Parametric curves

Geometry: Parametric curves

Parametric curves

So far we have we have not yet been able to describe geometric shapes different from a circle, triangle or a line. The theory of parametric equations allows us to do describe shapes of various kinds. We restrict our attention to parametric curves.

A parametric curve $\orange{C}$ is a figure in the plane which is described by two equations

$\orange C \colon \phantom{x}\begin{cases}\blue{x}&=\blue{x(t)}\\ \green{y}&=\green{y(t)} \end{cases}$

where $t$ varies over a specified $\text{interval}$ . These equations are called parametric equations. The curve $\orange{C}$ consists of all points $\ivcc{\blue{x(t)}}{\green{y(t)}}$ for $t$ in the interval.

Example

The parametric equations $\begin{array}{rcl}\blue{x(t)}&=&\blue{t^2}\\\green{y(t)}&=&\green{2-t},\end{array}$ with $t$ in the interval $\ivcc{-4}{2}$ define a parametric curve.

Example

Example

In the figure you see the curve in the example defined by the equations $\begin{array}{rcl}\blue{x(t)}&=&\blue{t^2}\\\green{y(t)}&=&\green{2-t},\end{array}$

on the interval $[-4,2]$ .

Figure

Interval

Two parametric curves with the same parametric equations can look very different if the intervals are different. In the example, the curve $\orange{C}$ was defined by $\begin{array}{rcl}\blue{x(t)}&=&\blue{t^2}\\\green{y(t)}&=&\green{2-t},\end{array}$ with $t$ ranging from $-4$ to $2$ .

When we change the interval to $t$ ranging from $0$ to $6$ , we get the following figure:

More dimensions

One can also define a parametric equation with more variables and using more equations. For example, the parametric equation given by a scaled version of $\begin{array}{rcl}x(t)&=&\cos(2t)+t\\y(t)&=&t+t^2-4\\z(t)&=&\cos(2t) + \sin(3t) +0.2t,\end{array}$ where $t$ ranges over the interval $\ivcc{-10}{10}$ defines a curve in three dimensional space.

We will not study these kinds of curves in this course.

Orbit

Orbit

Such parametric curves can be used to describe the orbit of a certain point $\orange P$ . If such an orbit is described by parametrical equations $\begin{cases}\blue x &= \blue{x(t)},\\ \green y & = \green{y(t)}. \end{cases}$ then one often writes $\orange{P_{t}}$ for the 'location' of $\orange P$ at time $t$ - i.e. $\orange{P_{t_0}}=\ivcc{\blue{x(t_0)}}{\green{y(t_0)}}$

where $t_0$ is a number in the interval. In the figure is an example where $\ivcc{ \blue{x(t)}}{ \green{y(t)}} = \ivcc{ \blue{\sin(t) + \cos(t)}}{ \green{\cos(t)}}$

Example

Directions

On a parametric curve there is a notion of direction. Informally the direction of the curve is determined by the direction the curve has when $t$ increases. One can reverse the direction of a parametric curve by "moving in the other direction". For example a parametric curve on an interval $[a,b]$ $\begin{cases}\blue{x}&=\blue{x(t)}\\ \green{y}&=\green{y(t)} \end{cases}$ can be reversed with the following curve: $\begin{cases}\blue{x}&=\blue{x(-t)}\\ \green{y}&=\green{y(-t)} \end{cases}$

on the interval $[-b, -a]$ .

These notions can be made precise using the derivative. We will do so later.

Intersections

The intersections of the curve with the axes of the $x,y$ -plane can be computed. To compute the intersection of the curve with the $y$ -axis, one should solve $\blue{x(t)}=0$ . The intersections with the $x$ -axis are computed by solving $\green{y(t)}=0$ .

Parametric curves are very useful to describe motion in physics.

Throwing a ball

The curve

$\orange P \colon \phantom{x}\begin{cases}\blue{x(t)}&=10 \cdot \cos( \theta ) t\\ \green{y(t)}&= 10 \cdot \sin(\theta) t - \frac{1}{2}g \cdot t^2 \end{cases}$

describes the orbit of an object being thrown from the origin with a varying angle $\theta$ and a speed of $10 \text{ } m/s$ with a gravitational constant $g$ . On most places on earth the gravitational constant is around $9.8 \text{ } m /s^2$ .

It is possible to adjust the values in the picture by using the sliders. Note that the distance thrown is maximal when the angle is $45$ degrees and that the value of $\theta$ in the slider is in radians.

Friction In the example we disregard friction for simplicity.

The maximal height of a parametric curve is attained whenever $\green{y(t)}$ is maximal within the specified interval. If we want to know the maximal height of the curve in the example, we should find the maximal value of

$\green{y(t)}= 10 \cdot \sin(\theta)\cdot t - \frac{1}{2}g \cdot t^2$

The time at which the maximum value is attained can be found by setting the derivative to zero:

$\frac{\partial y(t)}{\partial t}=10\cdot \sin(\theta) - g\cdot t = 0\quad\text{ so }\quad t= \frac{10\cdot \sin(\theta)}{g}$

The maximal height of the curve is therefore the height that is attained at time $t_h= \frac{10\cdot \sin(\theta)}{g}$ , so we substitute this value:

$\begin{array}{rcl}\green{y(t_h)}&=& 10 \cdot \sin(\theta)\cdot t_h - \frac{1}{2}g \cdot t_h^2\\ &=& 10\cdot \sin(\theta)\cdot \frac{10\cdot \sin(\theta)}{g} - \frac{1}{2}g\cdot \left(\frac{10\cdot \sin(\theta)}{g}\right)^2\\ &=& 100\cdot \frac{\sin^2(\theta)}{g}- 50\cdot \frac{\sin^2(\theta)}{g} \\ &=&50\cdot \frac{\sin^2(\theta)}{g} \end{array}$

This gives a maximal height of $50\cdot \frac{\sin^2(\theta)}{g}$ .

Every graph of a function can be described with a parametric curve. In this sense, the theory of parametric curves is a broader theory than the theory of functions and graphs.

If $f(x)$ is a function on a domain $\ivcc{a}{b}$ then we can define a parametric curve $\orange{C}$ as follows

$\begin{array}{rcl}\blue{x(t)}&=&\blue{t}\\\green{y(t)}&=&\green{f(t)},\end{array}$ where $t$ ranges over the interval $\ivcc{a}{b}$ . This curve coincides with the graph of the function $f(x)$ .

Using the definition of a parametric curve as above, it is actually not true that every graph of a function can be described with the parametric equation as in the example. There is a technicality that the domain of a function need not be an interval whereas the variable $t$ in a parametric curve should always vary over an interval. The precise statement therefore is: Every graph of a function defined on an interval can be described with a parametric curve.

Another way to get rid of this technicality is relaxing the definition of a parametric equation such that $t$ is also allowed to vary over any subset of the real numbers.

Not every parametric curve can be described as the graph of a single function. Take for example the unit circle, described with the parametric equation $\begin{array}{rcl}\green{x(t)}&=&\green{\cos (t)}\\ \blue{y(t)}&=&\blue{\sin(t)}.\end{array}$

This describes the circle with radius $1$ centered around the origin.

Since a function can only have a single $y$ -value corresponding to every $x$ -value, we need at least two functions to describe the unit circle. These would be one for the top half and one for the bottom half of the circle.

Given parametric equations $\ivcc{\blue{x(t)}}{\green{y(t)}}$ and an interval $\ivcc{a}{b}$ for $t$ it can be very useful to make a sketch of the curve. This can be done by picking some explicit values for $t$ in $\ivcc{a}{b}$ and substituting these in the parametric equations $\blue{x(t)}$ and $\green{y(t)}$ . After drawing these in the plane it will, in most cases, be clear what the corresponding curve should be.

Example

Consider the curve $\orange C$ given by $\ivcc{ \blue{x(t)}}{ \green{y(t)} }= \ivcc{ \blue{\frac{3t}{1+t^3}}}{ \green{\frac{3t^2}{1+3t^3}}}$ defined for $t \neq -1$ . We picked some values for $t$ and plotted the point $\orange{P_t}$ . The dotted line is the curve $\orange C$ itself.

Determine the maximal height $h$ of the parametric curve given by equations
$\begin{cases} x(t) &= \left(t-6\right)\cdot \left(t+6\right)-1, \\ y(t) &= 5\cdot \cos \left(\sqrt{t^2+5\cdot t+6}\right)+3. \end{cases}$
where $t$ lies in $\left[ -10 , 10 \right]$ .

The maximal height is $8$ .

The maximal height is attained whenever $y(t) = 5\cdot \cos \left(\sqrt{t^2+5\cdot t+6}\right)+3$ is maximal. The cosine is periodic and takes maximal value $1$ . This happens whenever $\sqrt{t^2+5\cdot t+6} = 0$ . Squaring the equation gives us $t^2+5\cdot t+6 = 0$ . We solve this by factorizing. We get $\left(t+2\right)\cdot \left(t+3\right) = 0$ and find solutions $\left[ t=-3 , t=-2 \right]$ . These lie in the desired interval. Consequently, the maximal height is given by substituting one of these values in $y(t)$ . We see that the height is given by $8$ .

New example