# derivative for parametric form

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

 $\displaystyle x=x(t),\quad y=y(t).$ (1)

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

 $\displaystyle\frac{y(t)-y(t_{0})}{x(t)-x(t_{0})}=\frac{\frac{y(t)-y(t_{0})}{t-% t_{0}}}{\frac{x(t)-x(t_{0})}{t-t_{0}}}.$ (2)

Let us assume that the functions (1) are differentiable   when  $t=t_{0}$  and that  $x^{\prime}(t_{0})\neq 0$. As we let  $t\to t_{0}$, the left side of (2) tends to the derivative $\frac{dy}{dx}$ and the side to the quotient $\frac{y^{\prime}(t_{0})}{x^{\prime}(t_{0})}$. Accordingly we have the result

 $\displaystyle\left(\frac{dy}{dx}\right)_{\!t=t_{0}}=\,\frac{y^{\prime}(t_{0})}% {x^{\prime}(t_{0})}.$ (3)

Note that the (3) may be written

 $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.$

Example. For the cycloid

 $x=a(\varphi-\sin{\varphi}),\quad y=a(1-\cos{\varphi}),$

we obtain

 $\frac{dy}{dx}=\frac{\frac{d}{d\varphi}(1-\cos\varphi)}{\frac{d}{d\varphi}(% \varphi-\sin\varphi)}=\frac{\sin\varphi}{1-\cos\varphi}=\cot\frac{\varphi}{2}.$
Title derivative for parametric form DerivativeForParametricForm 2013-03-22 17:30:48 2013-03-22 17:30:48 pahio (2872) pahio (2872) 9 pahio (2872) Derivation  msc 26B05 msc 46G05 msc 26A24 GoniometricFormulae CurvatureOfNielsensSpiral Parameter