Calculus 3: How to linearize a multivariable function

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bprp calculus basics

bprp calculus basics

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@ianfowler9340
@ianfowler9340 4 ай бұрын
An alternative: { pf/px is a stand-in for the partial derivative} 1) When x = 2 and y = 3 then z = 1 so P(2,3,1) lies on the surface. 2) Express the surface as: f(x,y,z) = 0 ==========> 1 + xln(yx-5) - z = 0 3) The gradient vector: N--> = < pf/px , pf/py , pf/pz > evaluated at (2,3,1) will be normal to the curve at (2 ,3,1) and hence be a normal direction for the tangent plane. 4) I'll leave out the details of partial derivatives as you have so kindly done most of them. Evaluated at (2,3,1) we get N--> = 5) The equation of the tangent plane is now : Ax + By + Cz = Ax0 + By0 + Cz0 { or in your form A(x - x0) + B(y - y0)+ +C(z - z0) = 0 } 6x + 4y - 1z = 6(2) + 4(3) -1(1) ====> 6x + 4y - z = 23 ====> z = 6x - 4y - 23 As an added note: The expression: A(x - x0) + B(y - y0)+ +C(z - z0) = 0 is just the dot product of the 2 vectors: N--> = < A , B , C > and PP_0 --> = < x - x0 , y - y0 , z - z0 > being FORCED to 0 in order to force the 2 vectors to be perpendicular as the vector < x - x0 , y - y0 , z - z0 > lies entirely on the plane. This was my method when I taught this in high school. I'm curious, is the gradient method ever taught? Seems pretty clean to me. At any rate, Cheers! and Well Done - Ian
@imperialisticvonhabsburg3149
@imperialisticvonhabsburg3149 4 ай бұрын
I love this channel
@amirhosseinrostamii
@amirhosseinrostamii 4 ай бұрын
Best of the bests of mathematics❤
@matematikgokseldir
@matematikgokseldir 4 ай бұрын
You beat calculus with these short videos and so we do! Thank you
@Asiago9
@Asiago9 4 ай бұрын
This made me wonder how you would use Euler's method in a multivariable case
@liamsegers832
@liamsegers832 4 ай бұрын
Dude that is a cursed idea but I am sure it must exist. Hope you get an answer
@JP-lz3vk
@JP-lz3vk 4 ай бұрын
Thank you. I learned something new in math today.
@dimBulb5
@dimBulb5 4 ай бұрын
Beautiful! Thanks!
@BRaleatoriedades
@BRaleatoriedades 4 ай бұрын
would be nice a demonstration of the formula
@iankr
@iankr 4 ай бұрын
When you say 'Calc 1', is this in the US education system?
@IoT_
@IoT_ 4 ай бұрын
Yes
@IoT_
@IoT_ 4 ай бұрын
Calculus 1,2,3 ,etc.
@evefroggo4755
@evefroggo4755 4 ай бұрын
In the US, Calc 1 is differential calculus with an introduction to integral calculus. Calc 2 is integral calculus along with analysis of infinite series. Calc 3 is Multivariable and vector calculus
@phill3986
@phill3986 4 ай бұрын
👍🎉👍
@kevinmadden1645
@kevinmadden1645 4 ай бұрын
The curve doesn't pass through (2,3). Therefore the question has no answer.
@IoT_
@IoT_ 4 ай бұрын
It's not a curve but the surface
@IoT_
@IoT_ 4 ай бұрын
And f(2,3) is obviously 1
@yuichiro12
@yuichiro12 4 ай бұрын
It touches f(x,y) at (2,3)
@bprpcalculusbasics
@bprpcalculusbasics 4 ай бұрын
The point on the surface is (2, 3, 1)
@Ninja20704
@Ninja20704 4 ай бұрын
Thats not how it works. The function has two variables x and y for its input. That means the input space is the x-y plane and the output will lie on a third axis (the z axis). So this is a 3d space we are talking about (2,3) is the point on the x-y plane that we are inputting into the function, and the output is z=1. So the function passes through the point (2,3,1) in the 3d space, which is the point we are interested in
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