Eliminating The Parameter From A Parametric Equation

About Pricing Login Subscribe Risk Free Eliminating the parameter from a parametric equation

There are three ways to eliminate the parameter from a parametric equation

Given a parametric curve where our function is defined by two equations, one for ???x??? and one for ???y???, and both of them in terms of a parameter ???t???,

???x=f(t)???

???y=g(t)???

we can eliminate the parameter value in a few different ways.

We can

1. solve each equation for the parameter ???t???, then set the equations equal to one another, or

2. solve one equation for the parameter ???t???, then plug that value into the second equation, or

3. solve each equation for part of an identity, then plug both values into the identity.

How to eliminate the parameter

Take the course

Want to learn more about Calculus 2? I have a step-by-step course for that. :)

Learn More

Eliminating the parameter using the second method

Example

Eliminate the parameter.

???x=2t^2+6???

???y=5t???

We’ll solve ???y=5t??? for ???t???, since this will be easier than solving ???x=2t^2+6??? for ???t???.

???y=5t???

???t=\frac{y}{5}???

Plugging this into the equation for ???x???, we get

???x=2\left(\frac{y}{5}\right)^2+6???

???x=\frac{2y^2}{25}+6???

Removing the fraction, we get

???25x=2y^2+150???

???25x-2y^2=150???

Example

Eliminate the parameter.

???x=e^t???

???y=e^{4t}???

We know that ???y=e^{ab}??? is the same as ???y=(e^a)^b???. If we use this property, we can take ???y=e^{4t}??? and rewrite it as ???y=(e^t)^4???. Since ???x=e^t???, we can substitute ???x??? into ???y=(e^t)^4??? for ???e^t???.

???y=x^4???

Remember, because we have ???e??? in the original parametric equations, and ???e??? requires that ???t>0???, we have to transfer this condition to our final answer, and say

???y=x^4???, where ???x>0???

Let’s try another example using the third method.

Example

Eliminate the parameter.

???x=2\cos{\theta}???

???y=3\sin{\theta}???

???0\le\theta\le2\pi???

Rearranging ???x=2\cos{\theta}??? and ???y=3\sin{\theta}??? to isolate the trigonometric functions, we get

???x=2\cos{\theta}???

???\cos{\theta}=\frac{x}{2}???

and

???y=3\sin{\theta}???

???\sin{\theta}=\frac{y}{3}???

Since we know that ???\sin^2{\theta}+\cos^2{\theta}=1???, we can substitute the values we just found for ???\cos{\theta}??? and ???\sin{\theta}???.

???\left(\frac{y}{3}\right)^2+\left(\frac{x}{2}\right)^2=1???

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

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

???4y^2+9x^2=36???

???9x^2+4y^2=36???

Get access to the complete Calculus 2 course

Get started Learn mathKrista KingSeptember 4, 2020math, learn online, online course, online math, calculus 2, calculus ii, calc 2, calc ii, parametric equations, polar and parametric curves, parametric curves, eliminating the parameter Facebook0 Twitter LinkedIn0 Reddit Tumblr Pinterest0 0 Likes Previous

Comparison theorem for improper integrals

Learn mathKrista KingSeptember 12, 2020math, learn online, online course, online math, calculus 2, calculus ii, calc 2, calc ii, improper integrals, comparison theorem, converging and diverging, convergence and divergence, converging integrals, diverging integrals, comparison series, integrals, integration Next

How to find all four second-order partial derivatives

Learn mathKrista KingSeptember 4, 2020math, learn online, online course, online math, calculus 3, calculus iii, calc 3, calc iii, partial derivatives, partial differentiation, second-order partial derivatives, higher-order partial derivatives, mixed partial derivatives, second-order partial derivative for x, second-order partial derivative for y