Gradient of a function.

Download the free PDF tinyurl.com/Eng...
A basic tutorial on the gradient field of a function. We show how to compute the gradient; its geometric significance; and how it is used when computing the directional derivative.
The gradient is a basic property of vector calculus. NOTE: I incorrectly write t as an angle to positive OX axis around 5 minutes. Apologies for any confusion caused.

[@DrChrisTisdell]

Thank you! I am very glad you have found these resources useful! :-) Please let me know when you graduate!!

[@danaricardo7]

I honestly like you Dr Chris, U are awesome. U just make everything very interesting and simple indeed. I am doing my 2nd year physics degree in one of the leading UK universities and so far have very much benefited from your outstanding youtube videos. I do promise to reference you sir, in my every single work. I am also gonna have to mention you in my graduation speech next year. We all love you Sir.

[@DrChrisTisdell]

It's my pleasure. I am so glad to hear of your great result. Well done.

[@lucreciaxb]

Thank you very much. You're a great teacher. I had love to have a teacher like you at the University.

[@DrChrisTisdell]

Hi - the form (4 cos t, 2 sin t) is a standard way of describing an ellipse. You can verify this by substitution in to the eqn x^2/16 + y^2/4 = 1.

[@josuanateba812]

Exactly the video I was looking for...Thank you very much Sir!!!

[@jordanboyce3082]

why is there a small picture moving at the bottom right

[@SkeleCrafteronYT]

That's the actual class. The big picture is what he is writing on.

[@Antrikshy]

That is some beautiful handwriting.

[@beckwilde]

Why is there a small tiny video in the corner? Just wondering! Lovely video!

[@isla2974]

Thanks professor. That is a nice video. But, I have a question. How do you know the directional vector, r(t) = (4cost, 2sint)?

[@DrChrisTisdell]

HI - you can find the answers in the free ebook. The download link is in the description. Good luck!

[@ruslan-pe3wx]

Great work. I didn't get one point. How did you parametrize C? Where did you get the r(t)?

[@shermusic8361]

Why does gradient point in the direction of maximum increase of a differentiable function from R2 to R.?

[@S05J11R92]

he uses basic trig. he gives r(t) a certain value, t, from the x axis and uses trig to describe the vector as if it were the hypotenuse part of a triangle. r(t) is obviously not the hypotenuse but it is in the same direction, which is what matters.

[@alehejazi92]

why did you parameterize it by cos(t) and sin(t), can any other parameters be substituted ?

[@Aluminiuminium]

Hello, I came across your youtube channel, and I find it very useful. I think the clips are very well done, explaining all the things. I want to thank you for your work. However I find a mistake in this clip, where you discuss the parametrization of an ellipse: t is NOT the same angle as the one made by the r(t), position vector, with the x-axis. I made myself this error some time ago.

[@DrChrisTisdell]

Thanks for the kind words. If t does not represent the angle to the positve x-axis then what do you think it represents?

[@Aluminiuminium]

This explains it very clear: www.mathopenref.com/coordparamellipse.html See here : The parameter t - section.

[@evangelialampiri9852]

very nice videos. Thank you very much!! I am searching a book about gradient flow systems, do y have any good in your mind? I would appreciate it if you could suggest me one! Thank you a lot

[@surajitju]

How come perpendicular to position vector on ellipse tangent to the ellipse also??????

[@DrChrisTisdell]

Hi - I dont claim that the position vector is perpendicular to tangent vector. What I have drawn is showing the tangent vector and normal vector are perpendicular.

[@surajitju]

Dr Chris Tisdell Thank you. I got it clear.

[@michaelDmambo]

Great Video thank you. Something I'm not getting at this point. Let's imagine a parabola y = x^2, so we would define f(x,y) = x^2 - y, and the parabola is defined for f(x,y) = 0 ? so Grad(f) = (2x,-1). That seems to give us the normal vector of the parabola at any point, great. Now it is also said that a critical point needs to satisfy Grad(f) = 0. Here we see that Grad(f) will never be equal to 0, but the parabola has a critical point at (0,0). Where am I making a mistake ? (just trying to self teach myself, so I apologize if there are some blatant mistakes here).

[@michaelDmambo]

I think I may have understood it now, Grad(f) = 0 for the dimensions except the one we're interested in find the extrema in, which in this case would be y. Or said differently if I had a line defined y = constant, what is the value of x such that Grad(f) = k.Grad(y=constant)

[@nirjharghosh4478]

Sir How Does That grad(F) become (8cos t,16sin t) ????

[@DrChrisTisdell]

Hi - check out 06:46. There you will see that x = 4 cos t and y = 2 sin t. Substitute those into the expression for grad(F).

[@DrChrisTisdell]

Wow - what an honour!

[@DrChrisTisdell]

Hi - good question. I do have a video on that. It is because the directional derivative is a dot product. /watch?v=gJmw3JpwPrQ and /watch?v=tNnnOv_3A2g t

[@asosdaxiao]

omg i duno how to solve the final example hope to see worked solutions to it.

[@2012daffyduck]

Thanks so much chris, if i have a kid some day - i'll name him chris.. :)

[@youspinmerightrounds]

That's correct alright

[@vincentmothapa2421]

i'm in a similar situation, that's why I am here