How to find the TANGENT PLANE | Linear approximation of multi-variable functions
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- čas přidán 18. 11. 2019
- How do you find the equation of a tangent plane to the graph of a function f(x,y)? This is the multi-variable analog of finding the equation of a tangent line to the single variable function f(x). At each point (if f is differentiable) the tangent plane is a linear approximation. To compute this we use the vector equation of a plane with respect to the normal vector. The components of the normal vector come from the partial derivatives so if you can find partial derivatives you can write down the equation of the tangent plane.
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If you change "Y-naught" to "why not?" this lecture becomes a lot more philosophical.
😂
Brilliant 😂
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Sir, please Never stop uploading videos.
Some of you, are a lamp of hope for university mathematics students like us.
If you leave, the lamp of our hope will be gone.
im physics major and its my first year i was lost in multivariable calc i used to watch khan academy and 3blue1brown they're perfect but i can't really understand the main topic when i found Dr Trefor Bazett you're one of the best math teachers i've ever seen please keep going you're the one that showed me the visuals of multivariable calc
this is literally one of the best mathematics channels i have seen, really helpful for a college student like me...
i have already given a like to all of his videos
I really appreciate that!
Really you're explaining beautifully
Dr Trefor .This multi-variable calculus course is very exciting; I press the like button every time.
Amounts to a shortcut, recognizing that the a and the b you have from dotting the normal vector with any vector in the plane, are equivalent to the partial derivatives evaluated at the chosen point. I actually checked your answer by doing the problem the long way, by using the partial derivatives to find other points in the plane, that I then made vectors with, did cross product to get a normal vector, , then dotted with a random vector in the plane, orthogonal to the normal vector (by definition), which made the dot product zero, and gave me the equation for the plane, x+y+z=5/2, which is also what you got in a more efficient manner!
The only thing that tripped me up a bit was a, b and c, with a and b not changing when c does. When I show this to my son, I will start with directional components p, q and r, which will become a, b and -1.
This sort of brings up the problem with math education, which is as interesting to me, if not more, than the math itself. Let’s hope that the smartest people are the teachers, but that’s also the problem. You are at one level, and pretty much everybody you’re teaching way below. So clarity is everything. It pays to connect all the dots. No imprecision, ideally.
I have to say, I found the ‘goal’ (1:18) confusing. The cool thing is just realizing that a and b are the partial derivatives evaluated at the point of tangency, which avoids quite a bit of work!
You motivate these ideas so cleanly and precisely! You are saving me in Calculus 3 thank you so much.
I thought it might be a value to those who are working on multivariable calculus to explain how and why I was confused about the gradient. I have rewatched videos and worked through various textbooks and finally come to the following understanding. The gradient, of a 3D surface that is plottable as z=f(x,y) in 3D, is the normal vector of a level curve. If more than one level curve is created and projected onto the xy plane they become a contour map. However the gradient of a 4D surface, which is geiven by w=f(x,y,z) and cannot be plotted in our 3D world, is the normal to a level surface in 3D and this then enables us to find the equation of a plane in 3D at a particular point on the level surface. That was my whole hang up at this point in my multi variable calculus studies and I hope this is useful to someone else.
This video literally saves my life. Thank you so much!
Great, thank you. you make my calculus studies much more interesting!
These videos are very helpful to not just understand but visualize things
I have a question. I still don't understand why we should use c= -1 instead of 1 if n is a normal vector.
Beautiful. Not just a like, I am gonna share it with as many people as possible..
thank youuu i was so confused where the equation came from, this really cleared it up for me :))
Great video!
It turns out to be the first order multivariable Taylor Polynomial as an approximation of a multivariable function.
You are a beautiful human. XD I My brain with multiple oil leaks appreciates the complete explanations from start to finish.
Explained Amazingly well!!!
great video Trefor, ur channel is super helpful, do you plan on making an introductory series on category theory?
@@DrTrefor perfect, ill gladly watch your algebraic topology videos as well!
@Dr. Trefor Bazett - Fantastic videos that I use in my lectures and encourage my Calc III students to watch! I've referred my math colleagues to your videos as well. At 8:04, I believe that you meant to write f(x, y) = 2 - x^2 - y^2, not f(x). Keep doing what you do!
Thank you for the kind words and for correcting the typo. I hope your students find them useful!
@@DrTrefor sir why you assumed C=-1 in linear approximation i didnt understant
Thank you for the good explanation. Could you please make a video about how to find the equation for a 3D graph?
Thanks a lot
Thanks a lot! Could you also explain matrix algebra? I find it sometimes hard to solve it
This is a great video series to get a little bit more intuition:
czcams.com/play/PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab.html
A very clear explanation.
You would have to make more videos.
Your work is for the peace of human beings, really!!!
for sure interesting video ever. I like it👍👍👍
Sir I can assure u no one in India can teach how tangent plane here was shown , sir really I love u and I follow only u for calculus , u are really great, i want to meet u sir , u can gimme some autograph of u
I had a question! I understand the mathematical ease that comes from setting C = -1, but why are we allowed to do that? if the other two vector components a and b are set to be partial derivatives, doesn't setting C = -1 destroy some information that we would have otherwise? If C is always negative 1, doesn't that force where the tangent plane is? I'm not sure I fully understand.
Thank you for the videos, I've been binging them on double speed to help me with my thermo-fluid dynamics and heat transfer courses and it's been great!
C is a component of a normal vector, setting C=-1 only constrains the magnitude of the normal vector and its direction is not changed. As long as is normal to , the plane is still specified by the equation.
@@trumanyao5637 thank you i was confused about this as well!
I think it in a slightly different way with the explanation above. We purposely make c = -1, so that we can obtain the equation a(x-xo) + zo = z at y=yo. Only with this equation we can say that a = fx, b=fy. The big deal is we wanna obtain a = fx, b = fy.
If we dont let c=-1, a might be some multiple of fx.
It is good. I love It. Keep it up
Flawless explanation👍, lovely derivation aswell
Thanks a lot 😊
Hi! Thanks! And could you make a video about total differentiability, with the definition(the limit that has to equal zero for a function to be differentiable)? Thanks!
I have such a video in the multivariable calculus playlist, check it out from my homepage
Man, you have such digestible videos, and you don't do any hand waving. When restricted to this online format at least, I think you're better than my professor. Now I feel dumb for not understanding linear approximations and what they even were. z=Just m(dx)+b+n(dy), adding up directional derivatives by how much they change buy to the initial height will obviously give you the height.
Keep it up!!
@Trefor Bazett Another question from me, superb set of videos again! Is it obvious that there always exists a plane tangent to any surface? Or is this somehow provable, or as a result of the definition of differentiability?
@@DrTrefor Thanks for that. I'm trying to figure out precisely how does the existence of a tangent plane follow from the definition and I can't seem to see it as obvious from the Increment theorem. For example I could stand at a point on a surface which is a function of 2 variables. If I stood there, I could look in any direction and specify the gradient of the slope in that direction and I can probably define it in such a way that the plane is tangent in some directions but not others, or am I completely mistaken?
Really helpful sir. your way of teaching is amazing.
Glad to hear that!
Thank you sir
I'm gonna share with everybuddy but these lectures is soooooooooo helpful!!!!!!
Thank you!
Thank you🙏
which recording app do you use?
You're amazing sir
Sir, please provide an explanation to Monge cones.
thanku so much😱😱😱😱😱 the explanation is on another level🔥🔥
keep it up pls!
Thank you, will do!
Would it be possible for you to retroactively develop a numbering system in the titles of your videos as I am not sure if youtube search brings your videos up in logical order? Great videos by the way. Tony Aimer
Check out the playlists on my homepage, they have everything in order and if you start there it will sequentially play them all:)
@@DrTrefor what is your homepage called?
I am from India i am very impressed by your way of teaching i have also subscribed your channels 👍👍
Thanks and welcome!
wondering how the idea of tangent plane can be used in defining differentiability
I got a question:
Prove that no two tangent planes on C are parallel. [Hint: two planes are parallel iff their normal vectors are parallel.]
If I prove any two tangent planes are equal is that enough to answer the question?
f(x, y) = x^2 + y^2
C is the contour curve of f at level 2
Previous question wanted the equation of of the tangent plane at a point (1, 1, 2) which I got as: 2x + 2y -z = 2
If I prove any two tangent planes are not parallel of f** ..sorry for the typo
At the beginning, a and b are the normal vector to the tangent plane, but when they become fx and fy, fx and fy are parallel to the tangent plane. Why? Hope someone helps clarify thanks!
Beautiful...
thanks
Hi Sir not sure if we can request solutions but I have this problem that I am stumped with. Would appreciate any help:
Find the tangent plane and normal line to the surface (x)^2+(y)^2+(z)^2=1 at the point (1/4, 1/4, 1/sqrt2)
Love the vids all the way from UK!
What is the best way for me to support you? I am to. This material is great.
Thank you, so glad it is helping! I love when people like and share with their classmates - and I DO have channel memberships if you really want to go that route:)
Dr. Bazett, there are tricky equations with tricky points that don't have a normal vector, even if they seem smooth and well behaved. For instance, consider z=(x^3-3xy^2)/(x^2+y^2). It is easy to see that z=0 at (0, 0) , since every limit converges to that value. It is also differentiable at that point. But it has three-fold symmetry around that point, so no plane fits as tangent. How should that point be treated? In particular, normal vectors define in optics how light reflects off a mirror. If a mirror was made with that shape, how would light be reflected at that point?
z is undefined at (0,0). lim (x,y) -> (0,0) f(x,y) = 0 does not mean f(0,0) = 0.
@@Andrea-fp2kj it is continuous and perfectly defined everywhere around 0, just not at exactly 0. There it is actually undefined, since the denominator equals to 0 at x=y=0. You could, however calculate the exact same things for the function z except the denominator now equals to x^2 +y^2 +k, where k is some real number. Now by taking smaller and smaller numbers for k, you could get closer and closer to the actual solution of how the mirror would reflect light at (0,0). My guess is that the calculations will go crazy and you get weird stuff out of it haha
How can we derive in the other form?
I love you man😘😘
Great explanation! Any reason why we are only talking about functions R^2 to R?
Only because I can plot stuff in 3 dimensions (2+1), but not higher. We absolutely can make all the same arguments for higher order functions.
@@DrTrefor Makes sense. I think it would be good - after explaining the diagram with R^3 - to describe a higher-order generalization and intuition as well. Great video!
Beautifully clear and engaging explanations. Like that you slip into the correct pronunciation of "z" as "zed". :-) *ducks*
Ha the problem of being a Canadian who spent three years teaching Americans:D
hmm never thought about it that way
I'm binge watching your videos😀😀
A healthy habit ha!
@@DrTrefor indeed!!!
Nice video, but why could you set C=-1 in the first place?
sir I am confused about c and I just think an explanation but cannot be sure. If you read and answer whether it is true I would be so appreciated. here is my explanation:
when we write a tangent plane for any function our first comment is c can be anything. ıf it is zero plane lives in xy plane and if not we can just manage it to -1 because it is useful for linearization. also, if our function F(x,y,z) is not differentiable at that point but its partials are, we can still use linearization by dividing our equation and arranging z to -1 (it does not matter coefficients of partials fx and fy)
ın other case, if our function F(x,y,z) is already differentiable at some point we cannot say c can be anything when we write the equation tangent plane because c cannot take any value other than -1 and the reason of this: we know that when F is differentiable then F(x,y,z)=0 defines z as a function of x and y. Also dz/dx=-Fx/Fy (its differentiability). so we can write F(x,y,z)=f(x,y)-z=0 and now Fz (our c) must be -1
Yohhhhh bless your ❤️
Nice
C can't be 0 because it is the component of the normal vector in the z direction, thus if it were 0, then the normal vector would be parallel to the xy plane, therefore it could only be normal to a vertical tangent plane. A vertical tangent plane has an undefined slope, therefore if the function is differentiable, we can't have any vertical tangent planes and thus the normal to the tangent plane can't be horizontal meaning the C component can't be zero.
The distinction here is between the graph of a function z=f(x,y) and a generic surface. For instance, the unit sphere can have vertical tangent planes and that is totally fine, but the unit sphere can't be written as the graph of a single function.
Clear: Occam’s Razor!
@8:24 shouldnt the f(x) be f(x,y)?
Sumptuous
I can sell the remote fragrance of mathematical analysis in this series.
I liked the video and I love your channel...
But I am still confused why we can just assume c=-1. Why is that? I didn't understand where that came from... :((
Correct me if I am wrong. I think this is just a way to transfrom the equation of a plane so the inputs are x and y but the output is z, so it has the format z=f(x,y). As to wether this assumption is mathematically correct or not, remember that a, b, and c are the components of the normal vector u, and setting c to -1 is just ultmately changing the length of u, because a and b will change accordingly to still make the plane represented by the equation tangent to the given point.
@@jamesjin1668 Hi, I don't understand this concept either. I don't understand how you know that c = -1 and why not c= 5 for example. Thank you!
Also, one question hehe. I have a surface like a sphere, which is not a function. And if I define a function in a higher dimension and I set for example F(x,y,z) = x^2 +y^2 +z^2. This will be a function right? Then, can I always convert any surface that is not a function into the level "curve" or level "surface"... of an actual function that lives in a higher dimension?Will this always happen?
rescued my exam
Btw I'm just hear to know how to find the "TANGENT LINE" of a sine function , could someone help me with this please
Just read a few comments, and no one seems to have mentioned it, so I will. I think your notation at the end of the video is slightly off. You have f(x) = 2 - x^2 - y^2... that should be f(x,y) = 2 - x^2 - y^2. Otherwise, nice job.
Good catch, quite right!
Good, good....
What will happen if c=0 ,is it even possible ?
What if c is zero? is the tangent plane just x-y plane?
yup!
when I see your beard I know that the topic is Done 😄
Thanks
😁
@UClUg89tl1SDRb2q1RZmP70g
@UClUg89tl1SDRb2q1RZmP70g
@
I meant for this comment to be realted to this video but youtube had advanced the video onto the next one without my noticing so I have done a double comment
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
There is grad f and grad F where F(x,y,z)=f(x,y)-z and the latter gradient is normal to a surface
@@DrTrefor So both are correct. Oh dear I need to go back and watch your series on multivariable calculus from the beginning again - I maust have missed something important!
Sorry but why C can't be zero ? is it because if c=0 then the plane is the same as (xy) ?
Yes of course
Why is equal to why not.
I have been a TA for all the calculus courses offered at my university. I have utilized your videos so often that I have resorted to liking them before I even watch them. If the students only knew that their TA was secretly plagiarizing the critical insight given by Dr. Trefor Bazett. Well, I'd be out of a job.
I'm kidding, of course. I refer students to your videos all the time.
My friend wants to know the secret about your beard.
you need a mic!
No hate, I love all your other videos but this was really fuzzy. You can be more concise than this.