Answer:
f(3.022 , 5.992 , 5.972)=80.7
Step-by-step explanation:
Linearization of Multivariable Functions
Let f be a function that depends on the independent variables (x,y,z) and assume the following partial derivatives exist:
[tex]\displaystyle \frac{\partial f}{\partial x}\ ,\ \frac{\partial f}{\partial y}\ ,\ \frac{\partial f}{\partial z}[/tex]
The function f can be linearized around a known point (xo,yo,zo) by the equation:
[tex]\displaystyle f(x,y,z)\approx f(x_o,y_o,z_o)+\frac{\partial f(x_o,y_o,z_o)}{\partial x} (x-x_o)+\frac{\partial f(x_o,y_o,z_o)}{\partial y} (y-y_o)+\frac{\partial f(x_o,y_o,z_o)}{\partial z} (z-z_o)[/tex]
Given
[tex]f(x, y, z) = x^2 + y^2 + z^2[/tex]
Evaluating in (3,6,6)
[tex]f(x, y, z) = 3^2 + 6^2 + 6^2=81[/tex]
The partial derivatives are
[tex]\displaystyle \frac{\partial f}{\partial x}=2x[/tex]
Evaluating at (3,6,6)
[tex]\displaystyle \frac{\partial f}{\partial x}=2(3)=6[/tex]
[tex]\displaystyle \frac{\partial f}{\partial y}=2y[/tex]
Evaluating at (3,6,6)
[tex]\displaystyle \frac{\partial f}{\partial y}=2(6)=12[/tex]
[tex]\displaystyle \frac{\partial f}{\partial z}=2z[/tex]
Evaluating at (3,6,6)
[tex]\displaystyle \frac{\partial f}{\partial z}=2(6)=12[/tex]
The linearization of f is
[tex]\displaystyle f(x,y,z)\approx 81+6 (x-3)+12 (y-6)+12 (z-6)[/tex]
Operating
[tex]\displaystyle f(x,y,z)\approx 81+6 x-18+12y-72+12z-72[/tex]
[tex]\displaystyle f(x,y,z)\approx 6 x+12y+12z-81[/tex]
Using the linearization to find f(3.022 , 5.992 , 5.972)
[tex]\displaystyle f(3.022 , 5.992 , 5.972)\approx 6 (3.022)+12(5.992)+12(5.972)-81=80.70000[/tex]
