derivative for parametric formDerivativeForParametricForm2007-08-29 17:33:252008-03-23 10:09:45Derivationderivativeparametric form% this is the default PlanetMath preamble. as your knowledge
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Instead of the usual way\, $y = f(x)$\, to present plane curves it is in many cases more comfortable to express both coordinates, $x$ and $y$, by means of a suitable auxiliary variable, the parametre. It is true e.g. for the cycloid curve.\\
Suppose we have the parametric form
\begin{align}
x = x(t),\quad y = y(t).
\end{align}
For getting now the derivative $\displaystyle\frac{dy}{dx}$ in a point $P_0$ of the curve, we chose another point $P$ of the curve. If the values of the parametre $t$ corresponding these points are $t_0$ and $t$, we thus have the points\, $(x(t_0),\,y(t_0))$\, and\, $(x(t),\,y(t))$\, and the slope of the secant line through the points is the difference quotient
\begin{align}
\frac{y(t)-y(t_0)}{x(t)-x(t_0)} = \frac{\frac{y(t)-y(t_0)}{t-t_0}}{\frac{x(t)-x(t_0)}{t-t_0}}.
\end{align}
Let us assume that the functions (1) are differentiable when\, $t = t_0$\, and that\, $x'(t_0) \neq 0$. As we let\, $t\to t_0$, the left side of (2) tends to the derivative $\frac{dy}{dx}$ and the \PMlinkescapetext{right} side to the quotient $\frac{y'(t_0)}{x'(t_0)}$. Accordingly we have the result
\begin{align}
\left(\frac{dy}{dx}\right)_{\!t=t_0} =\, \frac{y'(t_0)}{x'(t_0)}.
\end{align}
Note that the \PMlinkescapetext{formula} (3)
may be written
$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$$
\textbf{Example.} For the cycloid
$$x = a(\varphi-\sin{\varphi}),\quad y = a(1-\cos{\varphi}),$$
we obtain
$$\frac{dy}{dx} = \frac{\frac{d}{d\varphi}(1-\cos\varphi)}{\frac{d}{d\varphi}(\varphi-\sin\varphi)}
= \frac{\sin\varphi}{1-\cos\varphi} = \cot\frac{\varphi}{2}.$$