Think about z as f(x,y) instead of an independent variable, the same way that y is actually f(x) on a regular xy plane. Z is a function of both x and y.
y is held fixed. The partials wrt y can also be calculated in the same way by holding one of the other variables constant. So for py/pz you hold x constant, and for py/px you hold z constant.
F(x,y,z)is the function of three variables, you can calculate partial derivatives. If for some point F(Xo,Yo,Zo )=0, for very small surroudings of this point variables are dependent, this function f always exists for examp variables z, x , z=f(x), (you can write this dependence ) then dz/dx=-Fx/Fz...... or for z=g(y) is dz/dy=-Fy/Fz.......or y=h(x) ...... x=p(y)........(ifFz0.....). I think, there is some similar idea.
Can you do some videos on the fundamentals of Riemann integration? such as what it means for a function to be integrable, box integrals in R^n, and deducing whether some interesting functions are integrable over a box B, where the function is not continuous on B?
I saw the proof of the Theorem in my calculus 2 classes. Sadly, I wasn't able to apply it in the exam, but I passed the course! Thank you Dr Peyam for explain thing so well! Your calculus 3 videos are saving me in this quarantine because vector calculus is confusing
Since the original function is symmetric wrt x,y,z pXi/pXk = (2XjXk-Xi^2)/(Xk^2-2XiXj) where p is partial differentiation operator and i,j,k are subscripted indices that may be permutated. Or in other words you have also solved for px/py, px/pz, py/px, py/pz, pz/py.
Having a bit of a brain fart, but why is y allowed to be treated like a constant but z isn’t? If we wanted to find del y/del x would z be treated as a constant?
Yes it's because were differentiating z with respect to x so the value of y is not changing. You can think of a partial derivative as analysing vertical 2D slice of a surface and finding the gradient.
Good Dr, I have a question for you from another Dr (Aero Engr; math minor, retired). Although I am retired I still like playing with math and therefore greatly appreciate your presentations. So conceptually, I like to think of the process you described in method 1 as operating in an x-z plane while holding y constant, so therefore any delta-y (dy) is zero, you can then take directional derivatives (conceptually) constrained to such a plane.. Since you can choose any y-plane, you can span 3-space (except the origin in this case) and all is fine. But I wanted to try another method, akin to a directional derivative in the direction of a specified line, where I hold, say y and z constant, and am then constrained to operate on a line of varying x; dy and dz are now zero on that line and I proceed: F(x,y,z) = x^3 + y^3 + z^3 - 6xyz pF/px = 3x^2 - 6yz where p is the partial operator. Then holding x,y constant along a line of varying z (dx, dy = 0) pF/pz = 3z^2 - 6xy So I suppose I must choose an intersection of these two lines and say at that point: (pF/px) / (pF/pz) = (pF/px) (pz/pF) = pz/px = (3x^2 - 6yz)/(3z^2 -6xy) = (x^2 - 2yz)/(z^2 -2xy) and this is the answer provided except for a minus sign !?? So where is the flaw and where is the missing sign change, given that the magnitude is correct?
I had this theorem proven on my calculus lecture just 4 weeks ago, AGAIN your video was there just a ONE day earlier. Are you collaborating with my calculus professor to achieve that ?? O.o I really need to watch your videos as soon as they come out :D
I'm really really thankful! This was an awesome video, and tomorrow I have my Analysis 2 exam on implicit diff! so thanks a lot, a in-the nick on time video :)
Hi Dr. Sorry, but I think you are calculating dz/dx, it means total derivative of z respect to x... partial derivative of z respect to x is 0 by deffinition, it doesn't mean that it does not exist z(x,y)... but its derivatives are total. I know you have notice that. You are a genius. Anybody makes mistakes, even if the results are correct. Go on with maths... the best distraction for a quarintine.
I saw that, but only F(x,y,z) has partial derivatives respect to x, y and z. Given that definition. At fact, also exist dy/dx=-Fx/Fy, if Fy is not 0... don't be upset about, I learned that way 23 years ago. If I am not correct, I would like to konw. Even so, I think you are a total genius and I am very happy to connect you. You still owe me a challenge about spherical coordinates.
@@jesusalej1 Implicit differentiation. You are specifically and intentionally holding y constant. In other words, you are calculating the other derivatives in that y-plane only, but ... you will notice that y is in the answer so you may select whatever y value or y plane you want to evaluate those derivatives anywhere in 3-space (except the origin). The same process is applied in the other two (orthogonal) planes x and z for those respective derivatives. Think of it as a directional derivative, where the value of the derivative depends on the direction in n-space (here 3) the derivative is to be taken; choose a direction orthogonal to a coordinate plane if you choose.
Anything with Chen lu is good.
Agreed hahaha
And produlu
@@Happy_Abe Haha
"The Nike method: You just do it", hahaha. I loved it. Nice video. Gracias!
Gracias por este tema. Como siempre, claro, sucinto, explícito, suficiente,...etc.
USE THE CHEN LU... WHENEVER I SEE YOUR CHANNEL I AM REMINDED OF "USE THE CHEN LU"
I love your energy
Why doesn't y depend on x and z does ?
Think about z as f(x,y) instead of an independent variable, the same way that y is actually f(x) on a regular xy plane. Z is a function of both x and y.
y is held fixed. The partials wrt y can also be calculated in the same way by holding one of the other variables constant. So for py/pz you hold x constant, and for py/px you hold z constant.
F(x,y,z)is the function of three variables, you can calculate partial derivatives. If for some point F(Xo,Yo,Zo )=0, for very small surroudings of this point variables are dependent, this function f always exists for examp variables z, x , z=f(x), (you can write this dependence ) then dz/dx=-Fx/Fz...... or for z=g(y) is dz/dy=-Fy/Fz.......or y=h(x) ...... x=p(y)........(ifFz0.....). I think, there is some similar idea.
A good tune-up lesson!
Can you do some videos on the fundamentals of Riemann integration? such as what it means for a function to be integrable, box integrals in R^n, and deducing whether some interesting functions are integrable over a box B, where the function is not continuous on B?
Also maybe some stokes' theorem and differential form?
(I found these topics very interesting but I wished I could understand them better)
I saw the proof of the Theorem in my calculus 2 classes. Sadly, I wasn't able to apply it in the exam, but I passed the course! Thank you Dr Peyam for explain thing so well! Your calculus 3 videos are saving me in this quarantine because vector calculus is confusing
Nice Job Dr. Peyam, I learned a lot!!
Thanks!!
it appears to be a first integral of a system of euler-lagrange equations
I saw the proof of that theorm before, it's super difficult.
Since the original function is symmetric wrt x,y,z
pXi/pXk = (2XjXk-Xi^2)/(Xk^2-2XiXj) where p is partial differentiation operator and i,j,k are subscripted indices that may be permutated.
Or in other words you have also solved for px/py, px/pz, py/px, py/pz, pz/py.
Having a bit of a brain fart, but why is y allowed to be treated like a constant but z isn’t? If we wanted to find del y/del x would z be treated as a constant?
I asked the same question
Precisely! For dz/dx treat y as a constant, think z = f(x,y)
Yes it's because were differentiating z with respect to x so the value of y is not changing. You can think of a partial derivative as analysing vertical 2D slice of a surface and finding the gradient.
holy shit, that was good. thank you very much, sir
This always confuses me :/ Isn't y (or shouldn't we consider it as) also an implicit function of x?
Here no, x and y are independent variables, and z depends on x and y. It’s not quite the same thing as in single variable
I love the second method. What's the name of the second method btw?
Believe it or not it’s the Chen lu 😂
@@drpeyam Lol, nice job incorporating the Chen Lu into almost everything you do!! Stay safe, and keep up the amazing work Dr. Peyam!!
Good Dr, I have a question for you from another Dr (Aero Engr; math minor, retired). Although I am retired I still like playing with math and therefore greatly appreciate your presentations.
So conceptually, I like to think of the process you described in method 1 as operating in an x-z plane while holding y constant, so therefore any delta-y (dy) is zero, you can then take directional derivatives (conceptually) constrained to such a plane.. Since you can choose any y-plane, you can span 3-space (except the origin in this case) and all is fine.
But I wanted to try another method, akin to a directional derivative in the direction of a specified line, where I hold, say y and z constant, and am then constrained to operate on a line of varying x; dy and dz are now zero on that line and I proceed:
F(x,y,z) = x^3 + y^3 + z^3 - 6xyz
pF/px = 3x^2 - 6yz where p is the partial operator.
Then holding x,y constant along a line of varying z (dx, dy = 0)
pF/pz = 3z^2 - 6xy
So I suppose I must choose an intersection of these two lines and say at that point:
(pF/px) / (pF/pz) = (pF/px) (pz/pF) = pz/px = (3x^2 - 6yz)/(3z^2 -6xy) = (x^2 - 2yz)/(z^2 -2xy) and this is the answer provided except for a minus sign !?? So where is the flaw and where is the missing sign change, given that the magnitude is correct?
I had this theorem proven on my calculus lecture just 4 weeks ago, AGAIN your video was there just a ONE day earlier. Are you collaborating with my calculus professor to achieve that ?? O.o
I really need to watch your videos as soon as they come out :D
I'm really really thankful! This was an awesome video, and tomorrow I have my Analysis 2 exam on implicit diff! so thanks a lot, a in-the nick on time video :)
assuming y is non dependent on x is silly... just go full generality
Dini's theorem
your class was A +
Hi Dr. Sorry, but I think you are calculating dz/dx, it means total derivative of z respect to x... partial derivative of z respect to x is 0 by deffinition, it doesn't mean that it does not exist z(x,y)... but its derivatives are total. I know you have notice that. You are a genius. Anybody makes mistakes, even if the results are correct. Go on with maths... the best distraction for a quarintine.
???
I’m calculating partial z / partial x not the total derivative
I saw that, but only F(x,y,z) has partial derivatives respect to x, y and z. Given that definition. At fact, also exist dy/dx=-Fx/Fy, if Fy is not 0... don't be upset about, I learned that way 23 years ago. If I am not correct, I would like to konw. Even so, I think you are a total genius and I am very happy to connect you. You still owe me a challenge about spherical coordinates.
Thanks!
@@jesusalej1 Implicit differentiation. You are specifically and intentionally holding y constant. In other words, you are calculating the other derivatives in that y-plane only, but ... you will notice that y is in the answer so you may select whatever y value or y plane you want to evaluate those derivatives anywhere in 3-space (except the origin). The same process is applied in the other two (orthogonal) planes x and z for those respective derivatives. Think of it as a directional derivative, where the value of the derivative depends on the direction in n-space (here 3) the derivative is to be taken; choose a direction orthogonal to a coordinate plane if you choose.
7:07 2yz-x^2
Grax bro
Thanks Dr Peyam! :3
Is there a way to solve for dz/dx just in terms of x and y?
Not really since you cannot solve for z in terms of x and y