Question 868347: During an experiment, the intensity i(t) of the electric current of a device as a function of time t elapsed since the beginning of the experiment is given by:
i(t)=6sin((pi*t/12)+(2*pi/3))+6
The device emits a sound signal each time the current’s intensity is equal to 9.
The experiment lasts 120 seconds.
How many sound signals does the device emit during the experiment?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! The device emits a signal every time the intensity, i(t), is equal to 9. So we need to find out how many times it equals 9. This is within the interval t = 0 to t = 120 since the experiment lasts 120 seconds.
So that means t is in the interval [0,120]. This is important to selecting the right solutions.
Plug in i(t) = 9 and solve for t
i(t)=6sin((pi*t/12)+(2*pi/3))+6
9=6sin((pi*t/12)+(2*pi/3))+6
9-6=6sin((pi*t/12)+(2*pi/3))
3=6sin((pi*t/12)+(2*pi/3))
6sin((pi*t/12)+(2*pi/3)) = 3
sin((pi*t/12)+(2*pi/3)) = 3/6
sin((pi*t/12)+(2*pi/3)) = 1/2
(pi*t/12)+(2*pi/3) = arcsin(1/2) or (pi*t/12)+(2*pi/3) = pi - arcsin(1/2)
(pi*t/12)+(2*pi/3) = pi/6 or (pi*t/12)+(2*pi/3) = pi - pi/6
(pi*t/12)+(2*pi/3) = pi/6 or (pi*t/12)+(2*pi/3) = 5pi/6
(pi*t/12)+(2*pi/3) = pi/6 + 2pi*n or (pi*t/12)+(2*pi/3) = 5pi/6 + 2pi*n
pi*t/12 = pi/6 + 2pi*n - 2*pi/3 or pi*t/12 = 5pi/6 + 2pi*n - 2*pi/3
t = 12/pi(pi/6 + 2pi*n - 2*pi/3) or t = 12/pi(5pi/6 + 2pi*n - 2*pi/3)
t = 2 + 24n - 8 or t = 10 + 24n - 8
t = 24n - 6 or t = 24n + 2
So from all of that work above, we have determined that the solutions to i(t) = 9 are t = 24n - 6 or t = 24n + 2 where n is an integer.
Now let's plug in integral values for n into the two equations.
Let's pick on the first equation t = 24n - 6
If n = 0, then t is...
t = 24n - 6
t = 24(0) - 6
t = -6 ... out of range of the interval [0,120]
If n = 1, then t is...
t = 24n - 6
t = 24(1) - 6
t = 18
If n = 2, then t is...
t = 24n - 6
t = 24(2) - 6
t = 42
If n = 3, then t is...
t = 24n - 6
t = 24(3) - 6
t = 66
If n = 4, then t is...
t = 24n - 6
t = 24(4) - 6
t = 90
If n = 5, then t is...
t = 24n - 6
t = 24(5) - 6
t = 114
If n = 6, then t is...
t = 24n - 6
t = 24(6) - 6
t = 138 ... out of range of the interval [0,120]
If n = 7, then t is...
t = 24n - 6
t = 24(7) - 6
t = 162 ... out of range of the interval [0,120]
So we see that as n increases, we're getting larger and larger t values that will be outside the interval [0,120]. So we can stop for this equation.
So far, the solutions are: 18, 42, 66, 90, 114
Repeat for the other equation
If n = -1, then t is...
t = 24n + 2
t = 24(-1) + 2
t = -22 ... out of range of the interval [0,120]
If n = 0, then t is...
t = 24n + 2
t = 24(0) + 2
t = 2
If n = 1, then t is...
t = 24n + 2
t = 24(1) + 2
t = 26
If n = 2, then t is...
t = 24n + 2
t = 24(2) + 2
t = 50
If n = 3, then t is...
t = 24n + 2
t = 24(3) + 2
t = 74
If n = 4, then t is...
t = 24n + 2
t = 24(4) + 2
t = 98
If n = 5, then t is...
t = 24n + 2
t = 24(5) + 2
t = 122 ... out of range of the interval [0,120]
If n = 6, then t is...
t = 24n + 2
t = 24(6) + 2
t = 146 ... out of range of the interval [0,120]
The other set of solutions are: 2, 26, 50, 74, 98
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Put them all together to get this final set of solutions: 18, 42, 66, 90, 114, 2, 26, 50, 74, 98
Here is that list sorted: 2, 18, 26, 42, 50, 66, 74, 90, 98, 114
There are 10 solutions listed above, so it emits a signal exactly 10 times.
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