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

x=x(t),y=y(t). (1)

For getting now the derivativePlanetmathPlanetmath dydx in a point P0 of the curve, we chose another point P of the curve. If the values of the parametre t corresponding these points are t0 and t, we thus have the points  (x(t0),y(t0))  and  (x(t),y(t))  and the slope of the secant line through the points is the difference quotient

y(t)-y(t0)x(t)-x(t0)=y(t)-y(t0)t-t0x(t)-x(t0)t-t0. (2)

Let us assume that the functions (1) are differentiableMathworldPlanetmathPlanetmath when  t=t0  and that  x(t0)0. As we let  tt0, the left side of (2) tends to the derivative dydx and the side to the quotient y(t0)x(t0). Accordingly we have the result

(dydx)t=t0=y(t0)x(t0). (3)

Note that the (3) may be written

dydx=dydtdxdt.

Example. For the cycloid

x=a(φ-sinφ),y=a(1-cosφ),

we obtain

dydx=ddφ(1-cosφ)ddφ(φ-sinφ)=sinφ1-cosφ=cotφ2.
Title derivative for parametric form
Canonical name DerivativeForParametricForm
Date of creation 2013-03-22 17:30:48
Last modified on 2013-03-22 17:30:48
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 9
Author pahio (2872)
Entry type DerivationMathworldPlanetmath
Classification msc 26B05
Classification msc 46G05
Classification msc 26A24
Related topic GoniometricFormulae
Related topic CurvatureOfNielsensSpiral
Related topic Parameter