Question

Find an equation for the plane tangent to the given surface at the specified point. (just a moment)

Answer 1

(Thats impossible because their is no number length surface to complete it!)

Answer 2

\[x = u^2, y=usin({e^v}), z = \frac{1}{3}u\cos({e^v})\] at the point \[(13,-2,1)\]

Answer 3

The easiest way to do this is to find the normal vector to the surface at that point. That normal vector is also normal to the plane which is tangent to the surface at that point.

To find the normal vector, we take the cross product of two tangent vectors at the point.

And two tangent vectors at that point p = (13,-2,1) can be found by partially differentiating the position function
r(u,v) = (x,y,z)(u,v)
with respect to u and v.

Answer 4

r(u,v) = (x,y,z)(u,v) = ( x(u,v), y(u,v), z(u,v) )

Answer 5

I got it. I got afraid because \[v=\ln(\arcsin(\frac{2}{\sqrt{13}}))\]
and normal vector turned out to be very ugly :S.

Thank you!