Finding Normal Vector and Tangent Plane at a Given Point

[charmedbeauty]

help me find this normal vector!

Homework Statement

Find a normal vector n and the equation of the tangent plane to the surface S at the point x₀

S: z=ln(x²+3y²)

x₀=(2,-1,ln7)

Homework Equations

The Attempt at a Solution

ok so I have this general formula from the textbook.

z= z₀+F_x(x₀,y₀)(x-x₀)+F_y(x₀,y₀)(y-y₀)

and n=(F_x(x₀,y₀) , F_y(X₀,Y₀) , -1) respectively.

so using these

S: z= ln(x²+3y²)

let u=x²+3y²

z=ln(u)

dz/du=(1/u) (u')

F_x(x,y) = d/dx(u)/u = 2x/(x²+3y²) = 4

F_y = d/dy (u)/u = 6y/ (x²+3y²) = -6

z= ln7 + 4(x-2)-6(y+1)

z= ln7 +4x -6y -14

ln7 +4x -6y -14 -z = 0

so n =(4,-6,-1)^T

but they have 4x-6y-7z-14+7ln7=0

and n= (4, -6, -7)^T

?? I am not sure what I have done wrong but it looks very close to what they have.

any help is appreciated.

 

[uart]

F_x(x,y) = d/dx(u)/u = 2x/(x²+3y²) = 4

F_y = d/dy (u)/u = 6y/ (x²+3y²) = -6

Check your numerical results of the substitution of x0,y0 here. :)

 

[charmedbeauty]

Check your numerical results of the substitution of x0,y0 here. :)

ohh no (-1)²≠-1.

Thanks a bunch uart!

 

[charmedbeauty]

Do you think you could give me a hint in the right direction for this one?

z²+x²+y²=1

x₀=(1/3 , 1/2, √23/ 6)

If I do it the usual way and I solve for z, I get a negative sqrt term so I don't think that is the right approach.

there is some inequalities I can see but I don't think that helps
I think its a sphere but that doesn't really help me either.

please help. Thanks.

 

[HallsofIvy]

If $z^2+ x^2+ y^2= 1$ then $z= \pm\sqrt{1- x^2- y^2}$. That is not a "negative squareroot" it simply means that $x^2+ y^2$ cannot be larger than 1. And that must be true geometrically because this is the surface of a sphere of radius 1.

But it is not necessary to solve for z. If a surface is given by F(x, y, z)= constant, then the gradient, $\nabla F$ is a normal vector at each point on the surface. Here $F(x, y, z)= x^2+ y^2+ z^2$ so that $\nabla F= 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}$. At the given point, that is $(2/3)\vec{i}+ \vec{j}+ \sqrt{34}/3\vec{j}$ so the tangent plane at that point is $(2/3)(x- 1/3)+ (y- 1/2)+ \sqrt{34}{3}(z- \sqrt{34}/6)= 0$.

 

[charmedbeauty]

If $z^2+ x^2+ y^2= 1$ then $z= \pm\sqrt{1- x^2- y^2}$. That is not a "negative squareroot" it simply means that $x^2+ y^2$ cannot be larger than 1. And that must be true geometrically because this is the surface of a sphere of radius 1.

But it is not necessary to solve for z. If a surface is given by F(x, y, z)= constant, then the gradient, $\nabla F$ is a normal vector at each point on the surface. Here $F(x, y, z)= x^2+ y^2+ z^2$ so that $\nabla F= 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}$. At the given point, that is $(2/3)\vec{i}+ \vec{j}+ \sqrt{34}/3\vec{j}$ so the tangent plane at that point is $(2/3)(x- 1/3)+ (y- 1/2)+ \sqrt{34}{3}(z- \sqrt{34}/6)= 0$.

ok I think that should be a √23 /6 not √34 /6

but that is a really convenient way of doing that. thanks a lot.

but for some reason in their answer they have multiplied it by three i.e.


2x +3y+√23z -6 = 0

does this not change the normal vector in a geometric way?

since they have n=(2,3,√23)^T

and not n=(2/3,1,√23 /3)^T

??

Thanks.