Find the equation of the tangent line to the intersection curve of the two following surfaces at the point $(3,1,2)$:
$$x=2+\cos(\pi yz)$$
$$y=1+\sin(\pi xz)$$
The only way that I know to find the intersection curve is by plugging one of the surfaces into the other one:
$$x=2+\cos(\pi (1+\sin(\pi xz))z)$$
However I really doubt if this is how I should solve the problem.
Best Answer
We find normal vectors to the tangent planes of both surfaces at $(3, 1, 2)$.
$S1: f(x, y, z) = x - \cos(\pi yz) - 2 = 0$
$S2: g(x, y, z) = y - \sin(\pi xz) - 1 = 0$
$f'(x, y, z) = (1, \pi z \sin (\pi y z), \pi y \sin (\pi y z))$
$g'(x, y, z) = (- \pi z \cos (\pi x z), 1, -\pi x \cos (\pi x z))$
$f'(3, 1, 2) = (1, 0, 0)$
$g'(x, y, z) = (- 2 \pi, 1, - 3 \pi)$
The tangent line to the intersection curve of both surfaces at point $(3, 1, 2)$ should be on tangent planes of both surfaces. In other words, it is the intersection line of both planes.
To find direction vector of the intersection line, we take cross product of both normal vectors and that is $(0, 3 \pi, 1)$.
So the equation of tangent line is $ \ (3, 1, 2) + (0, 3 \pi, 1) t $.