How do you find the equations for the tangent plane to the surface #h(x,y)=cosy# through #(5, pi/4, sqrt2/2)#?

1 Answer
Dec 22, 2016

#qquad pi: qquad y + sqrt 2 z = pi/( 4) + 1#

Explanation:

for #h(x,y) = z = cos y#, if written as level surface #psi = z - cos y = 0#, then the gradient of #psi# is the normal vector to that surface:

#nabla psi = ((0),(sin y),(1)) = vec n#

The equation of a plane is

#(vec r - \vec r_o) cdot vec n = 0# where #vec r = < x,y,z>, vec r_o = < x_o,y_o,z_o> #

#implies \vec r cdot vec n = const#

So the tangent plane at any point (x,y,z) is:

#pi: (sin y) y + z = const #

we can determine the value of the constant as

#const = sin (pi/4) cdot pi/4 + (sqrt 2)/2 = 1/(sqrt 2) (pi/4 + 1)#

giving plane #qquad pi: qquad y + sqrt 2 z = pi/( 4) + 1#

#h(x,y)# is a very simple surface, it is just #z=cos y# extruded along the x axis so you can apply a reality check here quite easily.