Question

Find an equation for the tangent plane to the surface given by
Υ(u; v) = (u^2- v^2; u + v; u^2 + 3v)
at the point (3; 1; 1).

Answer 1

looks like you might need to use the chain rule ...

Answer 2

F = (x,y,z)

x = u^2 + v^2
x' = 2u u' + 2v v'

y = u+v
y' = u' + v'

z = u^2 + 3v
z' = 2u u' + 3v'

Answer 3

solve the system ...

u^2- v^2 = 3

u + v = 1

u^2 + 3v = 1

maybe ....

Answer 4

\[\frac{df}{du}=\frac{df}{dx}\frac{dx}{du}+\frac{df}{dy}\frac{dy}{du}+\frac{df}{dz}\frac{dz}{du}\]
\[\frac{df}{du}=\nabla f\cdot <x'(u),y'(u)>,z'(u)\]

Answer 5

...latex fail

Answer 6

regardless

x = u^2 + v^2
x'(u) = 2u
x'(v) = 2v

y = u+v
y'(u) = 1
y'(v) = 1

z = u^2 + 3v
z'(u) = 2u
z'(v) = 3

gradF = <1,1,1>

Fu = 4u+1
Fv = 2v+4