, $c=2$ for $0 \leq t \leq 4 \pi$.
    Give a brief explanation of how the values of $a$ and $c$ affect the graph.
  2. Consider again the circular helix $\mathbf{r}(t) = a \cos(t)
\mathbf{i} + a \sin(t) \mathbf{j} + ct \mathbf{k}$.
    1. Compute the curvature.
    2. Compute the normal and tangential components of the acceleration.

  3. Consider the function $\mathbf{r}(t)= t \cos(t)
\mathbf{i} + t \sin(t) \mathbf{j} + t \mathbf{k}$ that appeared as one of the examples.
    1. Compute the curvature, and plot it for $0 \leq t \leq 4 \pi$.
    2. Compute the normal acceleration, and plot it for $0 \leq t \leq 4 \pi$.
    3. Can you explain why the normal acceleration is increasing, even though the curvature is decreasing? (Hint - look at the speed.)


next up previous
Next: About this document ... Up: lab_template Previous: lab_template
William W. Farr
2001-11-09