I'm reviewing a paper and have run into a problem following the author's math, no doubt in part to my rusty skills with partial differential equations mixed with ordinary differential equations . The authors derives:

$$\dot r = \dot\lambda r^\prime$$

where r is a time varying three dimensional position vector, lambda is a time varying scalar parameter (arc length parameter) associated with a cubic spline representation of a desired position trajectory, the dot denotes the time derivative and the prime implies the partial derivative of r with respect to lambda.

The author then states that the second derivative wrt time of r is:

$$\ddot r = \ddot\lambda r^\prime + \dot\lambda^2 r^{\prime\prime} $$

Applying the chain rule to the r dot equation (I think you still can do that with mixed ODE and PDE) I understand how the first term for r double dot comes about but not the second. If someone can show me the steps that would be most helpful.

The author also states that $$\dot\lambda = \frac{V}{\left\|r^\prime\right\|}$$

Where V is a time varying scalar velocity and the double bars denote the Euclidean norm of r prime. Deriving the second and third time derivatives of lambda looks like a real nightmare. If anyone would like to take a shot at that, I can compare to the paper.


1 Answer 1


$$\ddot r = \ddot\lambda r^\prime + \dot\lambda^2 r^{\prime\prime}$$ Differentiate both sides: $$\dot r = \dot\lambda r^\prime$$

$$\ddot r=r^\prime\dfrac {d}{dt} \dot\lambda +\dot\lambda\dfrac {d}{dt} \dfrac {dr}{d \lambda}$$ $$\ddot r=\ddot\lambda r^\prime+\dot\lambda\dfrac {d}{d \lambda } \left (\dfrac {dr}{d \lambda} \right ) \dfrac {d \lambda}{dt }$$ $$\ddot r=\ddot\lambda r^\prime+\dot\lambda^2\dfrac {d^2r}{d \lambda^2 } $$ Finally: $$\ddot r=\ddot\lambda r^\prime+\dot\lambda^2r'' $$

  • $\begingroup$ Thank you ! A bit of rust came off my old neurons. $\endgroup$
    – dxhound
    Apr 21, 2020 at 13:37
  • $\begingroup$ Hi @dxhound you're welcome ... $\endgroup$ Apr 21, 2020 at 13:40

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