Find an equation of the tangent plane?

Question: Find an equation of the tangent plane to the parametric surface x=1rcosθ, y=4rsinθ, z=r at the point (sqrt(2),4sqrt(2),2) when r=2, θ=pi/4.

My attempt:

Let m = <rcosθ, 4rsinθ, r>, then
m_r = <cosθ, 4sinθ, 1>
m_θ = <-rsinθ, 4rcosθ, 0>

Normal = (m_r x m_θ) = [4sinθ*0-4rcosθ*1]i - [cosθ*0 - (-rsinθ*1)]j + [4rcosθ*cosθ - 4sinθ*-rsinθ]k = <4rcosθ, -rsinθ, 4r>
N at the point (sqrt(2),4sqrt(2),2) => <-4sqrt(2), sqrt(2), 8>

Now, <(x-sqrt(2), y-4sqrt(2), z-2> dot <-4sqrt(2), sqrt(2), 8> = 0
-4sqrt(2)((x-sqrt(2)) + (sqrt(2)(y-4sqrt(2))) + 8(z-2) = 0
z = (4sqrt(2)x - sqrt(2)y + 16) / 8

This is the wrong answer, could anyone point out where I went wrong or a better way to do this? Thanks!

Quote:

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Originally Posted by MrCryptoPrime

Question: Find an equation of the tangent plane to the parametric surface x=1rcosθ, y=4rsinθ, z=r at the point (sqrt(2),4sqrt(2),2) when r=2, θ=pi/4.

My attempt:

Let m = <rcosθ, 4rsinθ, r>, then
m_r = <cosθ, 4sinθ, 1>
m_θ = <-rsinθ, 4rcosθ, 0>

Normal = (m_r x m_θ) = [4sinθ*0-4rcosθ*1]i - [cosθ*0 - (-rsinθ*1)]j + [4rcosθ*cosθ - 4sinθ*-rsinθ]k = <-4rcosθ, -rsinθ, 4r> <--- here is your mistake
N at the point (sqrt(2),4sqrt(2),2) => <-4sqrt(2), sqrt(2), 8>

Now, <(x-sqrt(2), y-4sqrt(2), z-2> dot <-4sqrt(2), sqrt(2), 8> = 0
-4sqrt(2)((x-sqrt(2)) + (sqrt(2)(y-4sqrt(2))) + 8(z-2) = 0
z = (4sqrt(2)x - sqrt(2)y + 16) / 8

This is the wrong answer, could anyone point out where I went wrong or a better way to do this? Thanks!

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All your considerations are OK. You "only" made a tiny (but fatal) sign mistake.

According to my results the tangent plane has the equation:

$\displaystyle z = \frac{x \cdot \sqrt{2}}{2} - \frac{y \cdot \sqrt{2}}{8}$

My assignment was actually due at midnight last night, but thank you for the explanation of where I went wrong and for the awesome picture (what program do you use to generate that?). Vielen Dank! :)