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
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
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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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
You beat calculus with these short videos and so we do! Thank you
This made me wonder how you would use Euler's method in a multivariable case
Dude that is a cursed idea but I am sure it must exist. Hope you get an answer
Thank you. I learned something new in math today.
Beautiful! Thanks!
would be nice a demonstration of the formula
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When you say 'Calc 1', is this in the US education system?
Yes
Calculus 1,2,3 ,etc.
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
The curve doesn't pass through (2,3). Therefore the question has no answer.
It's not a curve but the surface
And f(2,3) is obviously 1
It touches f(x,y) at (2,3)
The point on the surface is (2, 3, 1)
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