- #1
Jhenrique
- 685
- 4
Helow!
For a long time I aks me if exist differentiation/integration with respect to vector and I think that today I discovered the answer! Given:
[tex]f(\vec{r}(t))[/tex]
So, df/dt is:
[tex]\bigtriangledown f\cdot D\vec{r}[/tex]
But, df/dt is:
[tex]\frac{df}{d\vec{r}}\cdot \frac{d\vec{r}}{dt}[/tex]
This means that:
[tex]\frac{d\vec{r}}{dt}=D\vec{r}[/tex]
and:
[tex]\frac{df}{d\vec{r}}=\bigtriangledown f[/tex]
So, analogously, the integration with respect to vector is:
[tex]\int f\;d\vec{r}=\bigtriangledown ^{-1}f[/tex]
What make I think that:
[tex]\frac{\mathrm{d} }{\mathrm{d} \vec{r}}=\bigtriangledown[/tex]
Then, would be possible to formalize the differintegration of vector with respect to vector too by:
[tex]\bigtriangledown \cdot \vec{f},\;\;\;\bigtriangledown \times \vec{f},\;\;\;\bigtriangledown^{-1} \cdot \vec{f}\;\;\;and\;\;\;\bigtriangledown^{-1} \times \vec{f}[/tex]
What you think about all this?
For a long time I aks me if exist differentiation/integration with respect to vector and I think that today I discovered the answer! Given:
[tex]f(\vec{r}(t))[/tex]
So, df/dt is:
[tex]\bigtriangledown f\cdot D\vec{r}[/tex]
But, df/dt is:
[tex]\frac{df}{d\vec{r}}\cdot \frac{d\vec{r}}{dt}[/tex]
This means that:
[tex]\frac{d\vec{r}}{dt}=D\vec{r}[/tex]
and:
[tex]\frac{df}{d\vec{r}}=\bigtriangledown f[/tex]
So, analogously, the integration with respect to vector is:
[tex]\int f\;d\vec{r}=\bigtriangledown ^{-1}f[/tex]
What make I think that:
[tex]\frac{\mathrm{d} }{\mathrm{d} \vec{r}}=\bigtriangledown[/tex]
Then, would be possible to formalize the differintegration of vector with respect to vector too by:
[tex]\bigtriangledown \cdot \vec{f},\;\;\;\bigtriangledown \times \vec{f},\;\;\;\bigtriangledown^{-1} \cdot \vec{f}\;\;\;and\;\;\;\bigtriangledown^{-1} \times \vec{f}[/tex]
What you think about all this?