(1 point) Let P(t) be the performance level of someone learning a skill as a fun
ID: 2856399 • Letter: #
Question
(1 point) Let P(t) be the performance level of someone learning a skill as a function of the training time t. The derivative dPdt 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
dPdt=k(MP(t))
where k is a positive constant.
Two new workers, Andy and John, were hired for an assembly line. Andy could process 12 units per minute after one hour and 15 units per minute after two hours. John 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.
Andy:..............??
John: .......... ???
Note: Please show clearly the final answer
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)
Andy could process 12 units per minute after one hour
12=M(1-e-k)
=>e-k=1-(12/M) ------------>(1)
15 units per minute after two hours
15=M(1-e-2k)
from (1)
15=M(1-(1-(12/M))2)
15=M(1-(1-(24/M) +(144/M2))
15=M((24/M) -(144/M2))
15=((24) -(144/M))
144/M=24-15
144/M=9
M=144/9
M=16
maximum number of units per minute that ANDY is capable of processing. =16
==================================
P(t)=M(1-e-kt)
John 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 JOHN is capable of processing. =25
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