How do you find the equations for the tangent plane to the surface #xy^2+3x-z^2=4# through #(2, 1, -2)#?

1 Answer
Dec 21, 2016

#x+y+z=1#

Explanation:

1) rearrange eqn to #" "f(x,y,z)=0#

so:#" "f(x,y,z)=xy^2+3x-z^2-4=0#

2) find normal by calculating#" "gradf(x,y,z), " "at" " (2,1,-2)#

#gradf(x,y,z)=(del/(delx)(xy^2+3x-z^2-4))hati+(del/(dely)(xy^2+3x-z^2-4))hatj+(del/(delz)(xy^2+3x-z^2-4))hatk#

remember when partially differentiating: differentiate with respect to the variable in question, treating the other variables as constant.

#gradf(x,y,z)=(y^2+3)hati+(2xy)hatj-2zhatk#

now evaluate this at #" "(2,1,-2)#

#gradf(2,1,-2)=(1^2+3)hati+(2xx2xx1)hatj-(2 xx-2)hatk#

#gradf(2,1,-2)=4hati+4hatj+4hatk=((4),(4),(4))#

3) the eqn of a plane in vector form is

#vecr * vecn=((x_1),(y_1),(z_1)) * vecn#

where# vecn=((4),(4),(4))# in this case and #"vecr=((x),(y),(z))#

so:

#((x),(y),(z)) * ((4),(4),(4))=((2),(1),(-2)) * ((4),(4),(4))#

# :. 4x+4y+4z=8+4-8#
# :. 4x+4y+4z=4#
# :. x+y+z=1#

We can confirm this result graphically:

enter image source here #