Parametric Equations

# Definition of a Parametric Equation

In the past, we have seen curves in two dimensions described as a statement of equality involving x and y.

(1)
\begin{equation} f(x,y)=0 \end{equation}

These are sometimes referred to as rectangular equations or Cartesian equations.
An alternative approach is two describe x and y separately in terms of a third parameter, usually t.

(2)
\begin{eqnarray} x=f(t)\\ y=g(t)\\ \end{eqnarray}

These types of equations are called parametric equations. There are several advantages that parametric equations have over Cartesian equations. We can illustrate these advantages through the following example. Here is an example of a cartesian equation and a similar parametric equation,

(3)
\begin{eqnarray} y=x^2\\ \mbox{or}\\ x=t\\ y=t^2\\ - \infty <t < \infty \end{eqnarray}

First, let us look at the graphs of these two equations. Notice that in the parametric equation $x=t$. So if we substitute $x$ in for $t$ into the parametric equation for $y$, we get $y=x^2$. So it would seem logical that their graphs would have the same shape since the relationship between $x$ and $y$ is the same in both equations. Below we have graphs of the two equations. They both have the same shape, but they have different arrow indicators. On the Cartesian equation's graph, the arrows indicate that the graph is continuing on in the established fashion. On the parametric equation, the arrows indicate the direction an object would move along the path as the parameter increases. This leads to our first advantage of parametric equations and that is they not only can show the path of an object they can indicate where the object will be at any given time. This leads to the ability to calculate velocity and acceleration as well. The Cartesian equation's graph cannot do that.
The second advantage to parametric equations comes from their utility in three dimensions. If we want to write the equation of a curve in three dimensions, the Cartesian approach fails completely. That is to say, you cannot write the equation of a curve in 3D using an equation of the form $f(x,y,z)=0$. In order to write the equation, we need to use a parametrization. This parametrization would be the same as in 2D, but with a parametric equation for $z$ added as follows.

(4)
\begin{eqnarray} x=f(t)\\ y=g(t)\\ z=h(t) \end{eqnarray}

An example of such a curve would be the helix below, The equation for this helix looks like

(5)
\begin{eqnarray} x=\cos(t)\\ y=\sin(t)\\ z=t \end{eqnarray}

# Converting Between Cartesian and Parametric

## Converting from Cartesian to Parametric

To convert a Cartesian equation to a parametric equation simply do the following.

1. Solve the equation for one variable, if possible
2. Let $t$ equal the other variable
3. Substitute t into your solved equation

Example
Convert 2x+3y=10 to a parametric form

## Converting from Parametric to Cartesian

To convert a parametric equation to a Cartesian equation, there is no one procedure that will always work. If you can solve for $t$ in one of the equations, you can do the following,

1. Solve for t in one of the parametric equations, if possible.
2. Substitute the result into the other parametric equation.

Example
Convert $x=t^2+5t-1$ , $y=t+1$ to Cartesian form

# Parametric Equation of a Line, Ray and Segment

## Equation of a Line

A parametric equation of the line passing through the point $(x_0,y_0,z_0)$ and parallel to the vector $\vec{v}=<v_x,v_y,v_z>$ is given by

(6)
\begin{eqnarray} x&=&x_0+v_xt\\ y&=&y_0+v_yt \hskip .5in ,-\infty < t < \infty\\ z&=&z_0+v_zt\\ \end{eqnarray}

Example
Find a parametrization for the line passing through the points $P(1,2,3)$ and $Q(0,5,3)$.

## Equation of a Ray

A parametric equation of the ray starting at the point $(x_0,y_0,z_0)$ and in the direction of the vector $\vec{v}=<v_x,v_y,v_z>$ is given by

(7)
\begin{eqnarray} x&=&x_0+v_xt\\ y&=&y_0+v_yt \hskip .5in ,0 \leq t < \infty\\ z&=&z_0+v_zt\\ \end{eqnarray}

Example
Find a parametrization for the ray starting at the point $P(10,12,13)$ and headed in the direction of $\vec{v}=<4,-8,7>$.

## Equation of a Segment

A parametric equation of the line segment from the point $(x_0,y_0,z_0)$ to the point $(x_1,y_1,z_1)$ is given by

(8)
\begin{eqnarray} x&=&x_0+(x_1-x_0)t\\ y&=&y_0+(y_1-y_0)t \hskip .5in ,0 \leq t \leq 1\\ z&=&z_0+(z_1-z_0)t\\ \end{eqnarray}

Example
Find a parametrization for the line segment from the point $P(-4,0,1)$ to the point $Q(3,-2,5)$

### LERP

In parametrizing a line segment you are creating a path between two points of constant velocity. This is referred to as a Linear Interpolation or LERP for short. If we conceptualize this as a path from one vector $\vec{p}_0=<x_0,y_0,z_0>$ to another vector $\vec{p_1}=<x_1,y_1,z_1>$, then we can rewrite our parametrization,

(9)
\begin{eqnarray} x&=&x_0+(x_1-x_0)t\\ y&=&y_0+(y_1-y_0)t \hskip .5in ,0 \leq t \leq 1\\ z&=&z_0+(z_1-z_0)t\\ \end{eqnarray}

as a vector equation,

(10)
\begin{align} L(\vec{p}_0,\vec{p}_1,t)=\vec{p}_0+(\vec{p}_1-\vec{p}_0)t \mbox{ }, 0 \leq t \leq 1 \end{align}

Which we can simplify to,

(11)
\begin{align} L(\vec{p}_0,\vec{p}_1,t)=(1-t)\vec{p}_0+t\vec{p}_1 , 0 \leq t \leq 1 \end{align}

# Parametric Equation of a Circle and an Ellipse

## Circle

To create a parametrization of a circle centered at the origin with radius a, we will use the pathagorean identity that $\sin^2t+\cos^2t=1$. We know that in rectangular form the equation of a circle of radius a centered at the origin is $x^2+y^2=a^2$ or for our purposes $(\frac{x}{a})^2+(\frac{y}{a})^2=1$. So by equating these two equations we see that we can use,

(12)
\begin{eqnarray} \frac{x}{a}=\sin t\\ \frac{y}{a}=\cos t\\ \end{eqnarray}

or simplified

(13)
\begin{eqnarray} x=a\sin t\\ y=a\cos t\\ \end{eqnarray}

This particular parametrization will start at time $t=0$ and the point $(0,a)$. As $t$ starts to increase past zero, the values of both $x$ and $y$ will be positive, so we will be moving into the first quadrant. So our overall motion will be clockwise. We could similarly use the parametrization.

(14)
\begin{eqnarray} x=a\cos t\\ y=a\sin t\\ \end{eqnarray}

Now we will be at the point $(a,0)$ at time $t=0$, but we will still move into the first quadrant afterwards, so we will be moving counter-clockwise. Example
Give a parametrization for the circle $x^2+y^2=9$

## Ellipse

Similarly, to create a parametrization of an ellipse centered at the origin with semi-major axis $a$ and semi-minor axis $b$, we will use the pathagorean identity that $\sin^2t+\cos^2t=1$. We know that in rectangular form the equation of an ellipse centered at the origin with semi-major axis $a$ and semi-minor axis $b$ has equation $(\frac{x}{a})^2+(\frac{y}{b})^2=1$. So by equating these two equations we see that we can use,

(15)
\begin{eqnarray} \frac{x}{a}=\sin t\\ \frac{y}{b}=\cos t\\ \end{eqnarray}

or simplified

(16)
\begin{eqnarray} x=a\sin t\\ y=b\cos t \end{eqnarray}

With this parametrization, we start at the point $(0,b)$ at time $t=0$ and as $t$ increases from zero, both $x$ and $y$ become positive moving us into the first quadrant. So our motion is clockwise. Example
Give a parametrization for the ellipse $\frac{x^2}{49}+y^2=1$, with counter-clockwise motion that completes two revolutions

# Exercises

1. From an initial point $(6, 7, –9)$ and direction $<–2, 5, 4>$ of a line, determine three other points on the line.
2. Write a parametric representation of the line determined by the points $(2, 5, 1)$ and $(7, 6, –3)$.
3. Write a parametrization for a line segment from $(1,6,2)$ to $(0,1,5)$
4. Now parametrize the same path but with twice the speed.
5. Write a parametric representation of the ray originating at the point $(4, 6, –9)$ that has a compass heading 244º and pitch 35º.
6. Find the heading and elevation of the ray $[5+7t,2-6t,4t]$ , when $t>0$ .
7. Give a parametrization for a circle of radius $4$, with a counter-clockwise motion
8. Give a parametrization for the circle $x^2+y^2=16$ with a clockwise motion.
9. Give a parametrization for the ellipse $\frac{x^2}{4}+\frac{y^2}{25}=1$, with a counter-clockwise motion.
10. Convert the Cartesian equation $x-y^2=3$ to a parametric equation.
11. Convert the parametric equation $x=t+4$ , $y=t^3-t$ to a Cartesian equation.
12. Can there be more than one parametrization for a curve?, Explain why or why not.
13. What are the advantages of parametric equations over Cartesian equations?
page revision: 76, last edited: 21 Aug 2019 20:45