Find the linearization of the function f(x,y)=√(129−3x^2−2y^2) at the point (5, -5).

Question

Find the linearization of the function f(x,y)=√(129−3x^2−2y^2) at the point (5, -5).

L(x,y)= ?

Use the linear approximation to estimate the value of f(4.9,−4.9) = ?

Answer 1

L ( x, y ) = ƒ ( a , b ) + ƒ_x( a, b) [ x - a ] + ƒ_y(a,b) [ y - b ]

( a, b ) = ( 5 , -5 )

ƒ_x(x,y) = -3x ( 129 -3 x² - 2 y² ) ^-1/2 ⇒ ƒ_x(5,-5) = -15 /2

ƒ_y(x,y) = -2x ( 129 -3 x² - 2 y² ) ^-1/2 ⇒ ƒ_y(5,-5) = 5

L ( x, y ) = ƒ ( a , b ) + ƒ_x( a, b) [ x - a ] + ƒ_y(a,b) [ y - b ]

L ( x, y ) = 2 - 15 /2 [ x -5 ] + 5 [y +5 ]

L ( x, y ) = -15 /2 [ x ] + 5 [y ] + 129 /2

Now using the linearization at ( 5, -5) we get the expected value of 2

Let us use it to approximate L (4.9, -4.9 ) = 3.25

when the actual value of the function f at (4.9, -4.9 ) is 2.9916

Answer 2

Hi,

As you know Lr0(x,y)=f(x0,y0)+∇f(x0,y0)⋅((x,y)−(x0,y0))

f(x,y)=√(129−3x^2−2y^2) at the point (5, -5).

f(5,-5) = √129 -3*25 -2*25 = 2

∇f(x,y)=(∂f/∂x, ∂f/∂y)-------> df/dx = (-6x / (2 √129 -3x^2-2y^2)) ----------> ( -30 / 2√4 ) = -15/2 = -7.5

df/dy = (-4y / (2 √129 -3x^2-2y^2)) ----------> ( 20 / 2√4 ) = 5

Lr0(x,y) = 2 + (-7.5 , 5 ) * ((x,y) - ( 5 , -5))

2 + (-7.5 , 5 ) * ( x-5 , y+5)

2 + -7.5( x-5) + 5 ( y+5)

2 -7.5x + 37.5 + 5y + 25

-7.5x + 5y + 64.5

I hope it is useful,

Minoo