Question

Find the equation of the tangent plane of z^2=x^2+y^2 at (1,1,1)

Answer 1

@phi @amistre64 @inkyvoyd

Answer 2

isn't the gradient of the function
f(x,y,z)= x^2+y^2-z^2 normal to the curve ?
\[ \nabla f \]= N \]
evaluated at 1,1,1
the equation of a plane is
\[ N \cdot p = b\]
where N is the normal and p is <x,y,z> and b is a scalar

Answer 3

Here is an example
http://www.ltcconline.net/greenl/courses/107/PartDeriv/tanplane.htm

Answer 4

I'm not sure? Is the formula:
f_x(x,y,z)(x-x0)+f_y(x,y,z)(y-y0)+f_z(x,y,z)(z-z0)=0?

Answer 5

I think you are thinking of the tangent vector as opposed to the tangent plane

Answer 6

It's the same thing. Once I get a normal to the plane, I can plug in a point that I know is on the plane to find b

gradient of f is (fx, fy, fz)
and if the known point is p= (x0,y0,z0)
N dot p = fx*x0 + fy*y0 + fz*z0 = b
so the equation would be
(fx, fy, fz) dot (x,y,z)= fx*x0 + fy*y0 + fz*z0 which can be written as
(x-x0)*fx + (y- y0)*fy + (z-z0)*fz =0

or just
\[ N \cdot p= b\]
where b is as above

Answer 7

Thanks!