Let $S$ be the piece of the cylinder $x^2 + z^2 = 1$ which is to the right of the $xz$–plane and to the left of the plane $y = 1 + x$.
Find the equation of the tangent plane to $S$ at the point $( \frac{1}{\sqrt{2}}, 1, \frac{1}{\sqrt{2}})$
Currently doing practice questions from old midterms and I dont know how to solve this problem. I understand how one would go about finding the equation of the tangent plane, but how do make the parametric equation given that you know the cylinder is bounded by $y = 1 + x$ and $y = 0$
This is how I learned it: by the equation of a plane (from the dot product of the normal vector $n=<A,B,C>$ and a point on the plane $(x_0,y_0,z_0)$),
$A(x-x_0)+B(y-y_0)+C(z-z_0)=0$
Rearranging this equation to obtain a function in terms of $z$ and point slope form
$z-z_0=-A/C(x-x_0)-B/C(y-y_0)$
where $z=\sqrt{1-x^2}$ (since the given $z$ is positive), $z_x=-A/C$, and $z_y=-B/C$ (by point slope form):
$z-z_0=z_x(x-x_0)-z_y(y-y_0)$
And plug in the given points yields:
$z-1/\sqrt{2}=-(x-1/\sqrt{2})$
Here's a visual:
https://www.math3d.org/tqxOnGGhK