# 8.6 Parametric equations  (Page 2/6)

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However, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. This will become clearer as we move forward.

## Parametric equations

Suppose $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is a number on an interval, $\text{\hspace{0.17em}}I.\text{\hspace{0.17em}}$ The set of ordered pairs, $\text{\hspace{0.17em}}\left(x\left(t\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\left(t\right)\right),\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}x=f\left(t\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=g\left(t\right),$ forms a plane curve based on the parameter $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ The equations $\text{\hspace{0.17em}}x=f\left(t\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=g\left(t\right)\text{\hspace{0.17em}}$ are the parametric equations.

## Parameterizing a curve

Parameterize the curve $\text{\hspace{0.17em}}y={x}^{2}-1\text{\hspace{0.17em}}$ letting $\text{\hspace{0.17em}}x\left(t\right)=t.\text{\hspace{0.17em}}$ Graph both equations.

If $\text{\hspace{0.17em}}x\left(t\right)=t,\text{\hspace{0.17em}}$ then to find $\text{\hspace{0.17em}}y\left(t\right)\text{\hspace{0.17em}}$ we replace the variable $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ with the expression given in $\text{\hspace{0.17em}}x\left(t\right).\text{\hspace{0.17em}}$ In other words, $\text{\hspace{0.17em}}y\left(t\right)={t}^{2}-1.$ Make a table of values similar to [link] , and sketch the graph.

$t$ $x\left(t\right)$ $y\left(t\right)$
$-4$ $-4$ $y\left(-4\right)={\left(-4\right)}^{2}-1=15$
$-3$ $-3$ $y\left(-3\right)={\left(-3\right)}^{2}-1=8$
$-2$ $-2$ $y\left(-2\right)={\left(-2\right)}^{2}-1=3$
$-1$ $-1$ $y\left(-1\right)={\left(-1\right)}^{2}-1=0$
$0$ $0$ $y\left(0\right)={\left(0\right)}^{2}-1=-1$
$1$ $1$ $y\left(1\right)={\left(1\right)}^{2}-1=0$
$2$ $2$ $y\left(2\right)={\left(2\right)}^{2}-1=3$
$3$ $3$ $y\left(3\right)={\left(3\right)}^{2}-1=8$
$4$ $4$ $y\left(4\right)={\left(4\right)}^{2}-1=15$

See the graphs in [link] . It may be helpful to use the TRACE feature of a graphing calculator to see how the points are generated as $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ increases.

Construct a table of values and plot the parametric equations: $\text{\hspace{0.17em}}x\left(t\right)=t-3,\text{\hspace{0.17em}}\text{\hspace{0.17em}}y\left(t\right)=2t+4;\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-1\le t\le 2.$

 $t$ $x\left(t\right)$ $y\left(t\right)$ $-1$ $-4$ $2$ $0$ $-3$ $4$ $1$ $-2$ $6$ $2$ $-1$ $8$

## Finding a pair of parametric equations

Find a pair of parametric equations that models the graph of $\text{\hspace{0.17em}}y=1-{x}^{2},\text{\hspace{0.17em}}$ using the parameter $\text{\hspace{0.17em}}x\left(t\right)=t.\text{\hspace{0.17em}}$ Plot some points and sketch the graph.

If $\text{\hspace{0.17em}}x\left(t\right)=t\text{\hspace{0.17em}}$ and we substitute $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ into the $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ equation, then $\text{\hspace{0.17em}}y\left(t\right)=1-{t}^{2}.\text{\hspace{0.17em}}$ Our pair of parametric equations is

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

To graph the equations, first we construct a table of values like that in [link] . We can choose values around $\text{\hspace{0.17em}}t=0,\text{\hspace{0.17em}}$ from $\text{\hspace{0.17em}}t=-3\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}t=3.\text{\hspace{0.17em}}$ The values in the $\text{\hspace{0.17em}}x\left(t\right)\text{\hspace{0.17em}}$ column will be the same as those in the $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ column because $\text{\hspace{0.17em}}x\left(t\right)=t.\text{\hspace{0.17em}}$ Calculate values for the column $\text{\hspace{0.17em}}y\left(t\right).\text{\hspace{0.17em}}$

$t$ $x\left(t\right)=t$ $y\left(t\right)=1-{t}^{2}$
$-3$ $-3$ $y\left(-3\right)=1-{\left(-3\right)}^{2}=-8$
$-2$ $-2$ $y\left(-2\right)=1-{\left(-2\right)}^{2}=-3$
$-1$ $-1$ $y\left(-1\right)=1-{\left(-1\right)}^{2}=0$
$0$ $0$ $y\left(0\right)=1-0=1$
$1$ $1$ $y\left(1\right)=1-{\left(1\right)}^{2}=0$
$2$ $2$ $y\left(2\right)=1-{\left(2\right)}^{2}=-3$
$3$ $3$ $y\left(3\right)=1-{\left(3\right)}^{2}=-8$

The graph of $\text{\hspace{0.17em}}y=1-{t}^{2}\text{\hspace{0.17em}}$ is a parabola facing downward, as shown in [link] . We have mapped the curve over the interval $\text{\hspace{0.17em}}\left[-3,\text{\hspace{0.17em}}3\right],$ shown as a solid line with arrows indicating the orientation of the curve according to $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ Orientation refers to the path traced along the curve in terms of increasing values of $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ As this parabola is symmetric with respect to the line $\text{\hspace{0.17em}}x=0,\text{\hspace{0.17em}}$ the values of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ are reflected across the y -axis.

Parameterize the curve given by $\text{\hspace{0.17em}}x={y}^{3}-2y.$

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

## Finding parametric equations that model given criteria

An object travels at a steady rate along a straight path $\text{\hspace{0.17em}}\left(-5,\text{\hspace{0.17em}}3\right)\text{\hspace{0.17em}}$ to $\text{\hspace{0.17em}}\left(3,\text{\hspace{0.17em}}-1\right)\text{\hspace{0.17em}}$ in the same plane in four seconds. The coordinates are measured in meters. Find parametric equations for the position of the object.

The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. The x -value of the object starts at $\text{\hspace{0.17em}}-5\text{\hspace{0.17em}}$ meters and goes to 3 meters. This means the distance x has changed by 8 meters in 4 seconds, which is a rate of or $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}\text{m}/\text{s}.\text{\hspace{0.17em}}$ We can write the x -coordinate as a linear function with respect to time as $\text{\hspace{0.17em}}x\left(t\right)=2t-5.\text{\hspace{0.17em}}$ In the linear function template $\text{\hspace{0.17em}}y=mx+b,2t=mx\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}-5=b.$

Similarly, the y -value of the object starts at 3 and goes to $\text{\hspace{0.17em}}-1,\text{\hspace{0.17em}}$ which is a change in the distance y of −4 meters in 4 seconds, which is a rate of or $\text{\hspace{0.17em}}-1\text{m}/\text{s}.\text{\hspace{0.17em}}$ We can also write the y -coordinate as the linear function $\text{\hspace{0.17em}}y\left(t\right)=-t+3.\text{\hspace{0.17em}}$ Together, these are the parametric equations for the position of the object, where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ are expressed in meters and $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ represents time:

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

Using these equations, we can build a table of values for $\text{\hspace{0.17em}}t,x,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y$ (see [link] ). In this example, we limited values of $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ to non-negative numbers. In general, any value of $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ can be used.

$t$ $x\left(t\right)=2t-5$ $y\left(t\right)=-t+3$
$0$ $x=2\left(0\right)-5=-5$ $y=-\left(0\right)+3=3$
$1$ $x=2\left(1\right)-5=-3$ $y=-\left(1\right)+3=2$
$2$ $x=2\left(2\right)-5=-1$ $y=-\left(2\right)+3=1$
$3$ $x=2\left(3\right)-5=1$ $y=-\left(3\right)+3=0$
$4$ $x=2\left(4\right)-5=3$ $y=-\left(4\right)+3=-1$

From this table, we can create three graphs, as shown in [link] .

can you not take the square root of a negative number
No because a negative times a negative is a positive. No matter what you do you can never multiply the same number by itself and end with a negative
lurverkitten
Actually you can. you get what's called an Imaginary number denoted by i which is represented on the complex plane. The reply above would be correct if we were still confined to the "real" number line.
Liam
Suppose P= {-3,1,3} Q={-3,-2-1} and R= {-2,2,3}.what is the intersection
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
All real x except 5 and - 3
Spiro
***youtu.be/ESxOXfh2Poc
Loree
how to prroved cos⁴x-sin⁴x= cos²x-sin²x are equal
Don't think that you can.
Elliott
By using some imaginary no.
Tanmay
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
Where do the rays point?
Spiro
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