next up previous
Next: About this document ... Up: lab_template Previous: lab_template


MA 1024 Lab 2: Curve Computations in Three Dimensions


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


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. 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)


\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.


  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. $a=3$, $c=1$ for $0 \leq t \leq 4 \pi$.
    2. $a=3$, $c=-1$ for $0 \leq t \leq 4 \pi$.
    3. $a=1$, $c=1$ for $0 \leq t \leq 4 \pi$.
    Give a brief explanation of how the values of $a$ and $c$ affect the graph.
  2. In the old days, students at WPI used to do projects in calculus as well as labs. One such project called for students to design a loop for a roller coaster. A group from the class of `95 came up with the following design, which they called the ``Vomit Comet''.

    \begin{displaymath}\mathbf{r}(t) = 6t \mathbf{i} + 30 \int_0^t \cos(w+w^2/8) \, dw
\, \mathbf{j} + 30 \int_0^t \sin(w+w^2/8) \, dw \, \mathbf{k}\end{displaymath}

    Here $\mathbf{r}(t)$ is the position in units of meters of the car on the loop and $t$ is the time in seconds. A Maple command you can use to define this function is given below. It is broken into two lines below, but you should type it in one line in Maple.
    > comet := t -> 
    1. Plot the position for $0 \leq t \leq 4$. The best way to view the loop is probably looking in along the $x$ axis. How high (approximately) is the top of the loop, in meters?
    2. The Speed command will give you the speed of the coaster car in units of meters per second. Given that one meter per second is equivalent to about $2.25$ miles per hour, compute the speed of the coaster car in miles per hour. Does it seem reasonable for a roller coaster?
    3. Plot the curvature of the loop for $0 \leq t \leq 4$.
    4. Plot the normal acceleration of the loop for $0 \leq t \leq 4$. To put it in perspective, you might want to plot the normal acceleration divided by $9.8$, the acceleration due to gravity in units of meters per second squared.
    5. Based on your results, do you think the loop lives up to its name?

next up previous
Next: About this document ... Up: lab_template Previous: lab_template
William W. Farr