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Think of a curve being traced out over time, sometimes doubling back on itself or crossing itself. Such a curve cannot be described by a function $y=f(x)$. Instead, we will describe our position along the curve at time $t$ by \begin{eqnarray*} x&=&x(t)\\ y&=&y(t). \end{eqnarray*} Then $x$ and $y$ are related to each other through their dependence on the parameter $t$.
Example
Suppose we trace out a curve according to
\begin{eqnarray*}
x&=&t^2-4t\\
y&=&3t
\end{eqnarray*}
where $t\geq 0$. The arrow on the curve indicates the direction of
increasing time or orientation of the curve. Drag the box
along the curve and notice how $x$ and $y$ vary with $t$.
The parameter does not always represent time:
Example
Consider the parametric equation
\begin{eqnarray*}
x&=&3\cos\theta\\
y&=&3\sin\theta.
\end{eqnarray*}
Here, the parameter $\theta$ represents the polar angle of the position on a
circle of radius $3$ centered at the origin and oriented
counterclockwise.
Differentiating Parametric EquationsLet $x=x(t)$ and $y=y(t)$. Suppose for the moment that we are able to re-write this as $y(t)=f(x(t))$. Then $\displaystyle \frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}$ by the Chain Rule. Solving for $\displaystyle \frac{dy}{dx}$ and assuming $\displaystyle \frac{dx}{dt}\neq 0$, \[\frac{dy}{dx}=\frac{~\frac{dy}{dt}~}{~\frac{dx}{dt}~}\] a formula that holds in general.
Example
If $x=t^2-3$ and $y=t^8$, then $\displaystyle \frac{dx}{dt}=2t$ and
$\displaystyle \frac{dy}{dt}=8t^7$. So
\begin{eqnarray*}
\frac{dy}{dx}&=&\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{8t^7}{2t}=4t^6\\
\frac{d^2y}{dx^2}&=&\frac{d}{dx}\left[\frac{dy}{dx}\right]=
\frac{~\frac{d[\frac{dy}{dx}] \strut}{dt}~}{~\frac{dx}{dt}~}=\frac{24t^5}{2t}=12t^4.
\end{eqnarray*}
Key Concepts
The derivative of $y$ with respect to $x$ (in terms of the parameter $t$) is given by $$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}. $$ |