Multi-variable Optimization & the Second Derivative Test
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- เผยแพร่เมื่อ 27 มิ.ย. 2024
- Finding Maximums and Minimums of multi-variable functions works pretty similar to single variable functions. First,find candidates for maximums/minimums by finding critical points. Critical Points are where the partial derivatives with respect to x and y are both zero. Then we classify each critical point using the second derivative test. In the multivariable case, there is a new option beyond max/min/neither, there is also the case of the saddle point. The second derivative test involves computing the Hessian, the determinant of a matrix that helps decide whether points are maximums/minimums/saddle or inconclusive. We sketch the geometric intuition behind the Hessian.
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Note that the conditions of the 2nd derivative test should also demand the 2nd partials are CONTINUOUS.
SO THAT'S WHAT THE MIXED PARTIALS ARE FOR THANK YOU SIR
This playlist is, without a doubt, FASCINATING! Thank you!
Thank you!!
Clear explanation of mixed partial derivatives. More videos on the Hessian matrix and its applications.
th-cam.com/video/XPCgGT9BlrQ/w-d-xo.html 👍💐💐💐💐
Where is the video "Hessian matrix and its applications"? Please give a link to it. Thanks.
THE BEST explanation.....I bet none of my teachers (in India) can even think of this...
Yes bhaai isilie India m koi maths m accha nhi karta bas rat Lete h underlying cheej koi nhi samjhaata ratwaa dete h
None😂
Kuch v bhai
Not quite true… my professor did an excellent job
U are from India and I am from India lets change it brother atleast in next generation....🔥🔥🔥
@@mohamedirshaathm32123 the correct mindset 💪
Thank you for the intuitive explanation of the mixed second partial derivative and what it is representing. I took some time sitting and contemplating on my own before I came to a similar conclusion. It's nice to hear some confirmation of that!
Please explain
Thank you so much! Your videos are helping me survive multivariable calculus :)
Exceptionally good and easy to follow explanation! I was having trouble understanding how to interpret the calculation for the 2nd derivative test and this definitely helped a lot.
th-cam.com/video/XPCgGT9BlrQ/w-d-xo.html 👍💐💐💐💐
The mixed partial derivative fxy=fyx is the concavity on the line x=y, explained in 10:30, is great.
Thank you.
Please explain
Dang, I posted two comments here, first one I thought I was so clever because your diagram and explanations were so clear I got the idea before you said the words ....... but then you said exactly the words ...... and the second comment was a question ....... and then you explained that question. You are an excellent teacher
Thank you so much! You are such a great teacher. This simplified this topic so much for me
Fantastic way of teaching!!! I recommend the classes here in Brazil!
This Video is awesome that I can clearly understand this concept.
I truly love the calculus!!!
so that is what it has application! thanks! you explain like 100x better my teacher.
thanks man! never knew what mixed partials are supposed to mean but you made it crystal clear! :D
Please explain
Excellent explanation!
Great video! Thanks for the explanation
Great video!! I really looking forward to your new video in this topic. Thank you!
th-cam.com/video/XPCgGT9BlrQ/w-d-xo.html 👍💐💐💐💐
Awesome video sir! Thank you!
great explanation. having visuals in 3D is sooo important to understanding multivariable calc
This is the best explanation I've ever seen. Thank you my man!
th-cam.com/video/XPCgGT9BlrQ/w-d-xo.html 👍💐💐💐💐💐
nice video, very straightforward, detailed explanation :)
Very helpful! Thanks 👏👏👏
Great, thank you very much for this explanation
Thanks so much Professor Bazett for this excellent, lively presentation! You convey your enthusiasm to your audience. They are helping me so much as I review multivariable calculus being offered on edX. I am very, very grateful.
Thank you So much... Very Easy and Perfect Explanation.. Really Helpful video
Thanks for that. always struggled to find an intuitive understanding for the mixed 2nd pd. and its' significance. Think you nailed it. Thank you.
th-cam.com/video/XQIbn27dOjE/w-d-xo.html
Please explain
For the people who don't understand the pink part, it is just the result of the Hessian 2x2 matrix. that you solve by doing, a.d-b.c
I nominate you for Nobel Prize.
im korean college student and thank you for your hing class lecture.
Great video hope youre having a great day
Sir you are the best really 😇
I love your intuitive explanation! :-D
Thank you!
the best as always
Really nice explanation... Waiting for more videos in this manner
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wow very informative and very clear wow !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
? from one of the comments below: Where is the video "Hessian matrix and its applications"? Please give a link to it. Thank You! Great job!
Thank you sir
Thanks!
You are just awesome!!!
I want to be this good in Maths... :)
Keep working at it, you'll get there!
@@DrTrefor thank you!
I am trying my level best.
Clear and simple
th-cam.com/video/XPCgGT9BlrQ/w-d-xo.html 💐👍👍👍
What is the connection for the formula to the derivative of the mixed partial matrix?
nice visualation
Great video, but I suggest to use convex/concave instead of down concave ....
What do the second derivative tests look like when the function has 3 or more independent variables?
Sir kasam s English m likh k feeling nhi aayega , mzaa AAA gyaa kasam s
Thank you 👍🏻
th-cam.com/video/XPCgGT9BlrQ/w-d-xo.html 💐💐💐💐💐💐
Hi
Will this work for 3 and 4 variable functions also? How to do in that case?
I love how well you explain the intuition, you have an extremely pedagogical way of teaching maths!
I had one doubt that I couldn't solve and confused me:
Assume the function x^2 + y^2 + 3 xy
It gives me fxx=2, fyy=2 and fxy=3 Hence fxx.fyy - fxy^2 = 4-9 = -5
Please can you make a video just on mixed partials - graphically what is going on. I still do not understand.
What should we do,if we encounter inconclusive condition?
Great video ❤️
th-cam.com/video/XPCgGT9BlrQ/w-d-xo.html 👍💐💐💐💐💐
Great Explanation!
I have a doubt though.. say my function is sinx+siny+sin(x+y) partial derivative w.r.t. x and y will be cosx+cos(x+y)=0 and cosy+cos(x+y)=0 respectively, now to find the critical points do I equate both the eqations and thus eliminate cos(x+y) or do I go by the general rule and expand the functions by addition formula? According to me, we should go by the second method as it is generalized but the first way also gives the correct answer...why is it so? Please Reply whenever you can.
Thank You :)
I think the first way is more of an trick, but totally works, both partial derivatives should be equal to 0 at some x,y, so in fact they will also be equal to each other. So you get cosx=cosy
Points where the partial derivatives don't exist also should be candidate points for the 1st derivative test right? For example, the point on a cone.
Exactly
I was reading another book on CAD/CAM where it is stated that the gradient vector gives the normal to a tangent plane at a point. I can provide the details of the book if required. I do not believe this to be true and so I am trying to resolve this apparent clash. Can you confirm whether or not the book statement is true or not to try and save me some time. Is there a special case where this might be true?
Thanks
Tony
Yes, it is true. Gradient vector always gives the normal to the plane which is why you use to find the tangent plane at any point. Why did you believe it is not true?
@@muthukumarr5217
I'm not Tony Aimer but I found it counter intuitive at first. I would have thought the gradient pointed in the direction of speepest ascent / descent but later learned about the whole "normal" thing.
wait can someone explain how at min 10:05 he gets the line x=y, f(x,x)=-x^2? Like idk if my brain is just fried but i do not see where that's coming from and now suddenly my mouse is hovering over the "withdraw from MATH 1320" option in my enrollment portal.
nevermind yall i just needed a nap i figured it out
th-cam.com/video/XPCgGT9BlrQ/w-d-xo.html 💐👍👍👍💐
Had to come back to this video lol, copied second derivative rule wrong with the equalities and it was a hard problem
WOW
I tried to figure out what fxy means so,
We think of first derivative as difference in function as in the limit formula for derivatives.
Think of the second derivative as difference IN DERIVATIVE.
For local minimum the derivative was negative then the derivative was positive and positive minus negative is positive so when fxx is positive it is a local minimum.
Now for fxy think of it as ok get the derivative with respect to x. Now we have a function fx (first derivative) so when we go in y direction, how the derivative of with respect to x changes.
More clearly as in the example in the video, the derivative at x=0 is 0 as z does not change. But if we go some small distance in y you will find that when you subsitute in the fx (first derivative) with new values of y and SAME X, you will find that z has decreased !!! So it wasnt local minimum as fx said and it is clear in the visualization that the slope IS DOWNWARDS. SO WE HAVE SLOPE = 0 THEN SLOPE = NEGATIVE. AND NEGATIVE - 0 = NEGATIVE as we found mathematically that fxy is negative.
It is like ok i am local minimum along x but am i higher in other directions? Like i am at the bottom of a house on a mountain, i can move only up my house or stay at the bottom but i cant move forward as i will fall (i am not at bottom at this point, there is a point more bottom than me).
But i really dont get the meaning of fxx times fyy, what does their result say. But i think it does not tell something special other than that it is compared with fxy to see which is larger. Also i dont get why someone larger meaning suddenly something. 😅
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Love u sir ❤️❤️❤️❤️❤️ MAA kasam pyaar ho gya sir aapse
It became complex from saddle and forward
great video but i feel mic needs improvement.
oh man, the office I was in for filming this was SO BAD for sound. Much better set up these days.
Agreed, the volume goes back and forth between too low and too loud
algo fuel
i want to understand marginal product
The content is good but the form of the presentation is destracting. Focusing back and forth on the presenter and the diagram and listening to a forceful voice is quite challenging for me.
Thank you very much sir keep growing up (◍•ᴗ•◍)❤♥╣[-_-]╠♥
Nice tutorial, but the acoustics is strange, it sounds like you are teaching from a bathroom or a cave.
Dear sir you are speaking too fast, it is difficult to understand your standpoint of your statements for we foreigners. Please speak more slowly, why are you in hurry!
th-cam.com/video/XPCgGT9BlrQ/w-d-xo.html 💐💐💐💐👍