- #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

Last edited: