Respuesta :
Answer:
See below in bold.
Step-by-step explanation:
f(x, y, z) = x^2 + y^2 + z^2 at (6, 6, 7)
Taking partial derivatives
f'x = 2x + y^2 + z^2 --> f'x(6,6,7) = 12+36+49= 97
f'y = x^2 + 2y + z^2---> f'y(6,6,7) = 36 + 12 + 49 = 97
f'z = x^2 + y^2 + 2z ----> f'z(6,6,7) = 36 + 36 + 14 = 86.
f(6,6,7) = 36 + 36 + 49 = 121
So our linear approximation is:
L(x,y,z) = 121 + 97(x - 6) + 97(y - 6) + 86(z - 7)
Substituting the given values
f(6.012, 5.972,6.982)
= L(6.012, 5.972, 6.982) ≈ 121 + 97(6.012 - 6) + 97(5.972 - 6) + 86(6.982 - 7)
= 117.9.
Linear approximation are used to get the approximated values of expressions.
The linear approximation of [tex]\mathbf{f'(6.012, 5.972, 6.982) }[/tex] is 117.90000
The given parameter is;
[tex]\mathbf{f(x, y, z) = x^2 + y^2 + z^2\ at\ (6, 6, 7)}[/tex]
Calculate [tex]\mathbf{f(6, 6, 7)}[/tex]
[tex]\mathbf{f(6, 6, 7) = 6^2 + 6^2 + 7^2}[/tex]
[tex]\mathbf{f(6, 6, 7) = 121}[/tex]
Differentiate (partial derivatives)
[tex]\mathbf{f_x = 2x + y^2 + z^2}[/tex]
[tex]\mathbf{f_y = x^2 + 2y + z^2}[/tex]
[tex]\mathbf{f_z = x^2 + y^2 + 2z}[/tex]
Substitute values for x, y and z in the above partial derivatives
[tex]\mathbf{f_x = 2 \times 6 + 6^2 + 7^2 = 97}[/tex]
[tex]\mathbf{f_y = 6^2 + 2 \times 6 + 7^2 = 97}[/tex]
[tex]\mathbf{f_z = 6^2 + 6^2 + 2 \times 7 = 86}[/tex]
The linear approximation is calculated as:
[tex]\mathbf{f'(x,y,z) = f(x,y,z) + f_x(x - 6) + f_y(y - 6) + f_z(z - 7)}[/tex]
This gives
[tex]\mathbf{f'(x,y,z) = 121 + 97(x - 6) + 97(y - 6) + 86(z - 7)}[/tex]
Substituting values for x, y and z
[tex]\mathbf{f'(6.012, 5.972, 6.982) \approx 121 + 97 \times (6.012 - 6) + 97\times (5.972 - 6) + 86\times (6.982 - 7)}[/tex]
[tex]\mathbf{f'(6.012, 5.972, 6.982) \approx 121 + 97 \times .012 + 97 \times -0.028 + 86\times -0.018}[/tex]
[tex]\mathbf{f'(6.012, 5.972, 6.982) \approx 121 + 1.164 -2.716 -1.548 }[/tex]
[tex]\mathbf{f'(6.012, 5.972, 6.982) \approx 117.90000}[/tex]
Hence, the linear approximation of [tex]\mathbf{f'(6.012, 5.972, 6.982) }[/tex] is 117.90000
Read more about linear approximations at:
https://brainly.com/question/6768038
