# 8.6 Parametric equations  (Page 4/6)

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Eliminate the parameter from the given pair of parametric equations and write as a Cartesian equation: $\text{\hspace{0.17em}}x\left(t\right)=2\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\left(t\right)=3\mathrm{sin}\text{\hspace{0.17em}}t.$

$\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1$

## Finding cartesian equations from curves defined parametrically

When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as $\text{\hspace{0.17em}}x\left(t\right)=t.\text{\hspace{0.17em}}$ In this case, $\text{\hspace{0.17em}}y\left(t\right)\text{\hspace{0.17em}}$ can be any expression. For example, consider the following pair of equations.

$\begin{array}{l}x\left(t\right)=t\\ y\left(t\right)={t}^{2}-3\end{array}$

Rewriting this set of parametric equations is a matter of substituting $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ Thus, the Cartesian equation is $\text{\hspace{0.17em}}y={x}^{2}-3.$

## Finding a cartesian equation using alternate methods

Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations.

$\begin{array}{l}\\ \begin{array}{l}x\left(t\right)=3t-2\hfill \\ y\left(t\right)=t+1\hfill \end{array}\end{array}$

Method 1 . First, let’s solve the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ equation for $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ Then we can substitute the result into the $y$ equation.

Now substitute the expression for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ into the $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ equation.

$\begin{array}{l}y=t+1\hfill \\ y=\left(\frac{x+2}{3}\right)+1\hfill \\ y=\frac{x}{3}+\frac{2}{3}+1\hfill \\ y=\frac{1}{3}x+\frac{5}{3}\hfill \end{array}$

Method 2 . Solve the $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ equation for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and substitute this expression in the $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ equation.

Make the substitution and then solve for $\text{\hspace{0.17em}}y.$

Write the given parametric equations as a Cartesian equation: $\text{\hspace{0.17em}}x\left(t\right)={t}^{3}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\left(t\right)={t}^{6}.$

$y={x}^{2}$

## Finding parametric equations for curves defined by rectangular equations

Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to represent $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ and then substitute it into the $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ equation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ as the domain of the rectangular equation, then the graphs will be different.

## Finding a set of parametric equations for curves defined by rectangular equations

Find a set of equivalent parametric equations for $\text{\hspace{0.17em}}y={\left(x+3\right)}^{2}+1.$

An obvious choice would be to let $\text{\hspace{0.17em}}x\left(t\right)=t.\text{\hspace{0.17em}}$ Then $\text{\hspace{0.17em}}y\left(t\right)={\left(t+3\right)}^{2}+1.$ But let’s try something more interesting. What if we let $\text{\hspace{0.17em}}x=t+3?\text{\hspace{0.17em}}$ Then we have

$\begin{array}{l}y={\left(x+3\right)}^{2}+1\hfill \\ y={\left(\left(t+3\right)+3\right)}^{2}+1\hfill \\ y={\left(t+6\right)}^{2}+1\hfill \end{array}$

The set of parametric equations is

$\begin{array}{l}\hfill \\ x\left(t\right)=t+3\hfill \\ y\left(t\right)={\left(t+6\right)}^{2}+1\hfill \end{array}$

Access these online resources for additional instruction and practice with parametric equations.

## Key concepts

• Parameterizing a curve involves translating a rectangular equation in two variables, $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y,\text{\hspace{0.17em}}$ into two equations in three variables, x , y , and t . Often, more information is obtained from a set of parametric equations. See [link] , [link] , and [link] .
• Sometimes equations are simpler to graph when written in rectangular form. By eliminating $\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}$ an equation in $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is the result.
• To eliminate $\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}$ solve one of the equations for $\text{\hspace{0.17em}}t,\text{\hspace{0.17em}}$ and substitute the expression into the second equation. See [link] , [link] , [link] , and [link] .
• Finding the rectangular equation for a curve defined parametrically is basically the same as eliminating the parameter. Solve for $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ in one of the equations, and substitute the expression into the second equation. See [link] .
• There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation.
• Find an expression for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ such that the domain of the set of parametric equations remains the same as the original rectangular equation. See [link] .

can I get some pretty basic questions
In what way does set notation relate to function notation
Ama
is precalculus needed to take caculus
It depends on what you already know. Just test yourself with some precalculus questions. If you find them easy, you're good to go.
Spiro
the solution doesn't seem right for this problem
what is the domain of f(x)=x-4/x^2-2x-15 then
x is different from -5&3
Seid
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
how do you provided cos⁴x-sin⁴x = cos²x-sin²x are equal
What are the question marks for?
Elliott
Someone should please solve it for me Add 2over ×+3 +y-4 over 5 simplify (×+a)with square root of two -×root 2 all over a multiply 1over ×-y{(×-y)(×+y)} over ×y
For the first question, I got (3y-2)/15 Second one, I got Root 2 Third one, I got 1/(y to the fourth power) I dont if it's right cause I can barely understand the question.
Is under distribute property, inverse function, algebra and addition and multiplication function; so is a combined question
Abena
find the equation of the line if m=3, and b=-2
graph the following linear equation using intercepts method. 2x+y=4
Ashley
how
Wargod
what?
John
ok, one moment
UriEl
how do I post your graph for you?
UriEl
it won't let me send an image?
UriEl
also for the first one... y=mx+b so.... y=3x-2
UriEl
y=mx+b you were already given the 'm' and 'b'. so.. y=3x-2
Tommy
Please were did you get y=mx+b from
Abena
y=mx+b is the formula of a straight line. where m = the slope & b = where the line crosses the y-axis. In this case, being that the "m" and "b", are given, all you have to do is plug them into the formula to complete the equation.
Tommy
thanks Tommy
Nimo
0=3x-2 2=3x x=3/2 then . y=3/2X-2 I think
Given
co ordinates for x x=0,(-2,0) x=1,(1,1) x=2,(2,4)
neil
"7"has an open circle and "10"has a filled in circle who can I have a set builder notation
x=-b+_Гb2-(4ac) ______________ 2a
I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
so good
abdikarin
this is an identity when 2 adding two angles within a cosine. it's called the cosine sum formula. there is also a different formula when cosine has an angle minus another angle it's called the sum and difference formulas and they are under any list of trig identities
strategies to form the general term
carlmark
consider r(a+b) = ra + rb. The a and b are the trig identity.
Mike
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator