Find an equation of the tangent plane to the parametric surface x = 4 rcosθ, y = -5 rsinθ...?

Question

Find an equation of the tangent plane to the parametric surface
x = 4rcosθ, y = -5rsinθ, z = r, at the point (4sqrt(2), -5sqrt(2), 2 when r = 2, θ = pi/2.

z = __________

Note: Your answer should be an expression of x and y; e.g. "3x - 4y"

Stuck... hope you can show me the answer and the steps how you got there.

Thanks

Answer 1

The point in question has θ = π/4, not π/2.

It can be shown that both

(∂x/∂r) i + (∂y/∂r) j +(∂z/∂r) k and (∂x/∂θ) i + (∂y/∂θ) j +(∂z/∂θ) k

evaluated at some point on the surface are tangent to the surface at that point. If they are nonzero and not parallel, they define the tangent plane at that point. So the cross product of those two vectors produces a vector normal to the tangent plane. You will find that this vector, evaluated at (4√2, -5√2, 2) is

5√2 i - 4√2 j - 40 k

If (x, y ,z) is any point in the plane distinct from (4√2, -5√2, 2),
then the vector

(x - 4√2) i + (y - (-5√2)) j + (z - 2) k

lies in the tangent plane and hence is perpendicular to 5√2 i - 4√2 j - 40 k, so the dot product of this vector with 5√2 i - 4√2 j - 40 k is zero:

[(x - 4√2) i + (y + 5√2) j + (z - 2) k] • [5√2 i - 4√2 j - 40 k] = 0

(To get the desired coefficient of 1 for z right away, you could also use
-√2/8 i + √2/10 j + k for the normal vector, since any nonzero multiple of the given normal vector is also normal to the plane.)

That will give you an equation for the tangent plane. If I did my arithmetic correctly, you should get

z = (√2 / 8) x - (√2 / 10) y