9.07 Parametric equations

Understanding parameters

A mathematical function  can be thought of as a set of ordered pairs $\left(x,y\right)$(x,y) on a given domain, and it is often the case that the dependent variable $y$y is related to the independent variable $x$x by some rule. 

As a simple example the function given by $y=4x^2$y=4x2  for real values of $x$x  allows us to generate as many ordered pairs for it as we like. Here are just a few:

$\left(-3,36\right),\left(0,0\right),\left(\frac{1}{2},1\right),\left(5,100\right),\left(1,4\right)...$(−3,36),(0,0),(12​,1),(5,100),(1,4)...

We could put these values into a table:

$x$x	$-4$−4	$0$0	$\frac{1}{2}$12​	$1$1	$\sqrt{8}$√8
$y$y	$64$64	$0$0	$1$1	$4$4	$32$32

 

We often refer to this type of equation, where both $x$x and $y$y are involved, the function's rectangular equation. 

There is, however, another way that we can generate the same set of ordered pairs.

We introduce the idea of a parameter, say $t$t, so that both $x$x and $y$y are functions of $t$t. Thus:

$x=g\left(t\right),y=h\left(t\right)$x=g(t),y=h(t) 

In this new format, ordered pairs are generated according to the single parameter $t$t. In that way, $t$t becomes the independent variable, and $x$x and $y$y vary according to the choice of $t$t made.

For the function $y=4x^2$y=4x2 above, a parametric form of the function (as it is known) becomes:

$x$x	$=$=	$2t$2t
$y$y	$=$=	$16t^2$16t2

Here is the same table of values as those above, generated parametrically:

$t$t	$-2$−2	$0$0	$\frac{1}{4}$14​	$\frac{1}{2}$12​	$\sqrt{2}$√2
$x=2t$x=2t	$-4$−4	$0$0	$\frac{1}{2}$12​	$1$1	$2\sqrt{2}$2√2
$y=16t^2$y=16t2	$4$4	$0$0	$1$1	$4$4	$32$32

This parametric form is certainly not unique for the function. 

 

Parametric forms are not unique

To develop a set of parametric equations for a given function is quite easy. We simply determine a parametric function for the $x$x variable, and then use that to determine what the parametric function is for the $y$y variable.

In the above example, we chose $x=2t$x=2t, and since $y=4x^2$y=4x2, we knew that $y=4\left(2t\right)^2=16t^2$y=4(2t)2=16t2.

We could have just as easily have chosen $x=t$x=t, from which we derive $y=4t^2$y=4t2. Similarly, we could choose, say $x=1-2t$x=1−2t, and this would result in $y=4\left(1-2t\right)^2$y=4(1−2t)2.

Worked examples

Question 1

Derive two parametric expressions for the line given by $y=2x-8$y=2x−8.

We have plenty of choices. Suppose we set $x=t+4$x=t+4. Then $y=2x-8=2\left(t+4\right)-8=2t$y=2x−8=2(t+4)−8=2t. This means that the parametric equations $x=t+4,y=2t$x=t+4,y=2t is a legitimate form.

Now suppose $x=2t$x=2t. Then $y=2\left(2t\right)-8=4t-8$y=2(2t)−8=4t−8. Again, $x=2t,y=4t-8$x=2t,y=4t−8 is another legitimate form.

Question 2

Use the parameter $t=\frac{1}{x}-5$t=1x​−5 to find a set of parametric equations that represent the line $x=\frac{y-1}{5}$x=y−15​.

If $t=\frac{1}{x}-5$t=1x​−5, we can rearrange to show that $x=\frac{1}{t+5}$x=1t+5​. Substituting this into the rectangular form, we have:

$x$x	$=$=	$\frac{y-1}{5}$y−15​
$\frac{1}{t+5}$1t+5​	$=$=	$\frac{y-1}{5}$y−15​
$\frac{5}{t+5}$5t+5​	$=$=	$y-1$y−1
$\therefore$∴    $y$y	$=$=	$1+\frac{5}{t+5}$1+5t+5​
	$=$=	$\frac{t+5}{t+5}+\frac{5}{t+5}$t+5t+5​+5t+5​
	$=$=	$\frac{t+10}{t+5}$t+10t+5​

 Hence, a parametric representation for the function is given by $x=\frac{1}{t+5},y=\frac{t+10}{t+5}$x=1t+5​,y=t+10t+5​.

Reversing the process

Suppose, on the other hand, a parametric form of a function has been provided, and we wish to know the function in the rectangular form $y=f\left(x\right)$y=f(x). 

We simply isolate $t$t in either of the two equations, and substitute into the the other equation.

 

Suppose a parametric form of an equation is given by:

$x$x	$=$=	$2at$2at
$y$y	$=$=	$at^2$at2

where $a$a is some constant.

Then since $x=2at$x=2at, we have that $t=\frac{x}{2a}$t=x2a​ and substituting this into the second equation, we have:

$y$y	$=$=	$at^2$at2
	$=$=	$a\left(\frac{x}{2a}\right)^2$a(x2a​)2
	$=$=	$a\left(\frac{x}{2a}\right)^2$a(x2a​)2
	$=$=	$a\left(\frac{x^2}{4a^2}\right)$a(x24a2​)
	$=$=	$\frac{x^2}{4a}$x24a​
$\therefore$∴     $x^2$x2	$=$=	$4ay$4ay

The two forms $x^2=4ay$x2=4ay and $x=2at,y=at^2$x=2at,y=at2 equivalently describe the function.

Here are some more examples to consider.

Worked examples

Question 3

Find the rectangular equation for the curve given by $x=t^3,y=2\ln t$x=t3,y=2lnt for $t$t in $\left(0,\infty\right)$(0,∞).

Since $x=t^3$x=t3, we have $t=x^{\frac{1}{3}}$t=x13​ and thus $y=2\ln\left(x^{\frac{1}{3}}\right)=\frac{2}{3}\ln x$y=2ln(x13​)=23​lnx. 

Note that the domain is restricted to positive reals.

Question 4

Find the rectangular form for the function defined parametrically as $x=2-t,y=\sqrt{4-t^2}$x=2−t,y=√4−t2 with $t$t defined on $\left[-2,2\right]$[−2,2]. 

Since $x=2-t$x=2−t, we have $t=2-x$t=2−x:

$y$y	$=$=	$\sqrt{4-x^2}$√4−x2
	$=$=	$\sqrt{4-\left(2-x\right)^2}$√4−(2−x)2
	$=$=	$\sqrt{4-4+4x-4x^2}$√4−4+4x−4x2
	$=$=	$2\sqrt{x\left(1-x\right)}$2√x(1−x)

Note that $x$x is defined on the domain $\left[0,1\right]$[0,1].

Since $y=\sqrt{4-t^2}$y=√4−t2 with $t$t defined on $\left[-2,2\right]$[−2,2], the maximum value of $y$y occurs when $t=0$t=0, so that $y=2$y=2, and the minimum value of $y$y occurs when $t=\pm2$t=±2, where $y=0$y=0. therefore the range of the function is given by $\left[0,2\right]$[0,2].  

We could also use the rectangular form $y=2\sqrt{x\left(1-x\right)}$y=2√x(1−x) to arrive at the same conclusion.

Question 5

Find the rectangular form for the function given parametrically by $x=3^t,y=3^{-t}$x=3t,y=3−t for $t$t any real number.

If $x=3^t$x=3t, then $\frac{1}{x}=3^{-t}$1x​=3−t, and this in turn means that $y=\frac{1}{x}$y=1x​. 

Note that the domain excludes $x=0$x=0.

Practice questions

Question 6

Question 7

Question 8

Question 9

Graphing parametric equations

The rectangular equation corresponding to a line or a curve, which is given parametrically as $x=g\left(t\right),y=h\left(t\right)$x=g(t),y=h(t) can be found by eliminating $t$t from the equations.   

For example, the function $x=2t,y=t^2$x=2t,y=t2 for real $t$t, shows $t=\frac{x}{2}$t=x2​ and thus $y=t^2=\left(\frac{x}{2}\right)^2=\frac{1}{4}x^2$y=t2=(x2​)2=14​x2. This graph shows how $t$t changes for various values of $x$x:

The elimination of $t$t from the parametric equations can require a knowledge of certain established identities, such as the Pythagorean identities $\sin^2\theta+\cos^2\theta=1$sin2θ+cos2θ=1, $\sec^2\theta=1+\tan^2\theta$sec2θ=1+tan2θ  and $\csc^2\theta=1+\cot^2\theta$csc2θ=1+cot2θ.   There are also shortcuts that can help you avoid unnecessary algebraic complexity.

Here are four more examples, along with their graphs.

Worked examples

Question 10

$x=t-4$x=t−4, $y=\frac{12}{t^2-8t+16}$y=12t2−8t+16​

Without thinking carefully before we proceed, we might set $t=x+4$t=x+4 and then substitute in to the second equation so that $y=\frac{12}{\left(x+4\right)^2-8\left(x+4\right)+16}$y=12(x+4)2−8(x+4)+16​. Simplification of this becomes cumbersome.

However, noticing first that $t^2-8t+16=\left(t-4\right)^2$t2−8t+16=(t−4)2, we simply write:

$y$y	$=$=	$\frac{12}{t^2-8t+16}$12t2−8t+16​
	$=$=	$\frac{12}{\left(t-4\right)^2}$12(t−4)2​
	$=$=	$\frac{12}{x^2}$12x2​

The graph shows a couple of values of $t$t adjacent to the corresponding coordinates:

 

 

Question 11

$x=12\sin t,y=-12\cos t$x=12sint,y=−12cost for $t$t in $\left[0,2\pi\right]$[0,2π].

Here, instead of isolating $t$t from either equation, we square both sides of each equation and add them together, so that:

$x^2$x2	$=$=	$144\sin^2t$144sin2t     $(1)$(1)
$y^2$y2	$=$=	$144\cos^2t$144cos2t     $(2)$(2)
$\therefore$∴   $x^2+y^2$x2+y2	$=$=	$144\sin^2t+144\cos^2t$144sin2t+144cos2t
	$=$=	$144\left(\sin^2t+\cos^2t\right)$144(sin2t+cos2t)
$\therefore$∴   $x^2+y^2$x2+y2	$=$=	$144$144

This is a circle, center origin and radius $12$12. The same result would have occurred had the parametric equations been given as $x=12\sin t,y=12\cos t$x=12sint,y=12cost, with the negative sign in the second equation omitted.

The interesting difference between the two is the starting position $t=0$t=0. In the original question, the circle is drawn from the point $\left(0,-12\right)$(0,−12), whereas in the case with the negative sign missing, the circle is drawn from $\left(0,12\right)$(0,12).

 

 

Question 12

Sketch the curve given by $x=\ln t,y=\frac{1}{t}$x=lnt,y=1t​ and within the restricted domain $x>0$x>0.

Note first that for $t$t to be defined, it must be positive, but this restriction may not effect the domain of the function itself. However, in this instance, an additional restriction of $x>0$x>0 has been placed on the function, and so the the graph has the domain given by $\left(0,\infty\right]$(0,∞].  

From $x=\ln t$x=lnt we have $t=e^x$t=ex, and therefore $y=\frac{1}{e^x}=e^{-x}$y=1ex​=e−x.

So for example, at $x=\ln e=1$x=lne=1, since $t=e$t=e we have $y=\frac{1}{e}$y=1e​. Similarly, at $x=\ln e^2=2$x=lne2=2, $y=\frac{1}{e^2}$y=1e2​.

 

 

Question 13

$x=5+3\cos t,y=3+5\sin t$x=5+3cost,y=3+5sint for $t$t in $\left[0,2\pi\right]$[0,2π].

From the first equation, we have $\frac{x-5}{3}=\cos t$x−53​=cost and from the second equation we have $\frac{y-3}{5}=\sin t$y−35​=sint. By adding the squares of both sides of these equations, we have:

$\frac{\left(x-5\right)^2}{9}+\frac{\left(y-3\right)^2}{25}$(x−5)29​+(y−3)225​	$=$=	$\cos^2t+\sin^2t$cos2t+sin2t
$\therefore$∴     $\frac{\left(x-5\right)^2}{9}+\frac{\left(y-3\right)^2}{25}$(x−5)29​+(y−3)225​	$=$=	$1$1

This is the equation of an ellipse whose center is located at the point $\left(5,3\right)$(5,3) and whose major and minor axis lengths are $10$10 and $6$6 respectively.

 

Practice questions

Question 14

Question 15

Question 16