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Find the equation of the tangent plane of each of the following surface patches at the indicated points:

$$σ(r,θ)=(r\cosh(θ),r\sinh(θ),r^2), (1, 0, 1).$$

I'm not sure what to do, any hints are greatly appreciated.

What I have:

$σ_r=(\cosh\theta,\sinh\theta,2r)$

$σ_\theta=(r\sinh\theta,r\cosh\theta,0)$

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1 Answer 1

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Compute $\sigma_r \times \sigma_\theta = (-2r\cosh\theta, 2r^2 \sinh \theta, r)$. Since $\sigma(1,0) = (1,0,1)$, the vector $\sigma_r \times \sigma_\theta(1,0) = (-2,0,1)$ is normal to the tangent plane at $(1,0,1)$. The equation of the plane is $(-2,0,1) \cdot (x - 1, y - 0, z - 1) = 0$, or $2x - z = 1$.

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  • $\begingroup$ I know $\sigma_r $ and $\sigma_\theta$ forms bases for tangent plane.But here $\sigma_r =(1,0,2)$ and $\sigma_\theta=(0,1,0)$ at (1,0) whose linear combination $(1,1,2)$ does not belong to tangent plane? $\endgroup$
    – math student
    Apr 1, 2020 at 16:42
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    $\begingroup$ @mathstudent the vectors $\sigma_r(1,0)$ and $\sigma_\theta(1,0)$ are vectors relative to the point $(1,0,1)$, not the origin. So the linear combination you have is a vector, not a point, that lies on the tangent plane. The plane passes through the point $(2,1,3)$, which is in the direction $\langle 1,1,2\rangle$ relative to the point $(1,0,1)$. $\endgroup$
    – kobe
    Apr 1, 2020 at 18:59

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