the productivity of a person at work is modeled by a cosine function: 5cos (pie/2 (t)) +5, where t is in hours. If the person starts work at t=0, being 8:00 am, at what times is the worker the least productive?
P = 5cos (pie/2 (t)) +5
the period of this function is 2π/(π/2) = 4
for the lowest productivity, we need the minimum of the function.
we know that cos (pie/2 (t)) has a minimum of -1
then 5cos (pie/2 (t))has a minimum of -5
and 5cos (pie/2 (t)) +5 has a minimum of 0
so when is 5cos (pie/2 (t)) +5 = 0 ?
5cos (pie/2 (t)) = -5
cos (pie/2 (t)) = -1
but cos π = -1
π/2t = π
t/2 = 1
t = 2
look at the graph:
http://www.wolframalpha.com/input/?i=plot+y++%3D+5cos+(%CF%80%2F2+(t))+%2B5
notice a minimum at t = 2
translate that into a time of 10:00 am
To determine when the worker is the least productive, we need to find the values of t where the cosine function reaches its minimum value.
The cosine function reaches its minimum value of -1 at t = π + 2kπ, where k is an integer.
In this case, the function is given as 5cos(π/2*t) + 5.
Setting this expression equal to -1, we have:
5cos(π/2*t) + 5 = -1
Subtracting 5 from both sides:
5cos(π/2*t) = -6
Dividing both sides by 5:
cos(π/2*t) = -6/5
To solve for t, we take the inverse cosine (also known as arccos) of both sides:
π/2*t = arccos(-6/5)
To find all the possible values of t, we need to consider both positive and negative values of arccos(-6/5) since the cosine function has symmetry.
t = (2/n) * arccos(-6/5), where n is an odd integer (to account for both positive and negative values of arccos)
Now, we substitute back the value of t into hours. Since the person starts work at t = 0 (8:00 am), we have:
t = (2/n) * arccos(-6/5) + 8:00 am
Note that the arccos function returns values in radians. To convert to hours, we can use the fact that π radians = 12 hours. Therefore, we multiply the value of arccos(-6/5) by (12/π) to convert it to hours.
The possible times when the worker is the least productive can be found by plugging in different odd integer values for n in the equation above and adding the result to 8:00 am.
To determine the times when the worker is least productive, we need to find the values of t for which the cosine function reaches its minimum value.
The general form of a cosine function is: y = A * cos(B * (t - C)) + D
In this case, the function representing productivity is: y = 5 * cos((π/2)t) + 5
To find the minimum values, we need to find the values of t that make the cosine function equal to -1.
For a general cosine function, we know that cos(θ) = -1 when θ = π + 2πk, where k is an integer.
By equating our function to -1 and solving for t, we have:
-1 = 5 * cos((π/2)t) + 5
-6 = 5 * cos((π/2)t)
cos((π/2)t) = -6/5
Now, to find the values of t for which the worker is least productive, we need to solve for t:
(π/2)t = arccos(-6/5)
t = (2/π)*arccos(-6/5)
Using a calculator, we can find the approximate value of t:
t ≈ (2/π)*(143.13°) ≈ 90.83
The worker is least productive at t ≈ 90.83 hours.
To convert this to a time, we need to consider that the worker starts at t = 0, which corresponds to 8:00 am.
So, 90.83 hours after 8:00 am would be approximately 6:50 am the next day.