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

Question

Christopher Allers · Accepted Answer

#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#

Emily Ammerman · Answer

undefined To find the linearization of $ f(x, y) $ at the point $ (-2, 1) $, we first need to find the partial derivatives of $ f $ with respect to $ x $ and $ y $ at that point. Then, we use these derivatives to construct the linear approximation formula:

$$ L(x, y) = f(-2, 1) + f_x(-2, 1)(x + 2) + f_y(-2, 1)(y - 1) $$

Where:
- $ f_x(-2, 1) $ is the partial derivative of $ f $ with respect to $ x $ evaluated at $ (-2, 1) $.
- $ f_y(-2, 1) $ is the partial derivative of $ f $ with respect to $ y $ evaluated at $ (-2, 1) $.

Let's find these derivatives:

$$ f_x = 2xy^3 - 4\cos(x + 2y) $$
$$ f_y = 3x^2y^2 - 8\cos(x + 2y) $$

Now, we evaluate these derivatives at $ (-2, 1) $:

$$ f_x(-2, 1) = 2(-2)(1)^3 - 4\cos((-2) + 2(1)) $$
$$ f_y(-2, 1) = 3(-2)^2(1)^2 - 8\cos((-2) + 2(1)) $$

After calculating these, we substitute them into the linear approximation formula to find $ L(x, y) $.