Find the equation of the tangent plane to the surface 4xe^yz+xz^2−7x3^y= 5 at (1,0,1).

Answer 1

The surface 4 x e ^{y z} + x z² - 7 x³ y = 5 is a level surface of the function

F ( x, y , z ) = 4 x e ^{y z} + x z² - 7 x³ y .

Therefore the normal vector of the tangent plane at the point Ω ( 1, 0, 1 )

is a scalar multiple of the gradient vector of F at the point Ω.

∇ F = [ ∂F / ∂x ] i + [ ∂F / ∂y ] j + [ ∂F / ∂z ] k

∇ F = [ 4 e ^{y z} + z² - 21 x² y] i + [ 4 x z e ^{y z} - 7 x³ ] j +[ 4 x y e ^{y z} + 2 x z ] k

∇ F( 1,0,1) = 5 i -3 j + 2 k

Hence the equation of the tangent plane is

5 ( x -1 ) -3 y + 2 ( z - 1 ) = 0