I really appreciate your videos. I have recently discovered your channel and it has helped me a lot. I’m going to take my calculus III and differential geometry exams in a few days. Keep up the good work!! Greetings from Greece 🇬🇷
awesome video dude, honestly, clearer explanation then my uni lecturers [Newcastle University Aus; The quality of education at this institution is horrible though admittedly, we have the WORST teachers, they basically don't have teaching skills and are so out of touch with reality/industry it's a joke!] Anyhow, thank you youtube teacher who is better then some of the guys i have paid thousands to over the years to!
First of all, thanks a lot for this superb series! Not trying to be a smart ass, but I would like to understand: a) At ~14:30 you talked about global maximum/minimum. Was that a mispronounciation, did you mean global in terms of "global on R", or did I get something wrong? b) What about the case where f(x, y, ...) = c (some constant value)? I *guess* that then we also have the case of f_xx * f_yy - f_xy^2 = 0, and would like to know if that is true, and how that is function is considered inside a fixed region?
Basically, you separate the two cos^2(theta) and one of them you use it to simplify cos^2(theta) + sin^2(theta) = 1. So, cos^2(theta) + [cos^2(theta) + sin^2(theta)] - sin(theta) = cos^2(theta) + [1] - sin(theta)
If I choose substitution instead of parametrization then will I have to substitute once y in terms of x and solve it and then x in terms of y and solve this also please reply to this doubt of mine and if not both then why.
If I use the second method (i.e. take df/dy after substituting 1-y^2 for x^2), I will only find the Maxima with y = -1/2. The nothing points at x = 0 don't show up. Do they just fall out of the computation depending on which parametrization you choose and nothing points like these can just come up from to time? Thanks a lot for your videos, very concise, they allow me to understand what I just memorized before!
Hi I had this same problem, but one thing I tried was substituting 1-x^2 for y^2 as well, which led me to x=0 and x=+-sqrt(3)/2. Perhaps you have to do both substitutions to find all solutions, but I am not sure.
Amazing content as usual ! Very clear and unbelievably helpful ! However anyone could help me to get this clearer ? I understand until the equation sin(theta)= -1/2 After that where does this triangle come from ?! And how did he solve it after that ? Cheers Pardon my french
The triangle he draws is from sin(theta)=1/2, not -1/2. If you know basic high school trigonometry you can easily solve sin(theta)=1/2. To solve for sin(theta)=-1/2, you need to know how sine acts in all 4 quadrants, using the definition of radians and unit circle. The graph he draws for sin comes from these 4 quadrant sine values. you can either see from that graph, or you can use rules for different quadrants, such as sin(theta)=-1/2 corresponds to sin(pi - theta)=1/2. this gives pi - theta = pi/6, hence theta = 7pi/6. These are the rules of inverse trigonometric functions, the fact that different trigonometric functions may be negative or positive in different quadrants requires that we have several tools to bring these angles back to our first quadrant (from which you may infer that values theta only goes from 0 to pi/2). Ill leave the other value of 11pi/6 up to you to calculate, because inverse trigonometry is very unpleasant to me!!
![[ ออกกอง ] วัยหนุ่ม 2544 โลกหลังลูกกรง..ทัวร์กองถ่ายหนังใน "คุกของจริง" | JUSTดูIT.](http://i.ytimg.com/vi/vB--aOkZCpw/mqdefault.jpg)
Please keep creating such great playlists..... very helpful while all of us study from home
Thank you I will!
I'm a TA for a multivariable calculus class and your videos are the best refreshers! Thank you!!
I really appreciate your videos. I have recently discovered your channel and it has helped me a lot. I’m going to take my calculus III and differential geometry exams in a few days. Keep up the good work!! Greetings from Greece 🇬🇷
Trefor Bazett Thank you!!
Smooth smooth smooth. Just fantastic
Thank you so much for these videos, they're a huge help!!!!
So glad they are helping!
Thank you from India 🇮🇳🙏❤.
I cleared my doubt.
awesome video dude, honestly, clearer explanation then my uni lecturers [Newcastle University Aus; The quality of education at this institution is horrible though admittedly, we have the WORST teachers, they basically don't have teaching skills and are so out of touch with reality/industry it's a joke!]
Anyhow, thank you youtube teacher who is better then some of the guys i have paid thousands to over the years to!
You are the goddamn best Dr. Bazett.
Nice there's actually a third method that's good for more complicated functions/boundaries.
Finally, a professor with diction!
You are awesome! Very clear explanations
Nicely explained
First of all, thanks a lot for this superb series! Not trying to be a smart ass, but I would like to understand:
a) At ~14:30 you talked about global maximum/minimum. Was that a mispronounciation, did you mean global in terms of "global on R", or did I get something wrong?
b) What about the case where f(x, y, ...) = c (some constant value)? I *guess* that then we also have the case of f_xx * f_yy - f_xy^2 = 0, and would like to know if that is true, and how that is function is considered inside a fixed region?
Hi how do you make these 3d graphs?
awesome super great explanation , great content!
good work keep going ahead
excellent video, very important subject :)
Glad it was helpful!
Great visualization!
imao, the solution presented is on the closed region x^2 +y^2 = 1, which is a hollow cylinder, not on the solid cylinder x^2 +y^2
Amazing, as always
how does it clean up to 1+cos^2-sin at 8:08 ?
using identities on the sin^2 term. 1-cos^2(theta)
Basically, you separate the two cos^2(theta) and one of them you use it to simplify cos^2(theta) + sin^2(theta) = 1. So,
cos^2(theta) + [cos^2(theta) + sin^2(theta)] - sin(theta)
= cos^2(theta) + [1] - sin(theta)
You earned a sub.
If I choose substitution instead of parametrization then will I have to substitute once y in terms of x and solve it and then x in terms of y and solve this also please reply to this doubt of mine and if not both then why.
Plzzz make videos on probability and statistics also
If I use the second method (i.e. take df/dy after substituting 1-y^2 for x^2), I will only find the Maxima with y = -1/2. The nothing points at x = 0 don't show up. Do they just fall out of the computation depending on which parametrization you choose and nothing points like these can just come up from to time?
Thanks a lot for your videos, very concise, they allow me to understand what I just memorized before!
Hi I had this same problem, but one thing I tried was substituting 1-x^2 for y^2 as well, which led me to x=0 and x=+-sqrt(3)/2. Perhaps you have to do both substitutions to find all solutions, but I am not sure.
What program are you using for your beautiful graphs
Nicely explained !!
Really appreciated
Omg thank you u saved me lotssss>< My professor sucks TT
Amazing content as usual !
Very clear and unbelievably helpful !
However anyone could help me to get this clearer ? I understand until the equation sin(theta)= -1/2
After that where does this triangle come from ?! And how did he solve it after that ?
Cheers
Pardon my french
The triangle he draws is from sin(theta)=1/2, not -1/2. If you know basic high school trigonometry you can easily solve sin(theta)=1/2.
To solve for sin(theta)=-1/2, you need to know how sine acts in all 4 quadrants, using the definition of radians and unit circle. The graph he draws for sin comes from these 4 quadrant sine values. you can either see from that graph, or you can use rules for different quadrants, such as sin(theta)=-1/2 corresponds to sin(pi - theta)=1/2. this gives pi - theta = pi/6, hence theta = 7pi/6.
These are the rules of inverse trigonometric functions, the fact that different trigonometric functions may be negative or positive in different quadrants requires that we have several tools to bring these angles back to our first quadrant (from which you may infer that values theta only goes from 0 to pi/2).
Ill leave the other value of 11pi/6 up to you to calculate, because inverse trigonometry is very unpleasant to me!!
Excellent!
sir, what is local maxima and local minima
Ty very much
Thank you!
Thank you
Thanks sir
thank u!
Maxima and minima
Matchless
Nice shirt
I thought this was langrange multipliers
That's a level constraint usually
global minimum is at (0,1/2), he forgot that point