dQ/dt = k(85-Q)
dQ/(85-Q) = k dt
ln(85-Q) = kt + c
85-Q = c*e^(kt)
Q = 85-c*e^(kt)
Q(10) = 50, so
85-c*e^(10k) = 50
c*e^(10k) = 35
c = 35e^(-10k)
Q(t) = 85-35e^(-10k)*e^(kt)
= 85-35e^(kt-10k)
Now, using Q(0) = 0,
85-35e^(-10k) = 0
e^(-10k) = 17/7
-10k = ln(17/7)
k = -0.0887
Q(t) = 85-35e^(0.887-0.0887t)
= 85-35e^(0.887)e^(-0.0887t)
= 85-84.97e^(-0.0887t)
or, more probably, knowing Q(0)=0,
Q(t) = 85(1-e^(-.0887t))
Hi, I can't seem to find out how to do this. My textbook has two Q(t) values in the example and I don't know how to interpret this problem.
An institute finds that the average student taking Elementary Machine Shorthand will progress at a rate given by
dQ/dt = k(85 − Q)
in a 20-week course, where Q(t) measures the number of words of dictation a student can take per minute after t weeks in the course. If the average student can take 50 words of dictation per minute after 10 weeks in the course, how many words per minute can the average student take after completing the course? (Round your answer to the nearest whole number. Assume
Q_0 = 0.)
1 answer
1 answer