Answer to Question #224057 in Calculus for Unknown346307

Question

(a) Find the linearization of f(x, y) = e^(x) cos y at the point (0, π/2)

Answer 1

Solution;

The linearization of a function f(x,y) at a point (x₀,y₀) is given by;

$L(x,y)=f(x_0,y_0)+( x-x_0)f_x(x_0,y_0)+(y-y_0)f_y(x_0,y_0)$

Given;

$f(x,y)=e^xcos y$ and $(x_0,y_0)=(0,\fracπ2)$

$f(0,\fracπ2)=e^0cos(\fracπ2)=0$

We can the derive;

$f_x=e^xcos y$

$f_y=-e^xsiny$

From which ;

$f_x(0,\fracπ2)=e^0cos(\fracπ2)=0$

$f_y(0,\fracπ2)=-e^0sin(\fracπ2)=-1$

Hence by substitution;

$L(x,y)=0+0(x-0)+(y-\fracπ2)(-1)=\fracπ2-y$

$L(x,y)=\fracπ2-y$