Posted by Shreya on Wednesday, January 23, 2013 at 10:53pm.
Last post for tonight, going to bed
diameter is 30 m, so a=15
So, (recall the last question with the 18 and 19)
the min value is going to be -15 , but we want it to be 1.5 above ground , (the x-axis)
so we have to add 16.5
sofar we have
y = 15 sin k(t + d) + 16.5 , t in minutes
it takes 3.5 minutes for one rotation
period = 3.5
but 2π/k = 3.5
3.5k = 2π
k = 2π/3.5 or 4π/7
getting there ....
y = 15 sin (4π/7)(t + d) + 16.5
when t= 0 , we have the lowest point ---> (0,1.5)
1.5 = 15 sin (4π/7)(0+d) + 16.5
-1 = sin ((4π/7)(d) )
I know sin 3π/2 = -1
so (4π/7)d = (3π/2)
d = 21/8
ok, how about
y = 15 sin (4π/7)(t + 21/8) + 16.5
testing:
if t=3.5, we should get 31.5
y = 15sin(4π/7)(3.5 - 21/8) + 16.5
= 15 sin (π/2) + 16.5 = 15(1) + 16.5 = 31.5 , Wow!!
-------------------------------
when t=25 seconds = 25/60 min or 5/12 min
y = 15sin (4π/7)(5/12 + 21/8) + 16.5
= 15 sin( 73π/42) + 16.5
= 5.5 m high
----------------------
Here comes the hard part:
we want y to be above 27
sketch a line y = 27 to cut our cosine curve, you will see two such places in each period
Let's find those two values of t
27 = 15sin(4π/7)(t+21/8) + 16.5
sin(4π/7)(t+21/8) = .7
set your calculator to radians and do inverse sin (.7)
I get .7754
but by the Cast rule, we know the sine is positive in quadrants I and II
so (4π/7)(t+21/8) = .7754 OR (4π/7)(t+21/8) = π - .7754
t+21/8 = .43193 OR t+21/8 = 2.366
t = -2.193 or t = -.2588
don't worry about the t values being negative, the show we are on the left part of the cosine curve. We could add 3.5 to both to make them positive
So the time interval = -.2588 - (-2.193) = 1.934 minutes or 116 seconds
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