How to find the plane to the surface with parametric equations below?

Question

Find the tangent plane to the surface with parametric equations x(u,v)=u²+2u, y(u,v)=3v-2u, z(u,v)=6v-10 at point (8,14,2).

Answer 1

Hi,

x(u,v)=u²+2u, y(u,v)=3v-2u, z(u,v)=6v-10 at point (8,14,2).

First: ∂x/∂u = 2u + 2 ∂y/∂u = -2 ∂z/∂u = 0

∂x/∂v = 0 ∂y/∂v = 3 ∂z/∂v = 6

r_u = ∂x/∂u i + ∂y/∂u j + ∂z/∂u k = (2u + 2) i -2 j + 0

r_{v =} ∂x/∂v i + ∂y/∂v j + ∂z/∂v k = 0 i + 3j + 6 k

r_u . r_{v =} i j k

2u +2 -2 0

0 3 6

after you find normal vector you will have : ( -12) i - ( 12u + 12) j + (6u + 6) k

x = u²+2u = 8 -----> u = -4, 2

y = 3v-2u = 14 ------> when u = -4 v = 2 and u = 2 v= 6

( -12) i - ( 12u + 12) j + (6u + 6) k -----> when u = 2 you have -12i - ( 12*2 + 12 ) j + ( 6*2 + 6 ) k = -12i -36j + 18k

Now we know (r-a) . n = 0-------> point is (8,14,2) ( x-8) ( -12) + ( y-14) (-36) + (z-2)(18) = 0 2x + 6y -3z =94

one more time you need to solve when u = -4

I hope it is useful,

Minoo