I was very confused when people said the gradient was "normal" to the curve. I thought they meant the function itself, not the "level curve". Now it makes complete sense! Thanks!
@@sumittete2804 rate of change of the function is minimum in the direction of the tangent vector i.e. in you move perpendicular to the gradient vector
@@Suyogya77 But if i move opposite to gradient vector i.e 180° I'm getting rate of change of function as negative which is less than 0. Moving along tangent vector the rate of change of function is zero. So how ??
Thank you so much. I read my textbook and understood about half of this material and watched this video a couple of times and now understand the gradient vector much better. You really helped me.
I am a civil engineer , now a days I am pursuing master's in structural engineering ,in structural engineering we use these concepts to find maximum stresses/strains , Before this video I tried a lot but couldn't get into the depth of concept but after watching this video ,I got the concept of it ,animations are very helpful .thankyou and keep up the good work.
It is a wonderful thing to see your passion about mathematics, I'm assure you it is contagious and I love you because of it. I wish best for you with my all heart. Please do continue to make videos like that.
Also really important how you pointed out grad f as in the x-y plane as that also can be very confusing initially thinking about it as the gradient itself but of course that’s why we need 😊
Man if my college calculus profs. were as articulate as Dr. Bazett..I would have gotten better grades in those classes. I'm now a retired software "geek" and really love watching these presentations. Very few folks who understand advanced math (and EE-Comp Sci) are good at teaching it to "undergrads". Of course I love the animations also.
Couldn't agree more, lot of lectures I have i can barely understand what they are trying to present. It's pretty funny that a you tuber can present ideas in a much more clear and straight forward matter.
great videos, Trefor, I have been looking for the explanations with geometrical insights vs just algebra on the board. This really helps to "see" the math. thanks!
I think that the direction you have shown for Gradient Vector is the path that a climber would have to take while the actual direction of Gradient at any point would be paralel to XY plane.
I usually don't comment on videos but that's the best explanation i've ever watched to understand... i had this confusing for a long time and this lecture cleared that up! you deserve more subs!
Thank you for this video. You are a very clear example of the fact that you don't need 3Blue1Brown levels of visual editing in order to explain something intuitively and clearly.
Great video, thanks ! The only point I don't get is how R(t) = x(t)î + y(t)^j rpz a curve in the plan (where the t come from). If someone has an explanation or some source I could read it would be awesome.
t is a parameter, R(t) - vector valued funciton represents a contour. When you change t - you change x(t) and y(t) - you change position of the vector R(t) that points to the particular dot on the curve. When you get gradient for (x(t), y(t)) it's always perpendicular to the tangent line of the curve - derivative of R(t).
I like to think of the normal vector to the contour plot as the shortest distance between two points is a straight line. And when the distance between them next contour is infinitesimal. It is a parallel line. And anything other than normal is going to be more than just going normal.
This was excellent, thank you. However, I am still left with a years old doubt. How do you know if the gradient is pointing into or away from the 'mountain'? I understand that it lies olong the line of greatest change of height, but what is it, in the expression of the gradient, that tells you that it's pointing uphill or downhill?
About gradient vector, can I think in this way? For one point, gradient vector is in 2D, and normal vector is in 3D. They are different vectors, but if i transfer normal vector to 2D (xy plane), they are same.
Hello sir MY QUESTION IS THAT : while you wrote the dot product of gradient and the tangent you have shown that the components of the gradient is in I and j hat directions but should that be true because the partials or the components of gradient are obtained in the z-x and z-y plane respectively according to which the vector can't be written in an absolute dxn of i or j hat instead it should be like: i + k and j +k for for both gradient components respectively...
The gradient vector is pointing in the x-y plane, and pointing in to the mountain. What it represents is the slope of the surface of the mountain. Another example that might help understand it, is to think of a temperature field as a function of space. A temperature gradient doesn't touch the temperature axis, it only resides in our x/y/z space. Temperature is a scalar, and it makes no sense for a vector to have a temperature component. But it points in the direction of the "steepest" rate of change of temperature, from cold to hot. Heat of course flows exactly in the opposite direction.
Of course I enjoyed it. For better understanding of the Gradient, I searched this subject, fortunately, I saw you and I just clicked! Thank you so much
My new favourite video of yours, the mountain example was great :) You taught me calc1 at Uvic last year and now you are teaching me calc 3. A true godsend, thanks Trefor!
This video is spot on! Very nice. You just clarified gradient, level curves and the directional derivative in an intuitive way. I know understand the meaning behind the math. Thank you so much!
@@sumittete2804 The direction of the gradient tells you the direction of steepest ascent, and the magnitude tells you the slope of that ascent. Opposite the gradient vector, is the direction of steepest descent. The rate of change of the function is minimized, if the input point travels perpendicular to the gradient vector. The contour lines are perpendicular to the gradient vector. The principle behind Lagrange multipliers comes from this idea. Given that the point in question follows a constrained path, the candidates for the local extreme value of the function's output will occur when the path and the contour line, share a common direction.
I have just one doubt about angles Here the angle between dr/dt and grad f is 90 so according to calculus it has no change since cos 90=0 So please make this doubt clear
Hii Im from India, and I really love how you explain such difficult concepts in such a simple and brilliant way, I never could have understood these concepts had I been dependent only on my college professor’s class, as he himself does’nt clearly know these concepts like you! However, I could still not connect with the idea of the directional derivative as a dot product of the gradient and the vector component, intuitively. Can you explain how is it actually helping in finding the slope of the function in any particular direction?
i think that it relates to what he said about cos(theta). If the dot product is equal to 0, the two components are orthogonal. And in this case you have gradient of f going up and the tangent going across (when you look at the mountain) this means that there is a 90° angle between them and cos(90°) = 0.
2:36 I don't get what f(x,y) is wrt r(t). Like, do you mean to say you slice a mountain, take its cross-section, then the f(x,y) is built on it which is at a height C from r(t) on the z-axis such that f(x,y) is also a level curve?
Why is the magnitude of the gradient vector said to be the RATE of maximum ascent? When I see "rate", I think slope. Why isn't the rate of ascent simply the partial of y divided by the partial of x.? Isn't this the slope of the gradient....i.e. change in y over the change in x? What am I missing? thanks
I was very confused when people said the gradient was "normal" to the curve. I thought they meant the function itself, not the "level curve". Now it makes complete sense! Thanks!
Same!
Is rate of change of function minimum in the direction of tangent vector or in the direction opposite to gradient vector ?
@@sumittete2804 rate of change of the function is minimum in the direction of the tangent vector i.e. in you move perpendicular to the gradient vector
@@Suyogya77 But if i move opposite to gradient vector i.e 180° I'm getting rate of change of function as negative which is less than 0. Moving along tangent vector the rate of change of function is zero. So how ??
@@sumittete2804 do you use telegram or something?
Thank you so much. I read my textbook and understood about half of this material and watched this video a couple of times and now understand the gradient vector much better. You really helped me.
Glad it was helpful!
I am a civil engineer , now a days I am pursuing master's in structural engineering ,in structural engineering we use these concepts to find maximum stresses/strains , Before this video I tried a lot but couldn't get into the depth of concept but after watching this video ,I got the concept of it ,animations are very helpful .thankyou and keep up the good work.
dude I love ya. that cleared everything about gradients in my mind. Thanks a lot bud.
Glad it helped!
Man, you really teach what's important to understand the concepts, and you explain yourself perfectly! Amazing! You've gained a new subscriber 😁
It is a wonderful thing to see your passion about mathematics, I'm assure you it is contagious and I love you because of it. I wish best for you with my all heart. Please do continue to make videos like that.
I am doing all possible steps to take this channel to a bigger audience
my mind is blown, finally I understand how the tangent unit vector gives a direction along which f(x,y) is constant, Thx alot Dr.
glad it helped!
Your demonstration is just amazing Sir....the best explanation of gradient vector on TH-cam....
Thanks a ton!
This video is amazing. First time I'm seeing these concepts clearly since I started taking this course.
Fantastic explanation. Now I understand the gradient for our purposes lies in the xy plane and that it points into the mountain.
My professor rambled on for 2 hours and I didn't understand anything. Here you are explaining it perfectly in 15 minutes. THANK YOU SO MUCH
Also really important how you pointed out grad f as in the x-y plane as that also can be very confusing initially thinking about it as the gradient itself but of course that’s why we need 😊
Beautifully presented! It's such a cool topic, and using mountains as an analogy makes everything so intuitive.
understood the beauty of multivariable calculus and gradient operator. Thanks a lot sir :)))
Happy to help!
Great video that gives a brilliant straight explanation for the gradient vector. Hope to have your class in UVic.
Man if my college calculus profs. were as articulate as Dr. Bazett..I would have gotten better grades in those classes. I'm now a retired software "geek" and really love watching these presentations. Very few folks who understand advanced math (and EE-Comp Sci) are good at teaching it to "undergrads". Of course I love the animations also.
Couldn't agree more, lot of lectures I have i can barely understand what they are trying to present. It's pretty funny that a you tuber can present ideas in a much more clear and straight forward matter.
@@codstary1015 LOL, Its pretty naive of you to assume that Dr.Trefor is just another youtuber!
You’re literally perfect
rewatched it several times, started losing hope but then it clicked and I was like 'wait that makes sense!'
A legend is living among us!!!!!!!!!
Amazing explanation. Thank you!
Great derivation and application! The derivation of the gradient-vector formula and its justification were both quite easy to follow!
thank you so much for clearing the doubt. The video was very helpful.
very informative
when i pass calc 3 and eventually become an electrical engineer itll be thanks to you thank you 😭😭
I suppose to drop out and join TH-cam University
thankyou so much! this one's great!!🌟
I have never seen such a beautiful explanation ever of Gradients love you
That's amazing zing zing
Man, you are Richard Feynman of our time
That is high praise!
@@DrTrefor more than him!
Literally the truth🫡☺️
Not high praise , the amount of hard work you have done to develop these is phenomenal ❤
@@Abdullahezzat1893of course not. What a stupid reply🤦🏽♂️
amazingly clear. thank you
Really enjoyed
great videos, Trefor, I have been looking for the explanations with geometrical insights vs just algebra on the board. This really helps to "see" the math. thanks!
I think that the direction you have shown for Gradient Vector is the path that a climber would have to take while the actual direction of Gradient at any point would be paralel to XY plane.
Excellent!
I usually don't comment on videos but that's the best explanation i've ever watched to understand... i had this confusing for a long time and this lecture cleared that up! you deserve more subs!
Amazing explanation! Thank you so much!!
Glad you enjoyed it!
what an amazing video!!! Please make a video on divergence on curl too...it would be very helpful
@@DrTrefor Wow!! Eagerly waiting for them
Thank God I found this channel 🙌
At 8:39, shouldn’t you say 0 slope instead of minimum slope? I think you get the minimum slope when theta is -π
Yess....Rate of change of function is minimum in the direction opposite to gradient vector that is at angle of 180°
Love you sir!
after visualizing these concepts it became easier for me to perform the mathematical formulas thank you so much sir for the valuable information
Ahhh, finally I fully got it, thx man:) The map help a lot. I knew what gradient is, but i strugled to get the geometric meaning
Glad it helped!
Thanks
Big Fan Sir
Thanks a lot sir...
This video is really helpful. Thank you so much,👏
you'll make us love calculus and maths!!
thanks for including practical example of Vancouver island
haha I had fun with that part!
I loved the video....I have a question...why does the gradient have to always point towards increasing values of f on the contour plot?
Thank you for this video. You are a very clear example of the fact that you don't need 3Blue1Brown levels of visual editing in order to explain something intuitively and clearly.
Amazing explanation!!!! thank you so much, you make great influence in the world..
When the background is black, the only two visible colors on it are white and yellow. Thankfully!
I love you, thank you!
please don't forget to subscribe !!! what you did and the number of subscription you have don't align together .
Best math channel on TH-cam!
I wonder if function of multivariable is study of mountains only, in my class they also gives example from mountains only ! ...Nice explanation though
great example, thank you for your video
Sir ,please recommend me a mathematics book for engineering, and for gate and IIT jam exams.
Great video, thanks ! The only point I don't get is how R(t) = x(t)î + y(t)^j rpz a curve in the plan (where the t come from). If someone has an explanation or some source I could read it would be awesome.
t is a parameter, R(t) - vector valued funciton represents a contour. When you change t - you change x(t) and y(t) - you change position of the vector R(t) that points to the particular dot on the curve. When you get gradient for (x(t), y(t)) it's always perpendicular to the tangent line of the curve - derivative of R(t).
I like to think of the normal vector to the contour plot as the shortest distance between two points is a straight line. And when the distance between them next contour is infinitesimal. It is a parallel line. And anything other than normal is going to be more than just going normal.
loved it
The great combination of theory and a real example of a mountain in Vancouver. I enjoy the lesson series of Calculus so much.
U are amazing ❤️
Example of a mountain was superb to explain gradient..thanks bro
This was excellent, thank you. However, I am still left with a years old doubt. How do you know if the gradient is pointing into or away from the 'mountain'? I understand that it lies olong the line of greatest change of height, but what is it, in the expression of the gradient, that tells you that it's pointing uphill or downhill?
Because the directional derivative is positive
amazing and so interesting! keep it up
Such a brilliant video, it truly heps
Perfectly explained. Thank you sir
About gradient vector, can I think in this way?
For one point, gradient vector is in 2D, and normal vector is in 3D.
They are different vectors, but if i transfer normal vector to 2D (xy plane), they are same.
Yes that works:)
Hey Professor which software you use for the video?
Hello sir MY QUESTION IS THAT : while you wrote the dot product of gradient and the tangent you have shown that the components of the gradient is in I and j hat directions but should that be true because the partials or the components of gradient are obtained in the z-x and z-y plane respectively according to which the vector can't be written in an absolute dxn of i or j hat instead it should be like: i + k and j +k for for both gradient components respectively...
thank you
excellent video as always, nice example :)
Math with visible imagination in 2d and 3D is wonderful. Math with only formulas is awful.
Oh god it helped me so much thanks sensai
I'm the first viewer
The Geometric element is fascinating. But the algebraic dot product provide a solid conclusion.
had seen the videos pop up in the search results and never found the time to have a look. Now just did : I'm a fan! Thanks Prof. Bazett! :)
Just amazing
If the gradient is "into the mountain" Then why is it called the "steepest" change?
The gradient vector is pointing in the x-y plane, and pointing in to the mountain. What it represents is the slope of the surface of the mountain.
Another example that might help understand it, is to think of a temperature field as a function of space. A temperature gradient doesn't touch the temperature axis, it only resides in our x/y/z space. Temperature is a scalar, and it makes no sense for a vector to have a temperature component. But it points in the direction of the "steepest" rate of change of temperature, from cold to hot. Heat of course flows exactly in the opposite direction.
Thank you so much🙏
You're amazing. 😊😎
Love your teaching sir...""LOVE""
Thanks a lot sir 🔥🔥🔥
Of course I enjoyed it.
For better understanding of the Gradient, I searched this subject, fortunately, I saw you and I just clicked!
Thank you so much
Excellent explanation
My new favourite video of yours, the mountain example was great :) You taught me calc1 at Uvic last year and now you are teaching me calc 3. A true godsend, thanks Trefor!
Thanks Conor, really appreciate that. Good luck with math 200!
Bro how did you differentiate both sides of an equation and equate them. LHS=RHS even after differentiating isn't always true , is it?
Think you sir,,,,,,,Respect from Bangladesh 🇧🇩🇧🇩🇧🇩🇧🇩🇧🇩
great job.
This video is spot on! Very nice. You just clarified gradient, level curves and the directional derivative in an intuitive way. I know understand the meaning behind the math. Thank you so much!
this video is one of the greatest one's that you can find on this topic
In starting, too many concepts to grasp. But the example in end was very interesting.
Thanku sir
Amazing lecture!
It essentially proves the notion that the gradient is orthogonal to the level set.
Thanks a lot Sir Trefor.
You are welcome!
Is rate of change of function minimum in the direction of tangent vector or in the direction opposite to gradient vector ?
@@sumittete2804 The direction of the gradient tells you the direction of steepest ascent, and the magnitude tells you the slope of that ascent. Opposite the gradient vector, is the direction of steepest descent.
The rate of change of the function is minimized, if the input point travels perpendicular to the gradient vector. The contour lines are perpendicular to the gradient vector.
The principle behind Lagrange multipliers comes from this idea. Given that the point in question follows a constrained path, the candidates for the local extreme value of the function's output will occur when the path and the contour line, share a common direction.
I have just one doubt about angles
Here the angle between dr/dt and grad f is 90 so according to calculus it has no change since cos 90=0
So please make this doubt clear
Hii Im from India, and I really love how you explain such difficult concepts in such a simple and brilliant way, I never could have understood these concepts had I been dependent only on my college professor’s class, as he himself does’nt clearly know these concepts like you!
However, I could still not connect with the idea of the directional derivative as a dot product of the gradient and the vector component, intuitively. Can you explain how is it actually helping in finding the slope of the function in any particular direction?
i think that it relates to what he said about cos(theta). If the dot product is equal to 0, the two components are orthogonal. And in this case you have gradient of f going up and the tangent going across (when you look at the mountain) this means that there is a 90° angle between them and cos(90°) = 0.
Vancouver, eh? I was just thinking you sounded like J.J. McCullough
I love you
2:36 I don't get what f(x,y) is wrt r(t). Like, do you mean to say you slice a mountain, take its cross-section, then the f(x,y) is built on it which is at a height C from r(t) on the z-axis such that f(x,y) is also a level curve?
Why is the magnitude of the gradient vector said to be the RATE of maximum ascent? When I see "rate", I think slope. Why isn't the rate of ascent simply the partial of y divided by the partial of x.? Isn't this the slope of the gradient....i.e. change in y over the change in x? What am I missing? thanks