Linearization of f(x,y) = sqrt(29 - 4x^2 - 4y^2) at (2,1)

[KTiaam]

Homework Statement

Find the linearization of the function ƒ(x,y) = sqrt(29 - 4x² - 4y²) at the point (2,1)

Homework Equations

[/B]
Point (a,b)
L(x,y) = Linearization
L(x,y) = ƒ(a,b) + ƒx(a,b)(x-a) + ƒy(a,b)(y-b)

The Attempt at a Solution

[/B]
ƒ(2,1) = 3

ƒx(x,y) = [1/2⋅-8x]/[sqrt(29-4x²-4y₂]
ƒx(2,1) = -8/3

ƒy(x,y) = [1/2⋅-8y]/[sqrt(29-4x²-4y₂]
ƒy(2,1) = -4/3

L(x,y) = 3 + [-8/3(x-2)] + [-4/3(y-1)]
= 3 + (-8/3x + 16/3) + (-4/3y+4/3)
= 3 -8/3x +16/3 - 4/3y + 4/3
Simplifying gives me
= -8x - 4y + 29

Which is wrong according to my Webwork

please any help would be amazing.

 

[BvU]

3 -8/3x +16/3 - 4/3y + 4/3 gives (-8x - 4y + 29)/3 You can't just strike the /3 ...

 

[KTiaam]

i multiplied everything by 3, can i not do that?

 

[BvU]

Nope: f = x + 2 is something quite different from f = 3x + 6 !

Lean back a little, relax, take a breath or a break and it'll be obvious...

:) If you don't believe me you can always evaluate f(2.1,1) and f(2,1.1) on a calculator or a spreadsheet...

 

[KTiaam]

Thank you BvU appreciate the help, more questions may be on the way, its been a rough day.