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# Parametric Equations

Top
 Sub Topics In plane, parametric equation is a pair of functions given by, $x$ = $f (s)$ and $y$ = $g(s)$ which is used to define the ‘$x$’ and ‘$y$’ coordinate graph of the given curve in the plane. In mathematics, a set of equations represent a set of quantities as unambiguous functions of a number of independent variables, said to be parameters. Parametric equations are the set of equations that express a set of quantities as explicit functions of a number of independent variables known as parameters. For example, for the circle equation the Cartesian coordinate is given by:$r^{2}$  = $x^{2}\ +\ y^{2}$, and the parametric equation is given by:$x$ = $r\ cos\ s$ and $y$ = $r\ sin\ s$; Basically the parametric equation is represented as non unique, so the same quantities are defined by number of different parameterizations. There are two dimensional and three dimensional parametric equations.Parabola and circle are two dimensional parametric equations.Parabola: The equation of parabola is given by: $\Rightarrow\ y$ = $x^{2}$;The given equation is parameterising using the other parameter, when we use another parameter then the equation of parabola is:$\Rightarrow\ x$ = $s$ and,$\Rightarrow\ y$ = $s^{2}$.In case of circle, the ordinary equation of circle is given by:$\Rightarrow\ x^{2}\ +\ y^{2}$ = $1$;The obtained equations are parametrized.

## Graphing Parametric Equations

In a coordinate plane, parametric equation is a pair of functions given by:

$\Rightarrow\ x$ = $f\ (a)$ and $y$ = $g\ (a)$; which is used to define the $x$ and $y$ coordinate graph of given curve in the plane.

Example 1: For the following set of parametric equations plot the parametric curve

$y$ = $2t^{2}\ +\ t$, $y$ = $5t\ -\ 1$;   $-2\ \leq\ t\ \leq\ 2$

Solution: Pick up values for t and plug them into the parametric equations and then plot the points.

 $t$ $y$ = $2t^{2} + t$ $y$ = $5t - 1$ $- 2$ $6$ $- 11$ $- 1$ $1$ $- 6$ $0$ $0$ $- 1$ $0.5$ $1$ $1.5$ $1$ $3$ $4$ $2$ $10$ $9$

From the above graph we can now only see the difference present on the limits on $t$. Any value of t outside the given range is not possible. In this case the curve starts at $t$ = $-\ 2$ and ends at $t$ = $2$.
Example 2: Suppose we have an equation $x\ (a)$ = $3\ sin\ 5a$ and $y\ (a)$ = $3\ sin\ 3a$ then we have to graph parametric equations.

The graph of this equation as:

This is how to graph parametric equations.

Let’s discuss three dimensional parametric equations. ‘Helix’ involve in the three dimensional parametric equation.

Parametric equation is used to define the curve in higher- dimensional space.

For example: $x$ = $a\ cos\ (p)$,

$\Rightarrow\ y$ = $a\ sin\ (p)$;

$\Rightarrow\ z$ = $bp$;

Given equation defines a three dimensional curve, the radius of Helix is $a$ and riser is $2\ \pi\ b$ unit per turn.

## Eliminating the Parameter

Principal way to change from parametric equations to Cartesian equations is to solve one of the equations for t and then substitute that into the other equation. This process is known as eliminating the parameter. Give below is an example for a better understanding of this concept.

### Solved Example

Question: x = 3t + 5 and y = 3t - 2 where t = 5
Solution:

Let's solve one of the x or y equations for t and then substitute that into the other equation, thereby eliminating the t's.
Solve the first equation for 't'.

t = $\frac{x - 5}{3}$

Plugging the value of 't' in the y equation we get

y  = 3( $\frac{x-5}{3}$ ) - 2

$\Rightarrow$ y = x - 7

Therefore, we have y = x - 7 as the cartesian equation for the given parametric equation.
We can even transform the parametric equations into an equation that is familiar.

## Parametric Equations Examples

### Solved Examples

Question 1: Find the equation obtained by eliminating the parameter t from the parametric equations x = t + 5, y = 3t + 7
Solution:

Given x = t + 5, y = 3t + 7

We solve for t from 'x'
$\Rightarrow$ t = x - 5
Plug in the value of t in y, we get
$\rightarrow$ y = 3x - 8

So, y = 3x - 8  is the required equation obtained by eliminating the parameter from the given equations.

Question 2: Find the equation obtained by eliminating the parameter t from the equations
x = $\sqrt{81 -t^{2}}$, y = t

Solution:

Given x = $\sqrt{81 -t^{2}}$, y = t

Substitute t = y in $\sqrt{81 - t^{2}}$
We get x = $\sqrt{81 -y^{2}}$
On squaring, x$^{2}$ = 81 - y2
$\rightarrow$ x$^{2}$ + y$^{2}$ = 81 is the required equation obtained by eliminating the parameter t.

Question 3: Which of the following points corresponds to t = 1 in the parametrization
x = t$^{2}$ + 5, y = 2t + 1, where t is a non zero real number?
1. (10, 2)
2. (-5, 4)
3. (-8, 14)
4. (6, 3)

Solution:

Given x = t$^{2}$ + 5, y = 2t + 1
Put t = 1 in x
We get x = 6
Similarly when t = 1 in y we get y = 3
So (6, 3) is a point corresponding to t = 1 in the parametrization given.
Therefore, 4 is the answer for the given equation.

Question 4: Find the parametric equation for the following equation $\frac{x^{2}}{9}$ + $\frac{y^{2}}{16}$ = 1
Solution:

Given $\frac{x^{2}}{9}$ + $\frac{y^{2}}{16}$ = 1

Instead of 1 we can write as cos$^{2}$t + sin$^{2}$t, since cos$^{2}$t + sin$^{2}$t = 1
The given equation reduces to

$\frac{x^{2}}{9}$ + $\frac{y^{2}}{16}$ = cos$^{2}$t + sin$^{2}$t

$\Rightarrow$ $\frac{x^{2}}{9}$ = cos$^{2}$t

$\frac{y^{2}}{16}$ = sin$^{2}$t

x$^{2}$ = 9 cos$^{2}$t
x = 3 cos t
Similarly y = 4 sin t
Therefore, the parametric equation is x = 3 cost  and y = 4 sin t