Find the slope of the tangent to the curve of intersection of the surface 2z = sqrt(9x^2 + 9y^2 − 36) and the plane y = 1 at the point (2; 1; 3/2).

Question

Find the slope of the tangent to the curve of intersection of the surface 2z = sqrt(9x^2 + 9y^2 − 36) and the plane y = 1 at the point (2; 1; 3/2).

Please help.

Answer 1

First thing I would do is find your partial derivatives:

Fz = 2

Fx = 1/2 * (9x^2 + 9y^2 -36)^(-1/2)) * 18x

Fy = 1/2 * (9x^2 + 9y^2 -36)^(-1/2)) * 18y

Evaluate at your point:

Fz = 2

Fx = 1/2 * 1/sqrt(9) * 36 = 6

Fy = 1/2 * 1/sqrt(9) * 18 = 3

Equation of your tangent line:

Fx (x - 2) + Fy (y -1) + Fz (z - 3/2) = 0

6(x - 2) + 3 (y - 1) + 2 (z - 3/2) = 0

6x - 12 + 3y - 3 + 2z - 3 = 0

6x + 3y + 2z = 18