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The paddle wheel of a boat measures 16 feet in diameter and is revolving at a rate of 20 rpm. The maximum depth of the paddle w...Asked by Jeremae
The paddle wheel of a boat measures 16 feet in diameter and is revolving at a rate of 20 rpm. The maximum depth of the paddle wheel under water is 1 foot. Suppose a point is located at the lowest point of the wheel at t=0.
1) Write a cosine function with phase shift 0 for the height above water at the initial point after t seconds.
Amplitude =16/2=8
Vertical translation= +7
Period = 3 sec
Equation (using radian measure)
H =-8 cos (2pi/3)t +7
or
H=7 -8 cos (2 pi t/3)
2) use your function to find the height of the initial point after 5.5 seconds
And then 3 feet
This question confuses me.
1) Write a cosine function with phase shift 0 for the height above water at the initial point after t seconds.
Amplitude =16/2=8
Vertical translation= +7
Period = 3 sec
Equation (using radian measure)
H =-8 cos (2pi/3)t +7
or
H=7 -8 cos (2 pi t/3)
2) use your function to find the height of the initial point after 5.5 seconds
And then 3 feet
This question confuses me.
Answers
Answered by
Dan
H=7 -8 cos (2 pi t/3) seems correct, I didn't check your math though.
If H is the height of the initial point after t seconds, simply set t = 5.5 seconds and solve for H.
H = 7 - 8 cos(2pi*5.5/3)
If H is the height of the initial point after t seconds, simply set t = 5.5 seconds and solve for H.
H = 7 - 8 cos(2pi*5.5/3)
Answered by
Reiny
Your amplitude, vertical translation and period are correct.
so lets look at the general shape of the curve so far.
we have y = 8cos (2pi/3)t
the minimum point of that is (1.5,-8)
but we want that minimum to occur at (0,-1)
so we have to translate 1.5 to the left and up 7
final equation
y = 8 cos (2pi/3)(t + 1.5) + 7
check for critical points
(0,-1) yes
(.75,7) yes
(1.5,15) yes
(2.25,7) yes
(3,-1) yes
so lets look at the general shape of the curve so far.
we have y = 8cos (2pi/3)t
the minimum point of that is (1.5,-8)
but we want that minimum to occur at (0,-1)
so we have to translate 1.5 to the left and up 7
final equation
y = 8 cos (2pi/3)(t + 1.5) + 7
check for critical points
(0,-1) yes
(.75,7) yes
(1.5,15) yes
(2.25,7) yes
(3,-1) yes
Answered by
Reiny
to find height after 5.5 sec
y = 8cos(2pi/3)(5.5+1.5) = 7
= 3 (I used my calculator)
y = 8cos(2pi/3)(5.5+1.5) = 7
= 3 (I used my calculator)
Answered by
Reiny
for a height of 3 feet?
3 = 8 cos (2pi/3)(t + 1.5) + 7
-4 = 8 cos (2pi/3)(t + 1.5)
-.5 = cos (2pi/3)(t + 1.5)
we know cos 2.09495 = -.5 and
cos 4.18879 = -.5
2.094595 = 2pi/3(t+1.5)
1 = t+1.5
t = -.5
or
4.18879 = 2pi/3(t+1.5)
t = .5
but the period of our curve is 3 sec
so adding multiples of 3 sec to our answers of t= -.5 and t= .5 will give us a height of 3 feet.
check t = .5 + 15 [15 is a multiple of 3]
t = 15.5
y = 8cos(2pi/3)(15.5+1.5) + 7
= 3
3 = 8 cos (2pi/3)(t + 1.5) + 7
-4 = 8 cos (2pi/3)(t + 1.5)
-.5 = cos (2pi/3)(t + 1.5)
we know cos 2.09495 = -.5 and
cos 4.18879 = -.5
2.094595 = 2pi/3(t+1.5)
1 = t+1.5
t = -.5
or
4.18879 = 2pi/3(t+1.5)
t = .5
but the period of our curve is 3 sec
so adding multiples of 3 sec to our answers of t= -.5 and t= .5 will give us a height of 3 feet.
check t = .5 + 15 [15 is a multiple of 3]
t = 15.5
y = 8cos(2pi/3)(15.5+1.5) + 7
= 3
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