differentiation puzzling me?

Question

i dont want the answer to this as much as i need a clear step by step method of how it is done, and why any important stages happen. my question is

y = (e^-t)sin 2t

show y is a maximum when t = 1/2 arctan 2

i know that dy/dt = 0 when y is a maximum
and i got dy/dt = (e^-t )(sin 2t + 2cos2t)

but dont know if that last bit is correct. assuming it is, i still don't know where to go from here?

first response - cheers you freak.
second response - i dont get how this helps...

aah cheers u have all helped out. thanks. apart from the first answer person.

Answer 1

y' = e^-t ( -sin2t + 2cos2t) = 0
-sin2t + 2cos2t = 0
tan2t = 2
t = 1/2 arctan 2
------
(e^-t)' = -e^-t. You missed the "-" sign.

Answer 2

dy/dt = e^-t *2cos 2t +sin 2t *-e^-t
dy/dt = e^-t(2cos 2t- sin 2t)
Now you need what value of t makes dy/dt =0, because that is where a max or min will occur.

So you can see that dy/dt = 0 if 2cos 2t = sin 2t
If you divide both sides by cos 2t , you get 2 = tan 2t

Now you can find t using the arctan function.

Answer 3

you missed out an important -ve sign:

dy/dt = (e^-t) (-sin 2t + 2 cos 2t) = 0 (for maxima)

e^-t cannot = 0 as t has to be infinite for that,

so, -sin 2t + 2 cos 2t = 0,

tan (2t) = 2

t = 1/2 arctan(2)

QED!!!