- #1
andresordonez
- 68
- 0
SOLVED
Show that:
[tex]
\sum_i \vec{\nabla_i}V \cdot \frac{\partial \vec{r_i}}{\partial q_j} = \frac{\partial V}{\partial q_j}
[/tex]
[tex]V=V(\vec{r_1},\vec{r_2},\vec{r_3},...\vec{r_N},t)[/tex]
[tex]\vec{r_i}=\vec{r_i}(q_1,q_2,q_3,...,q_n); i=1,2,3,...,N; n<N[/tex]
I'm not sure how to apply the chain rule when the function depends on vectors. I guess is something like this:
[tex]
\frac{\partial V}{\partial q_j}=\sum_i \frac{\partial V}{\partial \vec{r_i}} \frac{\partial \vec{r_i}}{\partial q_j}
[/tex]
However, the partial derivative of a scalar function with respect to a vector is something I'm not used to. Thanks in advance
Homework Statement
Show that:
[tex]
\sum_i \vec{\nabla_i}V \cdot \frac{\partial \vec{r_i}}{\partial q_j} = \frac{\partial V}{\partial q_j}
[/tex]
Homework Equations
[tex]V=V(\vec{r_1},\vec{r_2},\vec{r_3},...\vec{r_N},t)[/tex]
[tex]\vec{r_i}=\vec{r_i}(q_1,q_2,q_3,...,q_n); i=1,2,3,...,N; n<N[/tex]
The Attempt at a Solution
I'm not sure how to apply the chain rule when the function depends on vectors. I guess is something like this:
[tex]
\frac{\partial V}{\partial q_j}=\sum_i \frac{\partial V}{\partial \vec{r_i}} \frac{\partial \vec{r_i}}{\partial q_j}
[/tex]
However, the partial derivative of a scalar function with respect to a vector is something I'm not used to. Thanks in advance
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