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Okay so here's what I understand:

If we have a surface, then the directional derivate in the direction of a unit vector $\vec{u}$ at a point $P_{0}$ is the slope of the curve on the surface going through $P_{0}$ in the direction of $\vec{u}$

And $D_{\vec{u}} = f_{x}(x,y)a + f_{y}(x,y)b$ where $\vec{u} = \langle a, b \rangle$

and $\nabla f(x,y) = \langle f_{x}(x,y), f_{y}(x,y) \rangle $

So $D_{\vec{u}} = \nabla f(x,y) \cdot u$

But then $\nabla f(x,y)$ is in the direction of maximum ascent? how does that come to be? what is happening here?

proof-of-correctness
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    "$\nabla f(x, y)$ is in the direction of maximum ascent" is saying that for unit vectors $u$, $\nabla f(x, y) \cdot u$ is maximized for $u = \frac{\nabla f(x, y)}{|\nabla f(x, y)|}$. This is the Cauchy-Schwarz inequality. – Mason Sep 24 '22 at 19:13
  • I'm still a little confused, okay first things first. The directional derivative is in the direction of the tangent right? – proof-of-correctness Sep 24 '22 at 19:15
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    The directional derivative is $D_uf(x, y) = \frac{d}{dt}f((x, y) + tu)|_{t = 0}$. The utility of directional derivative is that it is a purely one-dimensional thing, so it allows you to express multidimensional calculus in terms of 1-d calculus. – Mason Sep 24 '22 at 20:04

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The gradient is the direction of maximum ascent, as you pointed out, in the underlying domain of the function. The directional derivative, on the other hand, is the rate of change in the function, in the direction of the given vector (again, in the underlying domain).

For example, imagine you have a function representing the height of a hill over some coordinate grid (e.g., easting and northing):

$$ f(x, y) = x+y $$

This hill slopes up to the northeast. Its gradient is given by

$$ \nabla f(x, y) = \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] $$

confirming that the hill slopes up to the northeast. On the other hand, if you wanted to know the extent to which the hill slopes up or down in an easterly direction, you want the directional derivative

$$ \left[ \begin{array}{c} 1 \\ 0 \end{array} \right] \cdot \nabla f(x, y) = \left[ \begin{array}{cc} 1 & 0 \end{array} \right] \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] = 1 $$

telling you that the slope is $1$ if you head due east. But you could also compute the directional derivative in other directions; for instance, you could compute it to the southeast (where the directional derivative would be $0$). Note that all these directions—east, northeast, southeast—are directions in the underlying coordinate grid, not in three dimensional space.

Does that help?

Brian Tung
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  • One thing I don't understand is, how is $D{\vec{u}} = \nabla f(x,y) \cdot u$. $\$ Why does multiplying the vector of the maximum slope with u give us the directional derivative? – proof-of-correctness Sep 24 '22 at 19:19
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    Well, it happens to give us the direction of maximum ascent, but that is not its only purpose, and interpreting it that way makes the directional derivative less intuitive (as you've noticed). The gradient is also a vector that tell us the slope of the function in the two basis directions (east and north, in this example). The directional derivative is a linear combination of those slopes; since the hill slopes up $1$ to both the east and north, the directional derivative tells us that when we walk in a given direction, the hill slopes up to the extent that we're walking north and east. – Brian Tung Sep 24 '22 at 19:23
  • oh okay, i think i sort of understand.

    del f is the maximum ascent, and when we go in the direction of u, we get the dot product of del f and u, which gives projection of del f in the direction of u

    and when theta is zero (u is in the direction of del f) then its maximum?

     – proof-of-correctness Sep 24 '22 at 19:27
  • I'm not sure what $\theta$ is here, but if $u$ is in the direction of $\nabla f$, then yes, the directional derivative is at its maximum (for a given magnitude of $u$). – Brian Tung Sep 24 '22 at 19:34