Parametric Equations

The Essentials

Parametric equations simplify an original equations with x and y to a more workable form. Parametric equations can also describe motion in space as functions of time y(t) and x(t). It essentially turns a regular position equation into a vector of functions of time.

Example

Example 1)

if you throw a ball with an initial velocity in the x direction of 3 \( \frac{m}{s} \). Because it of gravity, it experiences an acceleration of -9.81 \( \frac{m}{s^2} \). There is no resistance or losses, so the position equations can be denoted as...

\[ f(t) = < x(t),y(t) > \]
\[ x(t)=3t \]
\[ y(t)=\frac{-9.81}{2}t^2 \]

To find the velocity of the ball we can take the derivative of f(t).

\[ f'(t)=< 3,-9.81t > \]

The acceleration is the second derivative of the same equation.

\[ f"(t)=< 0,-9.81 > \]

Example 2)

When given the parameterization of an equation, we can convert a regular equation to a vector system of time. If we have a function \( y=3x+4 \), and we are given that \( x=t^2 \) . Then we can plug x(t) into the y(x) function to find the parametric system of \( f(t)=< t^2,3t^2+4 > \)

Example 3)

Sometimes we are given a parametric equation, and it is helpful to convert it back to a regular function. This can be done by solving for t and then substituting it back into the equation. If we have a given parameterization of \( f(t)=< t^2+3t,t+6 > \), we can solve for t and find an equation.

\[ y=t+6 \]
\[ t=y-6 \]

Once we have solved for t, we can plug it into the x part of the parameterized equations to find x as a function of y.

\[ x=t^2+3t \]
\[ x=(6-y)^2+3(y-6) \]
\[ x=y^2-9y+18 \]

Practice

Evaluate these expressions:

  1. Find the parameterization of the equation \( 9=x^2+y^2\)
  2. What is the equation that correlates with the parametric equation \( f(t)=< 6t,4t^3+2t > \)

Solutions:

  1. \( f(t)=< 3cos(x),3sin(x) > \)
  2. \( y=\frac{1}{54}x^3+\frac{1}{3}x \)

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