What is the Relationship Between Position and Velocity Vectors?

[Jhenrique]

If:
\vec{v}=\frac{\mathrm{d} \vec{r}}{\mathrm{d} t}
so:
\\ \vec{v} = \frac{d\vec{r}}{dt} \\ \\ \vec{v}\;dt = d\vec{r} \\ \\ dt = \frac{d\vec{r}}{\vec{v}} \\ \\ \int dt = \int \frac{d\vec{r}}{\vec{v}} \\ \\ t = \int \frac{d\vec{r}}{\vec{v}}
Is true?
Solving the equation for time t, is need divide the position vector r by velocicty vector v... But I don't know do division between vectors...

 

[D H]

Division by a vector is not defined.

It appears you are trying to solve some problem and you thought this might be a way to attack that problem. What is the problem that are you trying to solve?

 

[Jhenrique]

Given a vectorial velocity and a position vector, I want to calcule o time t.
In other words, I want solving the equation above for variable t.
\\ \vec{v} = \frac{d\vec{r}}{dt}

 

[HallsofIvy]

In general, if you are given \vec{v} and \frac{d\vec{r}}{dt}, there will NOT BE a variable t such that \vec{v}= \frac{d\vec{r}}{dt}!

You can try \int \vec{v}dt= \int\vec{dr} and then solve the resulting equation for t.

 

[ChrisVer]

\vec{r} \vec{u}= \vec{r} \frac{d\vec{r}}{dt}
\vec{r} \vec{u} =\frac{1}{2} \frac{dr^{2}}{dt}
dt= \frac{dr^{2}}{\vec{r} \vec{u}}=\frac{dr^{2}}{ru cosθ}

is that correct?

 

[Jhenrique]

In this case, the only solution is to consider the modulus of vectos: t = ∫ 1/v dr. Correct!?

 

[Jhenrique]

And more one thing, if no exist reciprocal of vector, so why exist derivative with respect to vector?
http://en.wikipedia.org/wiki/Matrix_calculus#Derivatives_with_vectors
http://en.wikipedia.org/wiki/Matrix_calculus#Derivatives_with_matrices