math problem?
the question is:
a water wheel rotates at a rate of 6 revolutions per minute. You start your stopwatch. After 2 seconds a pointt on the rim of the water wheel is at its lowest point 1 foot below the water s surface. Assume that the point s distance in feet from the water s surface varies sinisodally with time?
I m having trouble find my 3 main points, the two mins and one max
also, what would the equation of the word problem look like?
what would be the first point in time that the pointt is on the water s surface???!!!
Whats the initial height???
PLEASE HELP IM DESPERATE FOR HELP!!!! I ve been trying to solve this problem for the past 2 hours ;(
* I m a junior in high school
Comments
Are you told the diameter of the water wheel? Without it, we can find everything except the maximum height above the water. I will finish everything with a variable of M as the maximum distance above the water.
6 revolutions / minute
= 6 revolutions / 60 seconds
= 1 revolution / 10 seconds.
10 seconds is therefore the period of the sinusoid.
If (2,-1) is the first minimum, then one period later at (12, -1) will be the second minimum.
The maximum will occur after half a period － 5 seconds － after the minimum, Therefore, the max is (7,M). (Maybe that's why you've been spending 2 hours on this question, not knowing the maximum)
We need the max for the amplitude and vertical shift, so for now it will be in terms of the mystery M:
A = (Max - Min) / 2 = ½(M+1)
D = ((Max + Min) / 2 = ½(M-1)
We can find B and the horizontal shift.
B = 2π/period = 2π / 10 = π/5
Since the minimum occurs after 2 seconds, you can shift -cosx 2 to the right. Or you can use the fact that the maximum occurs after 7 seconds, making it a cosx shifted 7 to the right. Here are both:
s = -½(M+1) ∙ cos [π/5 (t-2)] + ½(M-1)
or
s = ½(M+1) ∙ cos [π/5 (t-7)] + ½(M-1)
The initial height is at t = 0.
To find the first time the point was at the water's surface, find the first positive t-intercept of the graph.