Calculus

So,y(t) = 2.5e^-t cos2t
I need to find the derivative, which is
y'(t)= -2.5e^-t(2sin 2t +cost 2t) .
And now I need to find t when y'(t) = 0
(I know 2.5 e^-t is never zero.)
I need to use trigonometry identity to find it ?

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  1. so you are solving

    2sin 2t + cos 2t = 0
    2sin 2t = -cos 2t
    2sin 2t/cos 2t = -1
    tan 2t = -1/2

    set your calculator to RAD
    and find 2nd function tan
    (tan^-1 (+1/2)
    to get .46365
    we know that the tangent is negative in II and IV
    so 2t = π - .46365 = 2.6779
    t = 1.339

    or

    2t = 2π - .46365 = 5.8195
    t = 2.9098

    We also know that the period of tan 2t = 2π/2 = π
    so by adding or subtracting multiples of π to any existing answer will yield as many answers as you want.
    e.g
    t = 1.339 + π = appr 4.4806
    t = 2.9098 + 5π = 18.6177

    checking with that last value
    2sin(2(18.6177)) + cos(2(18.6177))
    = very very close to zero

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  2. ok~ get it clearly..thanks^^

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  3. just recall that the period of tan(t) is pi, not 2pi, so

    tan(2t) has period pi/2

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