If f is a vector-valued function defined by <sin2t, cos2t> then what is f '' (pi/4)?
I think it is <-4, -√2/2>.
2 answers
Sorry...it is cost, not cos2t.
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 )
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 )
1 answer