# 10.6 Parametric equations  (Page 3/6)

 Page 3 / 6

## Eliminating the parameter

In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations.

## Eliminating the parameter from polynomial, exponential, and logarithmic equations

For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ We substitute the resulting expression for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ into the second equation. This gives one equation in $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$

## Eliminating the parameter in polynomials

Given $\text{\hspace{0.17em}}x\left(t\right)={t}^{2}+1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\left(t\right)=2+t,\text{\hspace{0.17em}}$ eliminate the parameter, and write the parametric equations as a Cartesian equation.

We will begin with the equation for $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ because the linear equation is easier to solve for $\text{\hspace{0.17em}}t.$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=2+t\hfill \\ y-2=t\hfill \end{array}$

Next, substitute $\text{\hspace{0.17em}}y-2\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ in $\text{\hspace{0.17em}}x\left(t\right).$

The Cartesian form is $\text{\hspace{0.17em}}x={y}^{2}-4y+5.$

Given the equations below, eliminate the parameter and write as a rectangular equation for $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ as a function
of $\text{\hspace{0.17em}}x.$

$\begin{array}{l}\\ \begin{array}{l}x\left(t\right)=2{t}^{2}+6\hfill \\ y\left(t\right)=5-t\hfill \end{array}\end{array}$

$y=5-\sqrt{\frac{1}{2}x-3}$

## Eliminating the parameter in exponential equations

Eliminate the parameter and write as a Cartesian equation: $\text{\hspace{0.17em}}x\left(t\right)={e}^{-t}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\left(t\right)=3{e}^{t},\text{\hspace{0.17em}}\text{\hspace{0.17em}}t>0.\text{\hspace{0.17em}}$

Isolate $\text{\hspace{0.17em}}{e}^{t}.\text{\hspace{0.17em}}$

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x={e}^{-t}\hfill \\ {e}^{t}=\frac{1}{x}\hfill \end{array}$

Substitute the expression into $\text{\hspace{0.17em}}y\left(t\right).$

$\begin{array}{l}y=3{e}^{t}\hfill \\ y=3\left(\frac{1}{x}\right)\hfill \\ y=\frac{3}{x}\hfill \end{array}$

The Cartesian form is $\text{\hspace{0.17em}}y=\frac{3}{x}.$

## Eliminating the parameter in logarithmic equations

Eliminate the parameter and write as a Cartesian equation: $\text{\hspace{0.17em}}x\left(t\right)=\sqrt{t}+2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\left(t\right)=\mathrm{log}\left(t\right).$

Solve the first equation for $\text{\hspace{0.17em}}t.$

Then, substitute the expression for $t$ into the $y$ equation.

$\begin{array}{l}y=\mathrm{log}\left(t\right)\\ y=\mathrm{log}{\left(x-2\right)}^{2}\end{array}$

The Cartesian form is $\text{\hspace{0.17em}}y=\mathrm{log}{\left(x-2\right)}^{2}.$

Eliminate the parameter and write as a rectangular equation .

$\begin{array}{l}\\ \begin{array}{l}x\left(t\right)={t}^{2}\hfill \\ y\left(t\right)=\mathrm{ln}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t>0\hfill \end{array}\end{array}$

$y=\mathrm{ln}\sqrt{x}$

## Eliminating the parameter from trigonometric equations

Eliminating the parameter from trigonometric equations is a straightforward substitution. We can use a few of the familiar trigonometric identities and the Pythagorean Theorem.

First, we use the identities:

$\begin{array}{l}x\left(t\right)=a\mathrm{cos}\text{\hspace{0.17em}}t\\ y\left(t\right)=b\mathrm{sin}\text{\hspace{0.17em}}t\end{array}$

Solving for $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}$ we have

$\begin{array}{l}\frac{x}{a}=\mathrm{cos}\text{\hspace{0.17em}}t\\ \frac{y}{b}=\mathrm{sin}\text{\hspace{0.17em}}t\end{array}$

Then, use the Pythagorean Theorem:

${\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1$

Substituting gives

${\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t={\left(\frac{x}{a}\right)}^{2}+{\left(\frac{y}{b}\right)}^{2}=1$

## Eliminating the parameter from a pair of trigonometric parametric equations

Eliminate the parameter from the given pair of trigonometric equations where $\text{\hspace{0.17em}}0\le t\le 2\pi \text{\hspace{0.17em}}$ and sketch the graph.

$\begin{array}{l}x\left(t\right)=4\mathrm{cos}\text{\hspace{0.17em}}t\\ y\left(t\right)=3\mathrm{sin}\text{\hspace{0.17em}}t\end{array}$

Solving for $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t,$ we have

$\begin{array}{l}\text{\hspace{0.17em}}x=4\mathrm{cos}\text{\hspace{0.17em}}t\hfill \\ \frac{x}{4}=\mathrm{cos}\text{\hspace{0.17em}}t\hfill \\ \text{\hspace{0.17em}}y=3\mathrm{sin}\text{\hspace{0.17em}}t\hfill \\ \frac{y}{3}=\mathrm{sin}\text{\hspace{0.17em}}t\hfill \end{array}$

Next, use the Pythagorean identity and make the substitutions.

$\begin{array}{r}\hfill {\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1\\ \hfill {\left(\frac{x}{4}\right)}^{2}+{\left(\frac{y}{3}\right)}^{2}=1\\ \hfill \frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1\end{array}$

The graph for the equation is shown in [link] .

prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
why {2kπ} union {kπ}={kπ}?
why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
what is complex numbers
give me treganamentry question
Solve 2cos x + 3sin x = 0.5
madras university algebra questions papers first year B. SC. maths
Hey
Rightspect
hi
chesky
Give me algebra questions
Rightspect
how to send you
Vandna
What does this mean
cos(x+iy)=cos alpha+isinalpha prove that: sin⁴x=sin²alpha
cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha
rajan
cos(x+iy)=cos aplha+i sinalpha prove that: sinh⁴y=sin²alpha
rajan
is there any case that you can have a polynomials with a degree of four?
victor
***sscc.edu/home/jdavidso/math/catalog/polynomials/fourth/fourth.html
Oliver
can you solve it step b step
give me some important question in tregnamentry
Anshuman
what is linear equation with one unknown 2x+5=3
-4
Joel
x=-4
Joel
x=-1
Joan
I was wrong. I didn't move all constants to the right of the equation.
Joel
x=-1
Cristian
y=x+1
gary
what is the VA Ha D R X int Y int of f(x) =x²+4x+4/x+2 f(x) =x³-1/x-1
can I get help with this?
Wayne
Are they two separate problems or are the two functions a system?
Wilson
Also, is the first x squared in "x+4x+4"
Wilson
x^2+4x+4?
Wilson
thank you
Wilson
Wilson
f(x)=x square-root 2 +2x+1 how to solve this value
Wilson
what is algebra
The product of two is 32. Find a function that represents the sum of their squares.
Paul
if theta =30degree so COS2 theta = 1- 10 square theta upon 1 + tan squared theta