- #1
charmedbeauty
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help me find this normal vector!
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)
ok so I have this general formula from the textbook.
z= z0+Fx(x0,y0)(x-x0)+Fy(x0,y0)(y-y0)
and n=(Fx(x0,y0) , Fy(X0,Y0) , -1) respectively.
so using these
S: z= ln(x2+3y2)
let u=x2+3y2
z=ln(u)
dz/du=(1/u) (u')
Fx(x,y) = d/dx(u)/u = 2x/(x2+3y2) = 4
Fy = 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.
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+Fx(x0,y0)(x-x0)+Fy(x0,y0)(y-y0)
and n=(Fx(x0,y0) , Fy(X0,Y0) , -1) respectively.
so using these
S: z= ln(x2+3y2)
let u=x2+3y2
z=ln(u)
dz/du=(1/u) (u')
Fx(x,y) = d/dx(u)/u = 2x/(x2+3y2) = 4
Fy = 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.