surface area?

Question

Find an equation of the tangent plane to the parametric surface
x=1r*cos(theta), y =-5r*sin(theta), z=r at the point (sqrt{2} , -5sqrt{2}, 2) when r = 2, theta = pi/4.

Answer 1

f(r,θ) = (r cosθ, -5r sinθ, r)
df/dr = (cosθ, -5 sinθ, 1)
df/dθ = (-r sinθ, -5r cosθ, 0)
Now df/dr and df/dθ are two vectors parallel to the surface.
So df/dr × df/dθ (cross product) will be a vector perpendicular to the surface.
At the point given:
df/dr = (1/√2,-5/√2,1)
df/dθ = (-√2,-5√2,0)
So
df/dr × df/dθ = (5√2,-√2,-10)
So the equation of the plane can be writen as:
5√2x - √2y - 10z = c for some constant c.
We can determine the constant easily as the points above must satisify the plane. So
5√2*√2 - √2(-5√2) - 10*2 = c
10 +10 - 20 = c
c = 0
So the equation of the plane is:
5√2x - √2y - 10z = 0

Answer 2

See if you can obtain your answer using the sketch of the solution to this very similar question:
http://answers.yahoo.com/question/index;_ylt=AnO5sIVjaOX.k2CjV.y6qLojzKIX;_ylv=3?qid=20071128233932AATRPvD