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 derivative 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 differentiable when t=t0 and that x′(t0)≠0. As we let t→t0, 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 |
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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 | Derivation![]() |
Classification | msc 26B05 |
Classification | msc 46G05 |
Classification | msc 26A24 |
Related topic | GoniometricFormulae |
Related topic | CurvatureOfNielsensSpiral |
Related topic | Parameter |