Jhenrique
- 676
- 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:
f(\vec{r}(t))
So, df/dt is:
\bigtriangledown f\cdot D\vec{r}
But, df/dt is:
\frac{df}{d\vec{r}}\cdot \frac{d\vec{r}}{dt}
This means that:
\frac{d\vec{r}}{dt}=D\vec{r}
and:
\frac{df}{d\vec{r}}=\bigtriangledown f
So, analogously, the integration with respect to vector is:
\int f\;d\vec{r}=\bigtriangledown ^{-1}f
What make I think that:
\frac{\mathrm{d} }{\mathrm{d} \vec{r}}=\bigtriangledown
Then, would be possible to formalize the differintegration of vector with respect to vector too by:
\bigtriangledown \cdot \vec{f},\;\;\;\bigtriangledown \times \vec{f},\;\;\;\bigtriangledown^{-1} \cdot \vec{f}\;\;\;and\;\;\;\bigtriangledown^{-1} \times \vec{f}
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:
f(\vec{r}(t))
So, df/dt is:
\bigtriangledown f\cdot D\vec{r}
But, df/dt is:
\frac{df}{d\vec{r}}\cdot \frac{d\vec{r}}{dt}
This means that:
\frac{d\vec{r}}{dt}=D\vec{r}
and:
\frac{df}{d\vec{r}}=\bigtriangledown f
So, analogously, the integration with respect to vector is:
\int f\;d\vec{r}=\bigtriangledown ^{-1}f
What make I think that:
\frac{\mathrm{d} }{\mathrm{d} \vec{r}}=\bigtriangledown
Then, would be possible to formalize the differintegration of vector with respect to vector too by:
\bigtriangledown \cdot \vec{f},\;\;\;\bigtriangledown \times \vec{f},\;\;\;\bigtriangledown^{-1} \cdot \vec{f}\;\;\;and\;\;\;\bigtriangledown^{-1} \times \vec{f}
What you think about all this?