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If I have a function $f(x,y)$ and take the gradient at $(a,b)$ then

$\vec{v} = \nabla f(a,b) = (\frac{\partial f(a,b)}{\partial x},\frac{\partial f(a,b)}{\partial y} )$

Clarifying Question 1: Because $\vec{v}$ points in the direction of "greatest ascent," then so does $\frac{\vec{v}}{\|\vec{v}\|}$, correct?

Clarifying Question 2: $\|\vec{v}\|$ gives the rate of change of $f$ in the direction of $\vec{v}$, so if $\|\vec{v}\| = 5$ for example, does that mean a one unit change in the direction of $\vec{v}$ would yield a 5 unit change in $f$?

Side question here: Would the gradient be the zero vector if moving in any direction would bring a decrease?

Clarifying Question 3: (Please let me know if my understanding of the concept is correct) if $\vec{u}$ is of the same dimension of $\vec{v}$, the directional derivative $D_{\vec{u}}f$ gives the rate of change of $f$ in the direction of $\vec{u}$, so if $\vec{u} = (3,2), (a,b) = (2,3)$ then the rate of change in the direction of $\vec{u}$ is the same as the rate of change in moving from the point $(2,3)$ towards the point $(5,5)$, also written as $$D_{\vec{u}}f(a,b) = \vec{v} \cdot \frac{\vec{u}}{\|\vec{u}\|}$$

and therefore $\|\vec{u}\|D_{\vec{u}}f(a,b)$ would give the change of $f$ when moving from $(2,3)$ to $(5,5)$

Clarifying Question 4 (last one): if Question 3 is correct, and $\vec{r}$ points in the same direction as $\vec{v}$ then is the following true?

$$\vec{v}\cdot\vec{r} = \|\vec{v}\|\cdot\|\vec{r}\|$$ which gives the change in $f$ when moving from $(a,b)$ to $(a + r_1, b + r_2)$

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    $\begingroup$ Do you mean $f(x, y)$ in the top line? $\endgroup$ Sep 23, 2021 at 2:58
  • $\begingroup$ yes, thank you... $\endgroup$
    – user848541
    Sep 23, 2021 at 3:00
  • $\begingroup$ For your side question, do you know what the gradient is at a local max or min (assuming differentiability there)? $\endgroup$
    – Joe
    Sep 23, 2021 at 3:07
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    $\begingroup$ “Does that mean a one unit change in the direction of $\vec v$ would yield a five unit change in $f$” Only approximately. It would be a better approximation to say that a $.01$ unit change in the direction of $\vec v$ would yield a $.05$ unit change in $f$. And it would be an even better approximation to say… $\endgroup$
    – littleO
    Sep 23, 2021 at 3:20
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    $\begingroup$ Pedagogically, I agree that saying "It would be a better approximation to say that a .01 unit change in the direction of $\vec{v}$ would yield a .05 unit change in $f$", is a good way to explain it at an introductory level, but technically this is not necessarily true. $\endgroup$
    – Joe
    Sep 23, 2021 at 11:31

1 Answer 1

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  1. Yes.

2, 3, 4. The linear approximation to the change, not the actual change. The linear approximation is only an accurate representation of the change in the limit as the distance one moves goes to $0$. This is just like the linear approximation you learned (perhaps using the word "differential") in your Calculus I class.

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  • $\begingroup$ @Joe Of course. $\endgroup$ Sep 23, 2021 at 3:12
  • $\begingroup$ I see - so if f was for example the equation for a plane, then this would hold entirely true? $\endgroup$
    – user848541
    Sep 23, 2021 at 3:17
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    $\begingroup$ Yes, @ilikemath3.14. To be picky, $f$ is the function. $z=f(x,y)$ is its graph (e.g., the plane). $\endgroup$ Sep 23, 2021 at 3:19
  • $\begingroup$ Thanks - can you explain the difference here? Why do we need z? $\endgroup$
    – user848541
    Sep 23, 2021 at 3:20
  • $\begingroup$ Go back to single-variable calculus. It’s always the graph $y=f(x)$ that you drew to see the tangent line, etc. For functions of two variables, it’s the tangent plane to the graph $z=f(x,y)$ at a point that gives the best linear approximation. $\endgroup$ Sep 23, 2021 at 3:23

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