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 ?
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
ok~ get it clearly..thanks^^
just recall that the period of tan(t) is pi, not 2pi, so
tan(2t) has period pi/2
To find the value of t when y'(t) = 0, we need to solve the equation -2.5e^(-t)(2sin(2t) + cost(2t)) = 0.
Since we know that 2.5e^-t is never zero, we can set the expression inside the parentheses equal to zero:
2sin(2t) + cost(2t) = 0
To solve this trigonometric equation, we can use the trigonometric identity for sin(a + b):
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
By comparing this identity with our equation, we can see that we have sin(a)cos(b) + cos(a)sin(b), which means the equation can be rewritten as:
sin(2t + π/4) = 0
Now we have sin(x) = 0, which means x must be an integer multiple of π:
2t + π/4 = nπ
where n is an integer. Now we can solve for t:
2t = nπ - π/4
t = (nπ - π/4)/2
So the value of t when y'(t) = 0 is t = (nπ - π/4)/2, where n is an integer.