Parametric Function

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math

Functions that use a single parameter to define a more complex shape such as a curve.

A classic example of a parametric-function is a [see page 9, sine] wave.

\begin{align*} x = r \sin t y = r \cos t \end{align*}

It can define the shape of a circle with two functions, one for each axis \((x,y)\) of the shape, using the single parameter \( t, \; 0 \leq t \leq 2 \pi \).

Cartesian Form

You can convert a parametric curve to a regular cartesian equation by defining \( t \) in terms of \( x \) (or \( y \)) and substituting it into the other equation.

\begin{align*} x &= t + 1 \
t &= x - 1 \
\
y &= 2t^2 \
y &= 2(x-1)^2 \end{align*}

Differentiation

Furthermore you can differentiate a parametric by differentiating each equation independtly and then combining them. For example:

\begin{align*} x &= 2t^2, & y &= 4t \
\frac{\,dx}{\,dt} &= 4t, & \frac{\,dy}{\,dt} &= 4 \
\frac{\,dy}{\,dx} &= \frac{(\frac{\,dy}{\,dt})}{\frac{\,dx}{\,dt}} \\

                &= \frac{4}{4t} \\\\
                &= \frac{1}{t}

\end{align*}

We can now find the gradient at any point of the curve by simply substituting a value for \( t \) (example: \( f'(3) = \frac{1}{3} \)).

Finding the $y$-Intercept

For example consider the parametric curve \( x = t+1, y = t^3 \).

The Y-intercept can be found by just solving the \( x \) equation to find \( t \) and then substituting into the \( y \) equation:

\begin{align*} t + 1 &= x \
t + 1 &= 0 \
t &= -1 \
\
y &= t^3 \\

&= (-1)^3 \\\\
&= 1

\end{align*}