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Vectors and Parametric Curves


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


Vector Computations

The examples below show how to enter a vector in Maple as well as syntax for some of the most basic vector commands.
> with(linalg):
> a:=[1,-2,2];
> b:=[4,0,-3];
> c:=[-1,5,4];
> 3*a-2*b+c;
> innerprod(a,b);
> crossprod(a,c);
> norm(a,2);
> ab:=b-a;
> ac:=c-a;
> solve(cos(theta)=innerprod(ab,ac)/(norm(ab,2)*norm(ac,2)),theta);

Area of a Triangle

The triangle created by connecting the terminal ends of the vectors a andb in standard form has area = $\frac{1}{2}\vert\vert\bf {a}X\bf {b}\vert\vert$

Parametric Curves

A parametric curve in the plane is represented by 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 $\mathbf{r}(t)$ 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} \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.

Plotting and animating curves in the plane

The ParamPlot command is in the CalcP package so you have to load it first. If you get an error from this command, ask for help right away.
The ParamPlot command produces an animated plot. To see the animation, execute the command and then click on the plot region below to make the controls appear in the Context Bar just above the worksheet window.
the direction of the motion on the curve can be reversed by simply changing the first component from t to -t, as shown below.
The ParamPlot command is nice for visualization, but its output doesn't always show up in a printout. To produce a printable plot, you can use the VPlot command as shown below.

Defining parametric curves and vector valued functions in Maple

The easiest way to define a vector function or a parametric curve is to use the Maple list notaion with square brackets[]. Strictly speaking, this does not define something that Maple recognizes as a vector, but it will work with all of the commands you need for this lab.
You can evaluate this function at any value of t in the usual way.
This is how to access a single component. You would use f(t)[2] to get the second component.

Derivatives and slope of parametric curves

The graph of a parametric curve may not have a slope at every point on the curve. When the slope exists, it must be given by the formula from class.

\begin{displaymath}\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \end{displaymath}

It is clear that this formula doesn't make sense if $\displaystyle
\frac{dx}{dt}= 0$ at some particular value of $t$. If $\displaystyle
\frac{dy}{dt} \neq 0$ at that same value of $t$, then it turns out the graph has a vertical tangent at that point. If both $\displaystyle
\frac{dx}{dt}$ and $\displaystyle \frac{dy}{dt}$ are zero at some value of $t$, then the curve often doesn't have a tangent line at that point. What you see instead is a sharp corner, called a cusp. The Maple syntax below can be used to find the slope of a parametric curve at the point $t=0$.
> slope:=diff(f(t)[2],t)/diff(f(t)[1],t);
> eval(slope,t=0);


  1. For the triangle whose vertices are $Q = (0,-1,5)$, $R = (8,14,-2)$, and $S =(-5,-5,-5)$, do the following
    1. Find the measure of the three angles in degrees. Remember to shift the vertex to the origin. Maple uses radians, so you will need to convert your calculations to degrees.
    2. Check that the measure of your angles add up to 180. Show the computation.
    3. Is the triangle a right triangle, acute triangle, or an obtuse triangle? State your reason.
    4. Find the area of the triangle.

  2. Use VPlot to plot the parametric curve

    \begin{displaymath}\mathbf{r}(t) = (5\cos(t)+\cos(5t)) \mathbf{i} + (5\sin(t)-\sin(5t)) \mathbf{j}\end{displaymath}

    for $0 \leq t \leq 2 \pi$. Find the formula for the slope of the parametric curve, then find $t$ values for two points on the graph where the slope does not exist or where there is a cusp and calculate the $(x,y)$ coordinate location for the two points.

  3. Find two different parametrizations of a semi-circle of radius 1, both with and without using trig functions for $x(t)$ and $y(t)$, and the necessary interval for $t$ and plot each of the parametric curves on separate graphs using ParamPlot. For each parametrization, find the $t$ value for when there be a horizontal tangent and evaluate the vector function at the $t$ value to find the $(x,y)$ coordinate location of the horizontal tangent. The $t$ values will be different for each parametrization, however, the $(x,y)$ coordinate location should be the same.

next up previous
Next: About this document ... Up: lab_template Previous: lab_template
Dina J. Solitro-Rassias