Find an equation of the tangent plane to the parametric surface?

Question

Find an equation of the tangent plane to the parametric surface x = −3r cos θ, y = 4r sin θ, z = r
at the point (-3sqrt(2), 4sqrt(2), 2) when r=2, θ=pi/4

z=

Answer 1

If we move from R = (x,y,z) to R + dR on this surface, we move tangent along the surface.

In the theta direction we have

dR1 = (3rsinθ|x> + 4rcosθ|y>)dθ

In the radial we have

dR2 = (-3cosθ|x> + 4sinθ|y> + |z>)dr

The normal to this surface is then

dR1xdR2 = (4rcosθ,-3rsinθ,12r)

Then a vector from the given point, (-3sqrt(2), 4sqrt(2), 2) to another point on the plane, R-R0, is orthogonal to the normal, or:

(x+3sqrt(2), y - 4sqrt(2), z-2)dot(n) = 0 or

(x+3sqrt(2))(4rcosθ) + (y - 4sqrt(2))(-3rsinθ) + (z-2)12r = 0

plugging in the values for theta and r at the given point R0 then will give you the equation of the tangent plane in x,y,z.