Quote;
"Now suppose we have a scalar function $f(\mathbf{r})$ and we want to consider the rate of change along a path $\mathbf{r}(u)$. A change $\delta u$ produces a change $\delta \mathbf{r} = \mathbf{r}' \delta u + o(\delta u)$, and $$ \delta f = \nabla f\cdot \delta \mathbf{r} + o(|\delta \mathbf{r}|) = \nabla f\cdot \mathbf{r}'(u)\delta u + o(\delta u). $$ This shows that $f$ is differentiable as a function of $u$ "
1) Looking at
$\delta \mathbf{r} = \mathbf{r}' \delta u + o(\delta u)$
How can we have vector = vector + scalar?
2) When we substitute in $\delta \mathbf{r}$, how do we get to the final result (i.e can someone expand the last equation to show workings in full and reasoning for each step)
1) Looking at
$\delta \mathbf{r} = \mathbf{r}' \delta u + o(\delta u)$
How can we have vector = vector + scalar?
A1) Read this not as vector + scalar, rather that the $o(\delta u)$ is a small vector. So rather read: vector + vector.
2) When we substitute in $\delta \mathbf{r}$, how do we get to the final result (i.e can someone expand the last equation to show workings in full and reasoning for each step)
A2) What have you tried? Where are you at in your investigation?