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anonymous
 4 years ago
let r(t)=(sint, cost,t) find k(t)
anonymous
 4 years ago
let r(t)=(sint, cost,t) find k(t)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0k(t)=r'(t)xr''(t) / r'(t)^3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0r'(t)= (cost,sint,1) r''(t)=(sint,cost,0) r'(t)= square root of 2

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.0r'(t)r''(t)=(sint cost, sint cost, 0) r'(t)r''(t)=root(sin^2t cos^2t+sin^2t cos^2+0) r'(t)r''(t)=root(2 sint^2t cos^2t)=sint.cost root(2) r't^3=2root (2) k(t)=(sint*cos t)/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0mm,,, in the cross product i got (cost, sint,1)

mathmate
 4 years ago
Best ResponseYou've already chosen the best response.1r(t)=(sint, cost,t) find k(t) r'(t)=<cos(t),sin(t),1> r"(t)=<sin(t),cos(t),0> r'(t)xr"(t) = <cos(t),sin(t),1> r'(t)xr"(t) = sqrt(2) r'(t) = sqrt(2) r'(t)xr"(t) / r'(t)^3 = sqrt(2)/sqrt(2)^3 = 1/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea thats what i have
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