Let $r:I \rightarrow \mathbb R^3$ be a continious differentiable curve. Prove the following formula $\frac{d}{dt}(\frac{r(t)}{\Vert r \Vert})=\frac{c\times r}{\Vert r \Vert^3}$ where $c:=r \times \dot r$

Just "calculating" the derivative I get the following:

$\frac{d}{dt}(\frac{r(t)}{\Vert r \Vert})=\frac{\dot r(t)}{\Vert r \Vert}-\frac{r(t)}{\Vert r \Vert^3}(r\cdot \dot r)\; \; \;$ where $\times$ is the cross product and $\cdot$ the scalar multiplication

I would appreciate some help/tips because I dont know how to go from here.


Hint $$c\times r=(r\times \dot r)\times r=(r.r)\dot r-(\dot r.r)r$$

  • $\begingroup$ Very neat formula. Didnt know this one. $\endgroup$ – XPenguen Jan 12 '17 at 22:01
  • $\begingroup$ @XPenguen you can search for "triple vector product" $\endgroup$ – Semsem Jan 12 '17 at 22:06

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