# Calculus

If f is a vector-valued function defined by <sin2t, cos2t> then what is f '' (pi/4)?

I think it is <-4, -√2/2>.

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1. Sorry...it is cost, not cos2t.

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2. x= sin 2 t
y = cos t
x' = 2 cos 2t
y' = -sin t
x" = - 4 sin 2t
y" = -cos t
at pi/4
sin pi/2 = 1
cos pi/4 = sqrt 2/2
so yes , (-4 , -sqrt 2 / 2 )

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