Subsections

• Purpose
• Background
  • Derivatives and the slope
  • Arc length
  
  
• Defining parametric curves and vector valued functions simply in Maple
• Exercises

Parametric Curves 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.

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 $\mathbf{r}(t)$ 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,

$$\mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j}$$

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.

Derivatives and the slope

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.

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

It is clear that this formula doesn't make sense if $$\frac{dx}{dt}= 0$$ at some particular value of $t$. If $$\frac{dy}{dt} \neq 0$$ at that same value of $t$, then it turns out the graph has a vertical tangent at that point.

Arc length

When we think of a parametric curve as representing motion, we need a way to measure the distance traveled by the particle. This distance is given by the arc length, $s$, of a curve. For a parametric curve $x=f(t)$, $y=g(t)$, the arc length of the curve for $a \leq t \leq b$ is given below.

$$s = \int_{a}^{b} \sqrt{ (f'(t))^2+(g'(t))^2} \, dt$$

While the concept of arc length is very useful for the theory of parametric curves, it turns out to be very difficult to compute in all but the simplest cases.

Defining parametric curves and vector valued functions simply 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.

>with(CalcP7):
>f:=t->[2*cos(t),2*sin(t)];

You can evaluate this function at any value of t in the usual way.

>f(0);

This is how to access a single component. You would use f(t)[2] to get the second component.

>f(t)[1]

One way to graph the function is with the VPlot command.

>VPlot(f(t),t=-Pi..Pi);

Notice how the derivative command works.

>diff(f(t),t);

The output shows the derivative of each component, therefore if you want the derivative of the parametric equation make sure you use the definition of the derivative along with the proper component syntax.

Exercises

1. Enter the parametric function $r(t)=[2cos(t)+4cos(t)^2,-2sin(t)]$.

   A)

   Plot the parametric equation with a domain of $0 \leq t \leq 2\pi$.

   B)

   Find the value of the function points at the following five t-values $0$, $\frac{\pi}{8}$, $\pi-arccos(\frac{1}{4})$, $\pi+arccos(\frac{1}{4})$, and $\frac{15\pi}{8}$.

   C)

   Plot the function and the five points.

2. A)

   By looking at the graph, which points will have a slope of zero, a positive slope, a negative slope, or an undefined slope.

   B)

   Find the derivative of the points with a positive or negative slope.

   C)

   What do the tangent lines look like at the points with undefined slopes? The equation of the derivatives will have what value in the denominator? Looking at the formula for the derivative of a parametric equation, where does the denominator come from? Evaluate the denominator at these points.