The Institute finds that the average student taking Elementary Shorthand will progress at a rate given by
dQ/dt=k(80-Q)
in a 20 -wk course, where Q(t) measures the number of words of dictation a student can take per minute after t wk in the course.
If the average student can take 60 words of dictation per minute after 5 wk in the course, how many words per minute can the average student take after completing the course?
Round the answer to the nearest word.
Assume a solution Q=Ce^-rt + K
To find the number of words per minute that the average student can take after completing the course, we need to solve the given differential equation:
dQ/dt = k(80 - Q)
We can solve this first-order linear differential equation using separation of variables. Rearranging the equation, we have:
dQ/(80 - Q) = k dt
Integrating both sides with respect to their respective variables, we get:
∫dQ/(80 - Q) = ∫k dt
The left side can be integrated using the substitution method, where u = 80 - Q, and du = -dQ. The integral becomes:
-∫(1/u) du = ∫k dt
-ln|u| = kt + C1
-ln|80 - Q| = kt + C1
To simplify further, we can remove the absolute value and exponentiate both sides:
|80 - Q| = e^(-kt - C1)
|80 - Q| = e^(-kt) * e^(-C1)
|80 - Q| = C * e^(-kt)
Here, C = e^(-C1), and we can write C as a positive constant, as e^(-C1) will always be positive.
Now, we can consider the initial condition given in the problem: after 5 weeks (t = 5), the average student can take 60 words per minute (Q = 60). Substituting these values into the equation:
|80 - 60| = C * e^(-5k)
|20| = C * e^(-5k)
Since the absolute value of 20 can be either 20 or -20, we have two possibilities:
20 = C * e^(-5k) or -20 = C * e^(-5k)
Solving the first equation for C:
C = 20 / e^(-5k)
Substituting this value of C into the second equation:
-20 = (20 / e^(-5k)) * e^(-5k)
-20 = 20
This equation doesn't hold, so it means that C = 20 / e^(-5k) is the correct solution.
Now, we need to find the value of Q when t = 20 (after completing the course). Substituting these values into the equation:
|80 - Q| = (20 / e^(-5k)) * e^(-20k)
Again, since the absolute value of 80 - Q can be either 80 - Q or -(80 - Q), we have two cases:
80 - Q = (20 / e^(-5k)) * e^(-20k) or -(80 - Q) = (20 / e^(-5k)) * e^(-20k)
We'll consider both cases:
Case 1: 80 - Q = (20 / e^(-5k)) * e^(-20k)
Simplifying and rearranging this equation:
80 - Q = 20 * e^(-25k)
Q = 80 - 20 * e^(-25k)
Case 2: -(80 - Q) = (20 / e^(-5k)) * e^(-20k)
Simplifying and rearranging this equation:
Q - 80 = 20 * e^(-25k)
Q = 80 + 20 * e^(-25k)
We have two possible solutions, so let's take the average of these two solutions to find the approximate value of Q after completing the course:
(Q1 + Q2) / 2 = (80 - 20 * e^(-25k) + 80 + 20 * e^(-25k)) / 2
(Q1 + Q2) / 2 = 160 / 2
(Q1 + Q2) / 2 = 80
Therefore, the average student can take approximately 80 words per minute after completing the course.