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Lec 12: Gradient; directional derivative; tangent plane | MIT 18.02 Multivariable Calculus, Fall 07
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- เผยแพร่เมื่อ 15 ม.ค. 2009
- Lecture 12: Gradient; directional derivative; tangent plane.
View the complete course at: ocw.mit.edu/18-...
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
Lecture 1: Dot Product
Lecture 2: Determinants
Lecture 3: Matrices
Lecture 4: Square Systems
Lecture 5: Parametric Equations
Lecture 6: Kepler's Second Law
Lecture 7: Exam Review (goes over practice exam 1a at 24 min 40 seconds)
Lecture 8: Partial Derivatives
Lecture 9: Max-Min and Least Squares
Lecture 10: Second Derivative Test
Lecture 11: Chain Rule
Lecture 12: Gradient
Lecture 13: Lagrange Multipliers
Lecture 14: Non-Independent Variables
Lecture 15: Partial Differential Equations
Lecture 16: Double Integrals
Lecture 17: Polar Coordinates
Lecture 18: Change of Variables
Lecture 19: Vector Fields
Lecture 20: Path Independence
Lecture 21: Gradient Fields
Lecture 22: Green's Theorem
Lecture 23: Flux
Lecture 24: Simply Connected Regions
Lecture 25: Triple Integrals
Lecture 26: Spherical Coordinates
Lecture 27: Vector Fields in 3D
Lecture 28: Divergence Theorem
Lecture 29: Divergence Theorem (cont.)
Lecture 30: Line Integrals
Lecture 31: Stokes' Theorem
Lecture 32: Stokes' Theorem (cont.)
Lecture 33: Maxwell's Equations
Lecture 34: Final Review
Lecture 35: Final Review (cont.)
Thank you bro
Thanku very much dear🙂
Yayy
¹!
blackboard duster action highlights - 28:28 and 43:30
This guy is the man!
43:20
"I see you guys are having fun"
I guess he was thinking something like "we'll see who has fun at the exam"
Actually 😂😂
@@vishalrajlodhi5822 hi bro 👋
Lol no
For people who are confused, try studying basics of 3-D geometry. You will easily understand the equation of tangent plane. It is basically asking us to find the equation of a plane whose direction ratios of its normal vector and the a point through which it passes is given.
"Ok, so let's talk about geometry, that should calm you down"
Lmao
classic
To all those who didn't exactly got why dr/ds=(u hat), try understanding this way of explanation:
In the X-Y plane lets just take R(r vector) to be the position vector of a point in X-Y plane. Now if we move a length ds along u vector(U), the position vector of the point will change by amount dR. Thus, if you draw a picture of what I just said, you will understand this that, (R +dR) - (R) = (ds)(U hat). That is dR=(ds)(u hat). And thus dR/ds=(u hat).
thank you
brilliant
Also I just realized that it is essentially the definition of unit vector after taking the limit since ds is the length of dr. And you divide the vector dr by its length you get the unit vector.
You can also prove dR/ds = (u hat) as follows. Observe that R = (x,y). Express x and y as functions of s, so x = x(s) = x0+s*(cos t) and y = y(s) = y0+s*(sin t). Notice that t is fixed but s varies. (s is the absolute value of delta R). Then dx/ds =cos t and dy/ds = sin t. Cos t and sin t are members of the unit vector so dR/ds = u hat.
This video is what is keeping me going in life, god bless calculus
god bless u man,nice clear lectures
We did this in class one week ago. My sucks ass. This is soooo much better and clearer. This guy should be given "best teacher in the world" award. He should win a noble prize in teacher. All other teachers should bow to him and call him "teacher god!"
someone needs to make a compilation of the students cheering for the professors erasing the blackboard!!
I find it annoying
Very well explained, and this chapter is quite important for you to understand gradient descent in machine learning. Thanks, professor
I am an engineering student at a two-year college that shall go unnamed. I can follow this man's MIT lecture far better than any other math professor I've had at junior college....Go figure.
Those who can't teach at universities, teach at CCs
what are you doing in life now ?
I love it when math gets graphic.
The alternative method that he used to find the tangent plane at at 29:45 shows his amazing geometrical intuition! I would not have thought of solving it so intuitively using linear approximation of the curve at a point.
im really glad MIT put up these series. im about to go into college and the summer breka and all gave me a chance to learn more classes, like single variable calculus which was fairly easy. but now that im learning this, I found that I had to really change my perspectve on certain fields of mathematics, particularily linear algebra. the whole beginning of the series with vectors took me a while to understand because id never thought of systems of equations in that way.
oh, these classes are so awesome, I wanted to clap at the end of the video.
me too lmao
let W = f(x,y) .i did not understand how dW/ds will give us the directional vector in the direction of u . From what I understood r(position vector) has been parametrised with arc length (s) so that the component of r can be given as relation of some c(initial point) + s*. Now I understand that the position vector obtained will be in the direction of u as when s = 1 we move in the direction u from initial point. Also from above both parametrised x(s) and y(s) have been obtained. But how this relates to dW/ds being the directional derivative in the direction of u (analogous to cutting a slice in the graph of W parallel to u at the initial point and getting the slope of the curve) is something I am not getting. dW/ds should be the rate of change of the function w.r.t arclength (s) not the directional derivative.
this is the most excellent math leasson i have learned, impressive and fantastic!
刘阳你说的太对了
、
I stumbled upon a concept that I am not able to understand regarding directional derivatives .The question is about how the prof had proved finding directional derivative would be dW/ds where w = f(x,y) and r(s) has been parametrised in terms of arc length such that dr/ds = unit vector u . I am not able to understand how dW/ds is giving us the directional derivative as it is w.r.t arc length rather than a particular direction u
What a great lecture. All hail Denis Auroux and MIT.
@pattttttrick You see the gradient is a sort of operator, when you apply it to a scalar field, it gives a vector field. The gradient it self is not the vector.
The way he parameterized the directional derivative was really confusing.
El profesor de universidad del que más he aprendido :')
1:20 change in z is cange in x and y
I love blackboards! I am glad they still use them.
Thank You.... Excellent lecture!
me: "So....What? Can you explain it again?"
him: "Which part?"
me: "All of it"
him: "MIT OCW...Okay, so next..."
I know its 8 years ago.. .But couldnt stop myself from laughing... Seems too natural!
@@gautamgopal3517 :)
I Really Like The Video Gradient; directional derivative; tangent plane From Your
Thanks dear Sir, love you. You are like God for us
It is my pleasure to learn from this lecture thanks a lot sir
This is one of the more difficult lectures for me so far. The fact that I didn't fully grasp parametric equations is coming back to bite me up the ass....
Way better than my online class (and way better then the textbook that is needed for the class)
Whew. Not easy. So the gradient vector is perpendicular to the level curves, so can be drawn on the contour plot of the function, NOT the graph of the function itself (?)...took me a while to realise this, so my thinking was at cross purposes with what he was saying. Also this business of "velocity vectors" and the vector r I found a bit confusing...think I've got the concept of a directional derivative though. Need to do some more reading on this...
I think the gradient vector would theoretically be able to be drawn on the graph of the function itself. Its just that with the case with three variables the graph could only be represented in four dimensions which is of course impossible to visualize. I say it would be possible to drawn the grad. vector on the curve because when you think of a three dimensional (2 variable) curve you can easily imagine the contours on the graph itself and the gradient vector is perpendicular to those contours and so is perpendicular to the tangent plane at a point on that contour in the 3 dimensional graph. In the case here with the 3 dimensional "contour' plot for a function with 3 variables imagine one of those "contours", the level surface. On this level surface defined by w=c (e.g. x^2+y^3-z=3) there are infinite 2-d curves and thus at any one point on the surface there are infinite tangents to these 2-d level curves, the fact that there are infinite tangents means that the velocity vector (dr/dt) lies in the tangent plane to the level curve. Since we set w equal to a constant to get a level curve, dw/dt is always 0 and hence [gradient vector DOT PRODUCT dr/dt] is always 0. So this means that the gradient vector is always perpendicular to the velocity vector, which lies in a tangent plane to the level surface and so the gradient vector is always perpendicular to the tangent plane to the level surface. If we now move to a lower dimension, the gradient vector is perpendicular to the tangent line to the contour, which is what you saw with the demonstration with that computer program.
Fantastic lecture series!
such a great lecuture!
16:16 If delw.V is zero because w is constant on a curve on the same level surface .. then delw . any other vector would be zero be it tangent or not .. because del w is always gonna be zero on the level surface.. isn't it ?
Watch it again
yes i am confused about that too
For all those who didn't understand, dw is 0 when we are moving along the tangent plane and not del w. Therefore the product del w • v is 0 which implies that del w is perpendicular to v.
No, because delW.dr/dt is a decomposition of dW/dt
Simply brilliant!!!
Thanks a lot to the person incharge of the camera...
If they would just zoom out a bit it would make it so much easier to follow...
This is awesome :D
This is helpful 🤍❤️
Thanks 🤍❤️
Im a bit confused when he talked that the gradient vector is perpendicular to the circle x^2 + y^2 = c ? did he mean that it's perpendicular to the tangent of the circle. And is that vector's direction parallel to the xy plane ?
1) Yes, he means that the gradient vector is perpendicular to the tangent to the circle at that point. If it was a different function and the contour lines / level lines were a different shape instead of a circle, then the gradient vector would be perpendicular to the tangent to that shape at that point.
2) The gradient vector has the same number of dimensions as the number of variables in your function. If you had f(x,y) and you plotted the result of the function in the z-axis, then the gradient vector would have 2 components (you can compute it) which means that it is 2 dimensional and parallel to the x-y plane. I.e. the gradient vector only points in the x-y direction of the fastest ascend; the arrow does not point up alongside the curve.
Woah!!! Thank you guys!!! I was kinda confused about the same.... Got my answer without even having to wait!!!
Here he used an example of w(x, y, z) = c, which is a constant. What if it is an actual equation that involves using the inputs of x, y, z which makes velocity nonzero?
These teachings resemble those I have been seeing in Africa the difference is who studies
If the gradient is a vector, why isn't there any arrow on top of it?
you can put an arrow over the del operator sign if you want. In most books it is written without an arrow but that doesn't change the definition i.e. it is a vector.
why we always consider a level surface to prove that gradient is perpendicular to tangent vector to the surface... what if the particle moves on a surface which doesn't has a constant value throughout its domain of definition
If gradient is always orthogonal to any tangent plane on the surface then how come the chain rule (grad f dot dr) doesn’t always equal zero? In other words for f(x,y) why doesn’t fxdx + fydy always equal zero.
While deriving the directional derivative, how did we approximate dr/ds=1 ( He said we're moving with unit speed. Firstly, why? And even if we are, how does that make dr/ds=1? )
making the vector unit length saves us from finding it's magnitude (always 1) as seen 44:14
Brilliant
Very nice.
Brilliant!
Where did he use in the proof of the calculation formula for directional derivative that u must be unit vector? Thank you
@joeglimmix Only if the professor can clean off the second chalk board while the 3rd is coming down.
Anyone knows how to access the tool that he is using to visualize all the graphs??
Can we say Wx + Xt ? The x and t are subscripts. Or... does that imply a partial derivative only?
What textbook do they use for MIT calculus?
The textbook listed for the course is: Edwards, Henry C., and David E. Penney. Multivariable Calculus. 6th ed. Lebanon, IN: Prentice Hall, 2002. ISBN: 9780130339676. See the course on MIT OpenCourseWare for more info and materials at: ocw.mit.edu/18-02F07. Best wishes on your studies!
Sleman, Yogyakarta, 21 April 2024
at 41:23 what kind of chain rule is that ? Isn't it too arbitrary?
If you haven't seen the lecture before this, this would not make sense. The way I think it works is that dw/ds = (∂w/∂x * ∂x/∂s) + (∂w/∂y *∂y/∂s). This is computationally identical to ∇w * v where ∇w = and v = = .
Can someone help me in f =cosxcoshy . We have to find gradient
Lovely
So HEEERE. As you can see. Something is appearing that we would note that way. Approximatively. These variables. We don't want these variables. OK. You drop it. What we want here it makes appear, on the plan, the derivatives of quantum mechanic (.)
So we're talking about 4D object😶
How did he get the equation x(s) = x(0) + as and similarly for y(s)??
well, a is the derivative in respect to s in the x direction. So a*s is the increase in x, and x(0) is the position where x is when s = 0. It's a special case of x(t) = x(to) + x'(to)(t-to), when the independent variable (in this case called s) is equal to 0.
It's just a vector in parametric form. You start at a point in the x-y plane, (x_0, y_0). Then, you go in some direction away from that point, (a, b). You travel exactly "t" away from the point, i.e., t(a,b). So your final position is (x_0, y_0) + t(a, b) = (x_0 + at, y_0 + bt).
I am not able to find the Functions of Two Variables Mathlets. MIT OpenCourseWare KINDLY HELP. Thanks!!
We are currently working on fixing this on the new site. You can find an older version here: ocw.mit.edu/ans7870/18/18.02/f07/tools/FunctionsTwoVariables.html. (You will need to have Java installed with applet support). We did find another resource that might help: www.geogebra.org/m/m9hvRWKQ. Best wishes on your studies!
P.S.- There are also mirrors of our older site that still have the mathlet linked: mit.usiu.ac.ke/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-a-functions-of-two-variables-tangent-approximation-and-optimization/session-24-functions-of-two-variables-graphs/MIT18_02SC_s24_applet.html
@@mitocw Thank you so much, MIT for the support!!
The guy is fast :D 28:21
i'm confused, he says at 12min that the "we know.. the function stays constant, but we can also know how function changes using the chain rule" didn't he just say it was constant? so confused.... T.T
it is constant because we chose a particular value for W(x,y,z) but in general it isn't constant so when you apply the differential operator on both sides of the equation W(x,y,z)=c
it becomes :
W'(x,y,z)=0
do directional derivatives only work in 3 dimensions?
No it's just harder to conceptualize more than that
If w is constant, how can its gradient vector point toward higher values?
+Michael Francesco Ala I believe in that case, the gradient vector would then point away from values around it because they're all part of the same level curve and give the same output, but the gradient vector points in the direction of greatest change in output. I was confused on the 2D xy-planer gradient vector vs 3D level plane gradient vector.
Setting w= to a value doesn't make it constant, w is still a variable.
assuming I know what part of the video you're asking about
Todd Nagel No matter what 2 variables you plug into w(x,y,z) the third (say, z) will change so that _w_ will always be *c*, by his definition. _w_ is a function whose output remains constant, so it's gradient will always be 0
harold:cube =12.137:9.
square 4y-65Q.
your area of circumfrence is 18.775. ^2SQ|MEGAPIXELS UNITS [LOARYIES HIDE-GO-SEEK÷(&).]
why are they laughing when the boards move?
I see you've been cutting on the previous lectures.
In the example x^2+y^2-z^2=4, why did the 4 disappear when he eventually wrote the equation of the plane? Shouldn't it be the constant 8+ the original 4?
we have the normal vector and a point. you plug the point into 4x+2y-2z=d to find d. He discusses this trick in lecture 4
31:28
@IamJacksColon4
So sill people are.
i need that applet to get a better understanding of tangent planes and gradients, where can i find it?
maybe he just appreciated guys having fun with math xD
This lecture was pretty confusing
really?? what part did you not understand
@@priyanksharma1124let W = f(x,y) .i did not understand how dW/ds will give us the directional vector in the direction of u . From what I understood r(position vector) has been parametrised with arc length (s) so that the component of r can be given as relation of some c(initial point) + s*. Now I understand that the position vector obtained will be in the direction of u as when s = 1 we move in the direction u from initial point. Also from above both parametrised x(s) and y(s) have been obtained. But how this relates to dW/ds being the directional derivative in the direction of u (analogous to cutting a slice in the graph of W parallel to u at the initial point and getting the slope of the curve) is something I am not getting. dW/ds should be the rate of change of the function w.r.t arclength (s) not the directional derivative.
The tutor was excellent but the students were childish.
so many chalkboards!!!
why the applause?
Peter Thiel on the Global Economy, the State of Our Technology, and Artificial Intelligence
44:46 independant non politician
i think his a french as " r" is not well pronounced
Somebody is coughing heavily . Please make a covid-19 test immediately 😂😂😂
*silly. seriously? Complaining about boards during a brilliant lecture?
@jphotguy
yeeting
why arent they using white boards? and why are they still using black boards when green boards is better for the eyes.
lol this guy's accent is hilarious.
why don't you have magnetic boards in classrooms? are you that poor?
25:26
31:20