27 | December | 2023 | Physics Town

p. 21

…

$\displaystyle{(Df(x)) b(x) \ne (Df(x)) \Delta x}$

Instead, $\displaystyle{(Df(x)) b(x)}$ is a generalization of $\displaystyle{(Df(x)) \Delta x}$ .

However, how to calculate $\displaystyle{(Df(x)) b(x)}$ ?

By this:

$\displaystyle{f(x + \Delta x) \approx f(x) + (Df(x)) \Delta x}$

Then:

$\displaystyle{(Df(x)) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}}$

So:

$\displaystyle{(Df(x)) b(x) = \left( \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \right) b(x)}$

$b$ is written as subscript to capture the meaning of “with respect to $b$ “. The original directional derivative uses the same convention:

So the spatial rate of change of $\displaystyle{f}$ along the direction of the vector $\displaystyle{\mathbf{v}}$ is

$\displaystyle{\begin{aligned} D_{\mathbf{v}}(f) &= \frac{\left(\delta f\right)_{\mathbf{v}}}{|\mathbf{v}|} \\ &= \frac{1}{|\mathbf{v}|} \left( \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial x} \delta y \right) \\ &= \frac{\partial f}{\partial x} \frac{\delta x}{\sqrt{(\delta x)^2 + (\delta y)^2}} + \frac{\partial f}{\partial x} \frac{\delta y}{\sqrt{(\delta x)^2 + (\delta y)^2}} \\ &= \left(\nabla f\right) \cdot \frac{\mathbf{v}}{|\mathbf{v}|} \\ &= \left(\nabla f\right) \cdot \hat{\mathbf{v}} \\ \end{aligned}}$

$\displaystyle{D_{\mathbf{v}}(f)}$ is called directional derivative.

— Me@2016-02-06 09:49:22 PM

— Me@2023-12-27 01:37:32 PM