Find the equation of the tangent plane of of the following surface patch at the indicated point: $$ σ(r, θ) = (r \cosh θ, r \sinh θ, r^2), (1, 0, 1).$$

I know that the tangent space of a surface patch is the vector space of the partial derivatives of the patch. So i took $$ dσ/dθ=(r \sinh θ,r \cosh θ,0) $$ and $$ dσ/dr=(\cosh θ,\sinh θ,2r). $$ Now in order to be at the point $(1,0,1)$, $r$ must be $r=1$ and $θ=0$. So the first partial der. is $(0,1,0)$ and th second $(1,0,2)$ now I can get the normal vector from the cross product and I know how to find the plane with a normal vector and a point on the plane. But what I get is a line, what am I doing wrong?

  • $\begingroup$ Your mistake is probably on the part you did not expose: calculating the equation of the orthogonal plane of the normal vector. $\endgroup$ – anderstood Jan 6 '16 at 20:27

You cross product is $N=(2,0,-1)$ and your point is $p=(1,0,1)$. The plane is defined by $$N\cdot (x-p)=0$$ where $x=(x,y,z)$. Expanding gives $$2(x-1)-(z-1)=0\\ 2x-z=1$$ which is the equation of your plane. This is not a line because you are working in $\mathbb{R}^3$, so $y$ can have any value.

  • $\begingroup$ i found the same eq but i thought it was a line because y wasnt there. So my other calculations are correct? $\endgroup$ – Manolis Lyviakis Jan 6 '16 at 20:41
  • $\begingroup$ Yes it is correct. $\endgroup$ – cheesyfluff Jan 6 '16 at 20:42
  • $\begingroup$ missing $y$ in the equation tells you that your plane in parallel to $y$ axis $\endgroup$ – user26977 Jan 7 '16 at 10:02

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