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Find the equation of the tangent plane?
For the surface with parametric equations r(s,t)=<st,s+t,s-t>, find the equation of the tangent plane at (2,3,1).
Also, find the surface area under the restriction s^2+t^2<=1
I thought I should take the derivative with respect to s and t and cross them to find a normal, but I still have everything in terms of s and t. I'm a little lost...
1 Answer
- kbLv 79 years agoFavorite Answer
1) n = r_s x r_t = <-2, t+s, t-s>
At (x,y,z) = (2,3,1) <==> (s,t) = (2,1), we have n = <-2, 3, -1>.
So, the equation of the tangent plane is
-2(x - 2) + 3(y - 3) - 1(z - 1) = 0.
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2) A = ∫∫ ||r_s x r_t|| dA
.......= ∫∫ √[4 + (t + s)^2 + (t - s)^2] dA
.......= ∫∫ √[4 + 2(s^2 + t^2)] dA, where the region of integration is s^2 + t^2 ≤ 1.
Converting to polar yields
A = ∫(θ = 0 to 2π) ∫(r = 0 to 1) √(4 + 2r^2) * r dr dθ
...= 2π ∫(r = 0 to 1) √2 * r√(2 + r^2) dr
...= π√2 * (2/3)(2 + r^2)^(3/2) {for r = 0 to 1}
...= π√2 * (2/3)[3^(3/2) - 2^(3/2)]
...= π√2 * (2/3)(3√3 - 2√2)
...= (2π/3)(3√6 - 4).
I hope this helps!
