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Consider a function $f:[a,b]^2\rightarrow[a,b]$.
What does the expression $\nabla f\cdot (t,s)$ mean? I read that it indicates "the rate of change of $f$ in the direction of $(s,t)$". I would appreciate if someone could elaborate this.
Moreover, suppose that $\nabla f\cdot (t,s) \le 0$. What is the intuitive meaning of this inequality?

Dot product of the gradient with a vector is not that clear to me, so any clarifications would be great.

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For a unit vector $(t,s)$, if $\nabla f \cdot (t,s)=10$ at a point $(a,b)$ this means that

Starting at $(a,b)$, if I walk a small distance $d$ in the direction $(t,s)$, I expect my function to increase roughly by $10d$.

You can have a similar decreasing interpretation if the quantity $\nabla f \cdot (t,s)$ was negative.

The quantity $\nabla f \cdot (t,s)$ is called the directional derivative of $f$ at $(a,b)$ in the direction of $(t,s)$.

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Don't focus on the dot product. Take a curve $c$ such that it's tangent vector at $x=0$ is equal to $(t,s)$, so $c:(-a,a)\rightarrow \mathbb{R}^2$ with $c(0)=p$ and $c^\prime(0) = (s,t)$. Then consider $$h:=f\circ c: (-a, a) \rightarrow \mathbb{R}$$

This is the function $f$ along the curve $c$. Now if you take the derivative $h^\prime(0)$ you get, by the chain rule, $$ h^\prime(0) = \langle \nabla f(c(0)), c^\prime(0)\rangle = \langle \nabla f(c(0)), (s, t)^T\rangle$$ which is the expression you want to understand (written a bit differently, but it's the same). So the expression tells you the slope of $f$ along $c$. If this has a sign it just means that $f$ is increasing (decreasing) along $c$, depending on the sign.

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An example: consider the paraboloid $f(x,y)=x^2+y^2$ with directional derivative,

$$ \nabla f \cdot (t,s) = 2xt + 2ys $$

The cross sections of the plot are parabolas themselves. So for example, when $(t,s) = (1,0)$. This is a parabola flat on the $x,y$ plane. So a normal 2-Dimensional parabola. Which we know $\frac{d}{dx}[x^2]=2x$ which is shown by the directional derivative,

$$ \nabla f \cdot (1,0) = 2x $$

Also notice this is only true for normalized $(t,s)$ (i.e $\frac{(t,s)}{||(t,s)||}$).

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