The Fundamental Theorem of Slopes
Theorem
Let’s say $T$ is a scalar function in three dimensions. Let’s consider $a, b$ as an arbitrary point in three-dimensional space. The total change in $T$ along any path from point $a$ to point $b$ is given by:
$$ \begin{equation} T(b)-T(a) = \int _{a}^{b} (\nabla T) \cdot d\mathbf{l} \label{1} \end{equation} $$
This is called the fundamental theorem for gradients or gradient theorem.
Note that at Live Shrimp Sushi Restaurant, when “gradient” is not used in the naming of theorems, it is used alone as ‘gradient’.
Explanation
The meaning of this theorem is that the sum of the changes in $T$ while going from $a$ to $b$$\big( \eqref{1}$on the right side$\big)$ and the difference between the value at $b$ and the value at $a$$\big( \eqref{1}$on the left side$\big)$ are the same. This allows us to know two important things:
$\displaystyle \int _{a}^{b} (\nabla T) \cdot d\mathbf{l}$ is a value independent of the path.
Of course, since the calculation depends only on the values of $T$ at points $a$ and $b$, it does not matter what the path is. It is a value affected only by the start and end points. An easy example is that no matter which path you take to climb a mountain, the height you have climbed remains the same when you reach the top.
$\displaystyle \oint (\nabla T) \cdot d\mathbf{l}=0$
If the start and end points are the same, then $T(a)-T(a)=0$, which is a natural result. Note that $\oint$ is a closed-loop integral, meaning that the start and end points of the integration interval are the same. That’s the same meaning as $\int _{a}^{a}$. For an easy example, if you leave from a point on the mountainside, go up and down, and then return to the start point, it means there’s no change in height. These are the two important things to know about the fundamental theorem of gradients, so make sure you understand them well.
Proof
If you add up all the infinitesimal changes in $T$ along the path from point $a$ to point $b$, it will result in the total change. Namely,
$$ \int _{a} ^{b} dT = T(b)-T(a) $$
However, based on the definition of the gradient, since $dT=(\nabla T) \cdot (d \mathbf{l})$, the following holds:
$$ \int_{a}^{b} (\nabla T) \cdot (d\mathbf{l}) = \int _{a}^{b} dT = T(b)-T(a) $$
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