Assume $\displaystyle{(x, y)}$ represents the position of an object and $\displaystyle{f(x,y)}$ is a scalar field on the $\displaystyle{x}$$\displaystyle{y}$ plane. Then $\displaystyle{\frac{\partial f}{\partial x}}$ represents the change of $\displaystyle{f}$ per unit length along the positive $\displaystyle{x}$ direction. In other words, it is the spatial rate of change of $\displaystyle{f}$ along the $\displaystyle{x}$ direction.

Similarly, derivative $\displaystyle{\frac{\partial f}{\partial y}}$ represents the spatial rate of change of $\displaystyle{f}$ along the $\displaystyle{y}$ direction.

For an arbitrary direction, due to the nature of displacement, the change of $\displaystyle{f}$ is $\displaystyle{\delta f = \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial x} \delta y}$ when the object has finished moving $\displaystyle{\delta x}$ in $\displaystyle{x}$ direction and then $\displaystyle{\delta y}$ in $\displaystyle{y}$ direction.

Then, the spatial rate of change of $\displaystyle{f}$ is

\displaystyle{ \begin{aligned} &\frac{\delta f}{\sqrt{(\delta x)^2 + (\delta y)^2}} \\ &= \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}} \\ \end{aligned} }

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For simplicity, denote the resultant displacement as $\displaystyle{\mathbf{v}}$:

$\displaystyle{\mathbf{v} = (\delta x, \delta y)}$

and define $\displaystyle{\nabla f(x)}$ as

$\displaystyle{\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)}$

Then, the change of the $\displaystyle{f}$ due to the displacement $\displaystyle{\mathbf{v}}$ is

\displaystyle{\begin{aligned} \left(\delta f\right)_{\mathbf{v}} &= \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial x} \delta y \\ &= \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial x}\right) \cdot (\delta x, \delta y) \\ &= \left(\nabla f\right) \cdot \mathbf{v} \\ \end{aligned}}

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So the spatial rate of change $\displaystyle{f}$ 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{\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

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This is the reason that $\displaystyle{\nabla f}$ is in the steepest direction.

If $\displaystyle{\hat{\mathbf{v}}}$ is chosen to be parallel to $\displaystyle{\nabla f}$, the directional derivative $\displaystyle{\left(\nabla f\right) \cdot \hat{\mathbf{v}}}$ would be maximized.

— Me@2021-08-20 05:20:02 PM

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# 注定外傳 2.4

Can it be Otherwise? 2.4 | The Beginning of Time, 7

（問：不會沒完沒了呀。只會追溯到「時間的起點」。）

（問：可能可以。所謂「時間的起點」，其實就即是「宇宙的開端」。）

（問：而物理學家知道，「字宙的開端」是「宇宙大爆炸」。所以我們知道，「時間的起點」，就是「宇宙大爆炸」。）

1. 「宇宙大爆炸」是一件事件，有一個過程，並不是時間上的「一點」，所以不算是「起點」。「宇宙大爆炸這件事的開始那刻」才算是起點。

2. 物理學家根據愛因斯坦的「廣義相對論」推斷，「宇宙開端」那一刻，開始發生的第一件事，是「宇宙大爆炸」。所以，如果「廣義相對論」不正確，「宇宙大爆炸」就未必為真。

3. 即使「廣義相對論」是可信的，普朗克時期（Planck epoch），即是開端後的頭$$10^{−43}$$秒之內，以現時的物理知識，是處理不到的。所以，物理學家推斷不到，那段時間內，發生了什麼事。

— Me@2016-02-15 07:04:56 PM