Hello here is my implicit function question

Question

F(x, y, z) = cos(xz)−sin(y+ z) + 1 = 0 and G(x, y, z) = x + y + z − 1 − π/2 = 0

Find the tangent line to the curve that is the intersection of the surfaces F(x, y, z) = 0 and G(x, y, z) = 0, where F and G are as in (a), at the point P = (1, 0, π/2).

Thanks a lot

Answer 1

Second equation is plane passing point (1,0,π/2), so tangent plane is the same.

Now tangent plane to second surface: F'_x = -zsin(xz), F'(1,0,π/2) = - π/2sin(π/2) = - π/2; F'_y = - cos(y + z),

F'_y(1,0,π/2) = - cos(0 + π/2) = 0; F'_z = -xsin(xz) - cos(y + z), F'(1, 0, π/2) = -1.

Equation of tangent plane: -π/2(x - 1) + 0(y - 0) -1(z - π/2) = 0 or - π/2x - z + π = 0, or πx + 2z - 2π = 0;

This 2 planes in intersection give tangent line for intersection line of this 2 surfaces.

x + y + z − 1 − π/2 = 0 and πx + 2z - 2π = 0.

This line passing point (1,0,π/2) and direction vector (m, n, p) which is perpendicular to the normal vectors to plane (1, 1, 1) and (-π/2,0,-1). So m + n + p = 0 and -π/2m - p = 0 , p = - π/2m, n = - m + π/2m = m(π/2 - 1).

Then equation of tangent line: (x - 1)/m = (y - 0)/m(π/2 - 1) = (z - π/2)/(-π/2m) = t

And wwe get after reducing by m parametric equation of line:

x = t + 1, y = (π/2 - 1)t, z = - π/2t + π/2, - ∞ < t < ∞