By differentiation and substitution, the answers to the four question are
a.) O'(4) = - 0.96 Tons/[tex]hr^{2}[/tex]
b.) O'(3) = - 0.85 tons/[tex]hr^{2}[/tex]
c.) U'(t) = - 71.98
d.) t = 3.125 hours
Given that on a given workday, the rate, in tons per hour, at which unprocessed ore arrives at a refinery is modeled by
O(t) = 20 + 10 cos (t^2/25)
where t is measured in hours and 0 ≤ t ≤ 10.
From the question, at t = 0, O(t) = 100 tons
And during the hour of operation, the refinery processes ore at a constant rate of 25 tons per hour.
a.) To find O'(4), we will differentiate the equation O(t) = 20 + 10 cos (t^2/25)
O'(4) = -20t/25Sin(
[tex]t^{2}[/tex]/25)
Substitute t for 4
O'(4) = - 40/25Sin(16/25)
O'(4) = - 1.6Sin(0.64)
O'(4) = - 0.96 Tons/ [tex]hr^{2}[/tex]
This means the rate at which unprocessed ore is arriving is decreasing by 0.96 per hour per hour at t = 4
b.) To calculate how much total ore has been processed by the refinery after the plant has been open for 3 hours, substitute t for 3 in the differential equation O'(t) = -20t/25Sin(
[tex]t^{2}[/tex]/25)
O'(3) = -20x3/25Sin(
[tex]3^{2}[/tex]/25)
O'(3) = -60/25Sin(9/25)
O'(3) = -2.4Sin(0.36)
O'(3) = - 0.85 tons/[tex]hr^{2}[/tex]
c.) At time t = 4
Let U(t) = amount of unprocessed ore and U'(t) = rate at which it is changing.
Then
U'(t) = O'(4) - 100
U'(t) = 20 + 10Cos(
[tex]4^{2}[/tex]/25) - 100
U'(t) = 20 + 10Cos(16/25) - 100
U'(t) = 20 + 10Cos(0.64) - 100
U'(t) = 28.021 - 100
U'(t) = - 71.98
U'(t) is negative. That means the amount of unprocessed ore is decreasing at time t = 4 hours.
d.) Since the refinery processes ore at a constant rate of 25 tons per hour, to determine the time at which the rate is constant, substitute O'(t) for 25 and look for t in equation O'(t) = -20t/25Sin([tex]t^{2}[/tex]
/25).
It is also constant at the beginning when the O(t) = 100 tons
25 = -20t/25Sin(
[tex]t^{2}[/tex]/25)
5 = -4t/25Sin(
[tex]t^{2}[/tex]/25)
125 = -4tSin(
[tex]t^{2}[/tex]/25)
Log both sides
log 125 = log -4tSin(
[tex]t^{2}[/tex]/25)
log 125 = log(-4t) + log Sin(
[tex]t^{2}[/tex]/25)
Graphically, log Sin(
[tex]t^{2}[/tex]/25) is between 0 and 1.
let assume log Sin(
[tex]t^{2}[/tex]/25) = 1
log 125 = log(-4t) + 1
log 125 - 1 = log(-4t)
2.0969 - 1 = log(-4t)
1.0969 = log(-4t)
4t = [tex]log^{-1}[/tex]
(1.0969)
4t = 12.5
t = 3.125 hours
Therefore, the time at which the rate is constant is 3.125 hours
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