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smallest value of speed
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v(t)= <3cos3t, -3sin3t, 4cos4t>
so |v(t)|= \[\sqrt{9\cos^2(3t)+9\sin^2(3t)+4\cos^2(4t)}\]
simplifing 9cos^2(3t) + 9sin^2(3t), what do you get?
uh.. nope
i am not sure
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lol i mean 9
good, so |v'(t)| = sqrt(9 + 16cos^2(3t)) yes? You forgot to square the 4
well, cos(0) = 1
Actually 5 is the maximum -.-
3 is indeed the minimum, but it's not at t = 0
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hmm but ans=3
yup
so t=pi/2
yes, i corrected my mistake. 3 is the minimum but not at t = 0
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