Let P(t) be the performance level of someone learning a skill as a function of t
ID: 2857189 • Letter: L
Question
Let P(t) be the performance level of someone learning a skill as a function of the training time 4. The derivative dp/dt represents the rate at which performance improves. If M is the maximum level of performance of which the learner is capable, then a model for learning is given by the differential equation dp/dt = k(M - P(t)) where K is a positive constant. Two new workers, Bill and Peter, were hired for an assembly line. Bill could process 11 units per minute after one hour and 15 units per minute after two hours. Peter could process 10 units per minute after one hour and 16 units per minute after two hours. Using the above model and assuming that P(0) = 0, estimate the maximum number of units per minute that each worker is capable of processing.Explanation / Answer
given
dP/dt=k(MP(t))
dP/(MP(t)) =k dt
dP/(M(1P(t)/M)) =kdt
dP/(1P(t)/M) =Mkdt
integrate on both sides
dP/(1P(t)/M) = Mkdt
-M(ln(1-P(t)/M))=Mkt +c
(ln(1-P(t)/M))=-kt +c
(1-P(t)/M)=e-kt +c
(1-P(t)/M)=Ce-kt
P(t)/M=1-Ce-kt
given P(0)=0
0/M=1-Ce-k0
0=1-C
C=1
P(t)/M=1-e-kt
P(t)=M(1-e-kt)
bill could process 11 units per minute after one hour
11=M(1-e-k)
=>e-k=1-(11/M) ------------>(1)
15 units per minute after two hours
15=M(1-e-2k)
from (1)
15=M(1-(1-(11/M))2)
15=M(1-(1-(22/M) +(121/M2))
15=M((22/M) -(121/M2))
15=((22) -(121/M))
121/M=22-15
121/M=7
M=121/7
M=17.28
maximum number of units per minute that bill is capable of processing. =17
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P(t)=M(1-e-kt)
peter could process 10 units per minute after one hour
10=M(1-e-k)
=>e-k=1-(10/M) ------------>(1)
16 units per minute after two hours
16=M(1-e-2k)
from (1)
16=M(1-(1-(10/M))2)
16=M(1-(1-(20/M) +(100/M2))
16=M((20/M) -(100/M2))
16=((20) -(100/M))
100/M=20-16
100/M=4
M=100/4
M=25
maximum number of units per minute that peter is capable of processing. =25
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