Thread: curvature
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teng125
teng125 is offline
#1
Jul18-06, 06:35 AM
P: 426
Find the curvature at the point P:

f : [0;2] IR^2 , f (t) = 2t,4 −2t^3 , P(2,2)

i subs 2x=0 then x=0 and 2y=2 then y=1

t=(0,1)

then i perform the curvature calculation.however,i'm confuse that in this case i have to subs t=1 to get a value instead of zero.
If the value t=(2 ,3) or others, which one should i choose??
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