Directional derivatives and slope

[@zbzb-ic1sr]

You know, I've watched this video more than three times already. When I finally understood however that the directional derivative is the slope of the line at that point through a mental image of the unit vector projecting onto the gradient, I got a hugely satisfying .ahhhhhhh.... God I love that feeling. Thanks Grant! Just a tip though, You might want to make this video better by showing the projection of the unit vector onto the gradient vector. This takes nothing away from you brilliant series and godlike videos on your channel however. Cheers!

[@zes7215]

wrg, no such thing as satisfx or lovx or understanx or not, doesn't matter, nonerx

[@jrremix5878]

@@zes7215 nice

[@ashayshah409]

Yo that made so much sense thanks

[@isaachester8475]

@@zes7215 ?

[@metuphys5611]

69

[@xoppa09]

I think its clearer to say that the directional derivative is the weighted sum of fx and fy, weighted by the unit vector v. e.g. v1 * fx + v2 * fy. This expression can be viewed as a type of total differential . v1 * ∂f/∂x is the change in height by walking parallel to the x axis a distance of v1, since v1 * ∂f/∂x = ∂f by algebra. Similarly v2* ∂f/∂x is the change in height walking parallel to the y axis a distance of v2. After these two walks you are at the tip of the vector of , and you have the total ∂f , or total change in height. This argument can also be recast using a unit circle, if you draw a unit circle and define v = < cos theta, sin theta>, then the directional derivative is cos Θ fx + sin Θ fy

[@yasseralg3928]

I really enjoy understanding the conceptual meaning as opposed to the "Plug & Chug" approach ,, :)

[@priyankkharat7407]

Thanks KA, What I understood from this video: Directional Derivative = [function's Induvidual dimension's derivative] ● [desired Direction of nudge in ip space]. Since the second vector is only meant to provide direction, we must normalize the result by magnitude of 2nd vector (v) making our directional derivative independant of magnitude of 2nd vector (v). Guys, do Correct me if I am wrong in understanding this.

[@agrajyadav2951]

Thanks Lord Grant! Helping me understand yet another obeject!

[@riteshbhartiya6155]

∇f ⋅ *v* gives the change in function value(moving along *v* ). | *v* | gives the change in function input(moving along *v* ). Thus, slope of the curve in plane containing *v* = ( ∇f ⋅ *v* )/| *v* | This makes sense to me.

[@spiderjerusalem4009]

AHHHHHH IF ONLY I'D FOUND THIS ERLIER. I WOULDN'T'VE THOUGHT SANDERSON TO DO THIS ON KHAN ACADEMY 😭😭😭😭😭😭😭😭

[@dogeteam2235]

who is sanderson lmao

[@spiderjerusalem4009]

grant sanderson 3Blue1Brown channel

[@catouncormery2995]

thank u from france, you are the best! loves your videos !

[@paulwbate]

Nice explanation, especially with your formal definition video - what graphing software are you using?

[@desudesudesu5326]

So is a non-normalised directional derivative a linear approximation to the change in the function in the direction of the vector? As in, the change along the tangent line to the point?

[@ΚωνσταντίνοςΛαζαρίδης-ξ9ι]

Thanks!

[@isaachester8475]

Vector calc is supposedly the hardest class, but so far 3B1B’s videos make it seem very intuitive and doable. I’m not in the class, but to anyone who has been, what makes it so difficult?

[@rchshi]

Currently enrolled in MV Calculus at my HS, I think what makes it difficult is that it's hard to conceptualize a lot of concepts (especially in 3D) when your teacher is not 3B1B or Sal Khan.

[@spiderjerusalem4009]

the visualization isn't clear/naked enough for me. Sure, it is all about the rate of change of a function with respect to distance, but most visualizations i've seen merely show the contour maps when pointing out to the steepest ascent. I'd rather have an elucidation for as if i were a teenager who just started his jhs. Anyone would? I'd appreciate it very much

[@mireazma]

Thanks for the well explained videos. At the end of the vid I still don't get if normalizing is arbitrary... why hasn't he normalized the vector in the previous video if the directional derivative result varies with normalization?

[@Raikaska]

It's not arbitrary, it comes up naturally when using the definition but, as we usually don't compute the df/dv using the definition and instead use faster methods like this, we need to make it unit length in order to use the theorem that guarantees that method. Hope I made myself clear..

[@effy1219]

I dont understand how you slice the graph with v[1, 1], there can be infinate number of plains through one vector, is the assumption here is that plain should be perpendicular to xy?

[@fofororo2070]

There is ONLY ONE plane that the inputs and their corresponding outputs fall on and that is the slicing plane. Remember the outputs are right above their corresponding inputs in this video.

[@carlitobrigante9740]

this video is brillant

[@Crawfly]

I am going through the multivariable calculus course on the Khan academy website and in some examples they normalize the vector when problems give the function, a point, and a vector while other questions provide the function, vector "a" and vector "v." Why is the vector normalized in some instances and not in others?

[@Postermaestro]

great video

[@numbercrunched]

thank you for making these videos :) I plan to keep studying tell Im at this math level :)

[@張海鷗-t2t]

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[@akashverma5756]

Directional derivatives :- It is rate of change of value of function in some certain direction.

[@BedrockBlocker]

Thats why we only defined the directional Derivative for unit vectors

[@danieljaszczyszczykoeczews2616]

thanks for a video

[@Festus2022]

What good is a directional derivative if you're NOT interested in the SLOPE of the function in a specific direction? That is, when would it be useful NOT to normalize the direction vector? My teacher taught us that the directional derivative required a "unit direction vector", but the Khan Academy in several videos implies that there is some useful reason to dot the gradient with a direction vector vector that is NOT normalized.

[@at-gv8sd]

This video made me giggle because of how easy it made it for me to understand 😂

[@sankaracharyadutta9408]

Really Helpful. Acknowledge the man behind this.... _/\_

[@sorinpanciuc5712]

really can't understand where the root(2)/2 is from in this video

[@asdfasdfkjllk856]

Pythagorean theorem: √[(√(2)/)^2+(√(2)/2)^2] =1, which is what Grant wanted to do: to make a unit vector in the direction of [1;1], or π/4 radians

[@sorinpanciuc5712]

Thanks ! I reviewed everything about vectors and now I understand :D

[@Tntpker]

Vector v = magnitude v * unit vector v --> unit vector v = v/magnitude --> [1 1] / √2 = [1/√2 , 1/√2] --> now multiply numerator and denominator by √2 --> [√2/√2*√2 , √2/√2*√2] = [√2/√4 , √2/√4] = [√2/2 , √2/2]

[@mohammadbinmahbub9160]

Why do you divide the directional derivative by the unit vector and not something infinitesimally small?

[@Mathieu-qx7bp]

Hello! The directional derivative, in its formal definition (the thing with the fraction and limit as h goes to 0), is indeed already divided by h, as h goes to 0 (not strictly speaking "infinitely small", but I think that's what you meant). It is already there, but written explicitly as already included inside of the "nabla" or Leibniz's notation. We simply additionally divide by the NORM of the vector v (aka magnitude, or simply put its "length"), written as ||v|| (NOT a unit vector). The reason as to why that is, is explained in this and the previous video in different ways, so I'd encourage you to rewatch them, but in short: Performing the computation of the limit with a vector whose norm ISN'T 1 will not return the derivative itself, but that derivative times the norm of v (properties of the derivative : (k*f)' = k*f')). So we then re-divide by the norm to counteract that effect and get ONLY the derivative. As for the notion of "unit vector", it simply means a vector whose norm is already 1. E.g. (sqrt = "square root of") [1, 1] has a norm ||v|| = sqrt(1^2 + 1^2) = sqrt2 => norm ||v|| not equal to 1, NOT a unit vector. We instead pick : [sqrt2, sqrt 2] which has the SAME direction, but ||v|| = 1 (you can verify that by performing the same calculation as above). Finally, as to how we can switch from one vector v, to one with the exact same direction but norm ||v||=1: we can simply calculate the initial norm ||v||, and then divide by that norm! The result can get pretty ugly (square roots and stuff), but it works on every vector! (except the 0-vector of course) Hoping that helped someone! Good day to you, dear reader!

[@shawon265]

I was wondering if it was necessary for the plane to go through origin... it's not necessary, right?

[@niccolodiana5612]

It shouldn't be a necessity. At least from what I can think a vector can have magnitude 1 and can still point somewhere else not in the same trajectory as the origin.

[@JohnnyDoeDoeDoe]

What graphing software is used here?

[@muqrizadnan]

I think he used built-in apps in mac. It is called grapher

[@JohnnyDoeDoeDoe]

Ah that's too bad, I'm on Windows atm, thanks though!

[@merttok4291]

As far as I know he coded it himself using Python

[@smolytchannel5062]

I think it's his very own creation called manim

[@owenhe5070]

how to understand matrix dot products of 2x1 times 2x1? When he computed the directional derivative of point f(-1, -1).

[@legacies9041]

That was just a dot product of two vectors. You multiply corresponding components and sum them up

[@RadomName3457]

Hi everyone, i kinda understood that the whole concept is about the slope of a point in certain vector direction. But i don't intuitive understand why dot product between vector of derivative and vector of direction gives the slope result. Can anyone clear my doubt, pls?

[@shawaizhaider3978]

He talks about it in his first video about directional derivatives.

[@coreynoodles6876]

is the plane arbitrarily selected? why?

[@danielcs6489]

whats and example of a normalised direction vector that does not run through the origin?

[@asdfasdfkjllk856]

Every vector! Vectors are independent of position unless specified by a function, which is not the case in this video.

[@cauchyschwarz3295]

I don't much like the del f / del v notation. You can't divide by a vector. The symbolism feels unclean and makes the concept appear more complicated than it is.

[@pat6595]

Why (((2)^1/2)/2)?

[@thatskychu]

Since a vector is composed of its magnitude and direction, we could consider it as vector V = magnitude of vector V x unit vector of vector V. By simple algebra, you can find that unit vector of vector V = vector V / magnitude of vector V. The magnitude of vector V is just the sqrt(1^2 + 1^2) given the vector V is (1 i-head + 1 j-head) in 2:41, giving sqrt(2). (1 i-head + 1 j-head) / (sqrt of 2) gets ( 1/sqrt(2) i-head + ( 1/sqrt(2) j-head ). Rationalizing it leaves you ( sqrt(2)/2 i-head + sqrt(2)/2 j-head ), then you get it. In addition, the magnitude of the any unit vector always equals to 1. Taking the vector V as an example, the magnitude of the vector V is sqrt( (sqrt(2)/2)^2 + (sqrt(2)/2)^2 ) = 1, which Mr. Grant has written in 2:21 just in case you wonder.

[@pat6595]

@@thatskychu Thank you!

[@WittyComm3nter]

@@thatskychu Not sure what you mean by "rationalizing it"

[@asdfasdfkjllk856]

In all the Directional Derivatives videos in the playlist, Grant doesn't explain why the directional derivative is equal to the dot product of the gradient and the vector. Is there a proof? There seems to be a serious lack of rigor here for such a huge statement. Its unlike Grant to take things for granted (no pun intended). How are the rates of change in the x and y plane related to rates of change in any other plane? It seems to me like the planar cross-section of the 3D graph in the vector v direction is completely unrelated to the planar cross-section of the 3D graph in the x and y plane. For just a random example to explain what I am asking, in an arbitrary 3D interval, lets say the curve in the x direction follows the curve a^2, the curve in the y direction follows the curve 1/a, and the curve in the vector v direction follows sin(a), how does the partial derivatives for x and y determine how the curve's rates will behave in the vector v direction?

[@matiassandacz9145]

If u still need a proof just respond to this message and i'l show you!

[@merttok4291]

"... why the directional derivative is equal to the dot product of the gradient and the vector. ..." The equation of the plane passing through (x0,y0) in the direction of v = u1 + u2 is: x = x0 + s . u1 y = x0 + s . u2 Basically what we are trying to find is how does change in v affect F's value.The only thing that can change v is the variable s so df / dv can be written as df / ds Using chain rule df / ds = (del f / del x) . (dx / ds) + (del f / del y) . (dy / ds) We know the values of dx/ds and dy/ds so the equation becomes, Directional derivative of f = (delf / del x) . u1 + (delf / del x) . u2 = (delf / del x + delf / del y) . ( u1 + u2) u1 + u2 is the directional vector we had at the start! Direc. der. = (delf / del x + delf / del y) . (directional vector) after mathematicians derived this formula the expression (delf / del x + delf / del y) was very new to them so they named it "The Gradient" and that's how the gradient was born. At least that's how I know :) Instead of showing a method like this Grant chose to introduce the gradient the other way around, which is totally understandable but then it might leave something unexplained.I hope I managed to describe it well. Have a good day and take care of yourself :)

[@MrMicheal8181]

not sure if this is what ur looking for but kzbin.info/www/bejne/hJC9g5aCncqBrJI

[@whitewalker608]

You are correct he hasn't shown the proof. However it's pretty simple. Convert the equation into parametric form with t and apply partial derivative chain rule. Here: math.stackexchange.com/a/1450160/407209 Also he hasn't shown why gradient is perpendicular to the level surface/contours. See this MIT OCW's simple derivation here. Again chain rule. www.google.co.in/url?sa=t&source=web&rct=j&url=ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-b-chain-rule-gradient-and-directional-derivatives/session-36-proof/MIT18_02SC_notes_19.pdf&ved=2ahUKEwip8Ja91q7dAhWM6Y8KHUCZDpcQFjAUegQIAxAB&usg=AOvVaw3VUAcETuGofmq-ijtu9bSv

[@TheReligiousAtheists]

I've watched a lot of Grant's videos and one thing I noticed is that he's a great teacher (obviously), but if you want to learn from him, you have to let go the rigour and just take his word for it, and if you're REALLY itching for the rigour, you either need to do it yourself or look for some other teacher. Not related to your question, but just saying.

[@curtpiazza1688]

😊

[@karancontractor586]

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[@almasaabdulla8259]

I love you

[@victormuchina4865]

3blue1Brown

[@benniebijwaard3092]

CAN SOMEONE PLEASE HELP ME?! Great vid! 1 question though: What if the slope in the x direction is 0, and the slope in the y direction is 0, But the slope in a direction between the x and y direction is not 0? Then the whole concept fails. I imagining a rectengular tablecloth naild to the table in 3 corners, and is held up in the 4th corner. (One corner is the point, the sides are the axes) Can anyone help me?

[@NNOTM]

I've thought about this for a few minutes, and I think the answer is that the sort of function you're thinking about wouldn't be differentiable in that point, because there would be a sharp crease, stretching from the 4th corner to the nail on the opposite corner. The derivative in that direction would still exist, but you can't compute it with the formula presented in the video, because that only applies to points in which the function is differentiable.

[@Rubikorigami]

Bennie Bijwaard Look at the function f(x,y)=sqrt(x*y), at the origin : it’s constant in the x and y directions but not in the direction of the vector (1,1) for example.

[@Rubikorigami]

that said, I think that technically it’s not differentiable there.

[@asdfasdfkjllk856]

+Nnotm Your answer doesn't seem satisfactory to me. No one said there is going to be a sharp crease, it could be a crease of any smooth shape imaginable. The lifted corner need not be taut in the way you imagine it. This ultimately leads to my question that I asked in a separate comment: why is the directional derivative equal to the dot product of the gradient and the vector v? How do the 2D functions in the x and y crossection dictate the nature of the 2D function in the vector v direction?

[@paulluk1671]

because functions are used to plot the graph, each of the domain can only link to one co-domain, so the corner will not appear on the graph. as each set of (x,y) corresponds to one set of f(x,y) only.

[@prabalsrivastava4945]

From where does he get such weird graphs? 😂😂😂

[@nowavelength982]

Imagine getting reply after 3 year

[@austinjohnson4551]

Imagine getting a reply after 3 day

[@raphaelbob8093]

first