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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 x and y . 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 t 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 t . We substitute the resulting expression for t into the second equation. This gives one equation in x and y .

Eliminating the parameter in polynomials

Given x ( t ) = t 2 + 1 and y ( t ) = 2 + t , eliminate the parameter, and write the parametric equations as a Cartesian equation.

We will begin with the equation for y because the linear equation is easier to solve for t .

y = 2 + t y 2 = t

Next, substitute y 2 for t in x ( t ) .

x = t 2 + 1 x = ( y 2 ) 2 + 1 Substitute the expression for  t  into  x . x = y 2 4 y + 4 + 1 x = y 2 4 y + 5 x = y 2 4 y + 5

The Cartesian form is x = y 2 4 y + 5.

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Given the equations below, eliminate the parameter and write as a rectangular equation for y as a function
of x .

x ( t ) = 2 t 2 + 6 y ( t ) = 5 t

y = 5 1 2 x 3

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Eliminating the parameter in exponential equations

Eliminate the parameter and write as a Cartesian equation: x ( t ) = e t and y ( t ) = 3 e t , t > 0.

Isolate e t .

x = e t e t = 1 x

Substitute the expression into y ( t ) .

y = 3 e t y = 3 ( 1 x ) y = 3 x

The Cartesian form is y = 3 x .

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Eliminating the parameter in logarithmic equations

Eliminate the parameter and write as a Cartesian equation: x ( t ) = t + 2 and y ( t ) = log ( t ) .

Solve the first equation for t .

            x = t + 2      x 2 = t ( x 2 ) 2 = t Square both sides .

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

y = log ( t ) y = log ( x 2 ) 2

The Cartesian form is y = log ( x 2 ) 2 .

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Eliminate the parameter and write as a rectangular equation .

x ( t ) = t 2 y ( t ) = ln t t > 0

y = ln x

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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:

x ( t ) = a cos t y ( t ) = b sin t

Solving for cos t and sin t , we have

x a = cos t y b = sin t

Then, use the Pythagorean Theorem:

cos 2 t + sin 2 t = 1

Substituting gives

cos 2 t + sin 2 t = ( x a ) 2 + ( y b ) 2 = 1

Eliminating the parameter from a pair of trigonometric parametric equations

Eliminate the parameter from the given pair of trigonometric equations where 0 t 2 π and sketch the graph.

x ( t ) = 4 cos t y ( t ) = 3 sin t

Solving for cos t and sin t , we have

x = 4 cos t x 4 = cos t y = 3 sin t y 3 = sin t

Next, use the Pythagorean identity and make the substitutions.

cos 2 t + sin 2 t = 1 ( x 4 ) 2 + ( y 3 ) 2 = 1 x 2 16 + y 2 9 = 1

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

Graph of given ellipse centered at (0,0).
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Questions & Answers

how to understand calculus?
Jenica Reply
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
rachel Reply
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
Reena Reply
what is foci?
Reena Reply
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
Bryssen Reply
i want to sure my answer of the exercise
meena Reply
what is the diameter of(x-2)²+(y-3)²=25
Den Reply
how to solve the Identity ?
Barcenas Reply
what type of identity
Jeffrey
Confunction Identity
Barcenas
how to solve the sums
meena
hello guys
meena
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
Shakeena Reply
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
Rhudy Reply
what is a complex number used for?
Drew Reply
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
They are used in any profession where the phase of a waveform has to be accounted for in the calculations. Imagine two electrical signals in a wire that are out of phase by 90°. At some times they will interfere constructively, others destructively. Complex numbers simplify those equations
Tim
Is there any rule we can use to get the nth term ?
Anwar Reply
how do you get the (1.4427)^t in the carp problem?
Gabrielle Reply
A hedge is contrusted to be in the shape of hyperbola near a fountain at the center of yard.the hedge will follow the asymptotes y=x and y=-x and closest distance near the distance to the centre fountain at 5 yards find the eqution of the hyperbola
ayesha Reply
A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. To the nearest hour, what is the half-life of the drug?
Sandra Reply
Practice Key Terms 1

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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