How do you find the linearization at (-2,1) of $f(x,y) = x^2y^3 - 4sin(x+2y)$ ?

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Answer 1 · 2016-06-17T12:21:19

$8 (2 + x) - 4 ( y-1) +z-4=0$

Let us consider the surface

$f(x,y,z) = z-x^2y^3 + 4sin(x+2y)=0$

$f(x,y,z)$ is analytical, and at point $p_0={-2,1,4}$

the surface normal vector is given by

$grad f(x,y,z) = {f_x,f_y,f_z} = vec n$

or

$vec n={-2 x y^3 + 4 Cos(x + 2 y), -3 x^2 y^2 + 8 Cos(x + 2 y), 1}$

at point $p_0$ we have

$vec n_0 = {8,-4,1}$

The tangent plane at point $p_0$ is given by

$<< vec n_0,p-p_0 >> = 0$ with $p = {x,y,z}$

So

$<< vec n_0,p-p_0 >> =8 (2 + x) - 4 ( y-1) +z-4 = 0$

How do you find the linearization at (-2,1) of f(x,y) = x^2y^3 - 4sin(x+2y) f(x,y) = x^2y^3 - 4sin(x+2y) ?