Use Taylor's formula to find the requested approximation of f(x, y) near the origin.
-Cubic approximation to f(x, y) = sin(6x + y)
A) $6 x + y - 36 x ^ { 3 } - 18 x ^ { 2 } y - 3 x y ^ { 2 } - \frac { 1 } { 6 } y ^ { 3 }$
B) $6 x + y - 36 x ^ { 3 } - \frac { 1 } { 6 } y ^ { 3 }$
C) $6 x + y - 72 x ^ { 3 } - \frac { 1 } { 3 } y ^ { 3 }$
D) $6 x + y - 72 x ^ { 3 } - 18 x ^ { 2 } y - 3 x y ^ { 2 } - \frac { 1 } { 3 } y ^ { 3 }$

Question

Use Taylor's formula to find the requested approximation of f(x, y) near the origin.
-Cubic approximation to f(x, y) = sin(6x + y) 
A) $6 x + y - 36 x ^ { 3 } - 18 x ^ { 2 } y - 3 x y ^ { 2 } - \frac { 1 } { 6 } y ^ { 3 }$
B) $6 x + y - 36 x ^ { 3 } - \frac { 1 } { 6 } y ^ { 3 }$
C) $6 x + y - 72 x ^ { 3 } - \frac { 1 } { 3 } y ^ { 3 }$
D) $6 x + y - 72 x ^ { 3 } - 18 x ^ { 2 } y - 3 x y ^ { 2 } - \frac { 1 } { 3 } y ^ { 3 }$

Quizplus · Accepted Answer

The cubic approximation of $f(x, y) = \sin(6x + y)$ using Taylor's formula involves calculating partial derivatives of $f$ up to the third order at the origin (0,0) and forming a polynomial. The first-order partial derivatives are $f_x = 6\cos(6x+y)$ and $f_y = \cos(6x+y)$, which at the origin give 6 and 1, respectively, contributing to the linear terms $6x + y$. Higher-order derivatives involve terms with $\cos(6x+y)$ and $\sin(6x+y)$, but evaluated at the origin, many simplify, leading to the coefficients in front of the cubic and mixed terms as shown in option A. This option correctly accounts for the contributions of all relevant derivatives up to the third order, including mixed terms like $x^2y$ and $xy^2$, which are essential for an accurate cubic approximation.

Use Taylor's Formula to Find the Requested Approximation of F(x 6x+y−36x3−18x2y−3xy2−16y36 x + y - 36 x ^ { 3 } - 18 x ^ { 2 } y - 3 x y ^ { 2 } - \frac { 1 } { 6 } y ^ { 3 }6x+y−36x3−18x2y−3xy2−61​y3

Use Taylor's Formula to Find the Requested Approximation of F(x $6 x + y - 36 x ^ { 3 } - 18 x ^ { 2 } y - 3 x y ^ { 2 } - \frac { 1 } { 6 } y ^ { 3 }$