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.)
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))
To solve this problem, we need to integrate the given differential equation and use the initial condition to find the value of the constant of integration.
1. Start with the given differential equation:
dQ/dt = k(85 − Q)
2. Separate the variables by moving all terms involving Q to one side and terms involving t to the other side:
dQ/(85 − Q) = k dt
3. Integrate both sides of the equation:
∫dQ/(85 − Q) = ∫k dt
The left side can be solved using the substitution method:
Let u = 85 - Q, then du = -dQ
∫-du/u = ∫k dt
After integration, we have:
-ln|u| = kt + C1
4. Substitute back for u:
-ln|85 - Q| = kt + C1
5. Since the initial condition is given as "the average student can take 50 words of dictation per minute after 10 weeks in the course," we can write Q(10) = 50. Substitute this value into the equation:
-ln|85 - 50| = k(10) + C1
-ln|35| = 10k + C1
6. Solve for C1:
-ln|35| = 10k + C1
C1 = -ln|35| - 10k
7. Substitute the value of C1 back into the equation:
-ln|85 - Q| = kt - ln|35| - 10k
8. Now, we need to find the value of k to solve for the final value of Q. We can use the second initial condition, which states that "after completing the course," Q(T) = ? where T = 20 weeks.
Plugging in T = 20:
-ln|85 - Q| = k(20) - ln|35| - 10k
9. Solve for k by substituting the known values Q(10) = 50 and T = 20 into the equation above and solving for k.
-ln|35| = 10k - ln|35| - 10k
0 = 0
We find that the equation for k is inconclusive, suggesting that there may be an error in the problem statement or given information.
10. At this point, we cannot determine the value of Q after completing the course without the correct value of k.
To interpret this problem, let's break it down step by step.
1. First, let's understand the given equation: dQ/dt = k(85 − Q). This equation represents the rate of change of the number of words per minute (Q) with respect to time (t). It tells us that the rate of improvement of a student's shorthand ability is proportional to the difference between 85 and the current number of words per minute. The constant k represents the rate of improvement.
2. We are told that Q(t) represents the number of words of dictation a student can take per minute after t weeks in the course. So, Q(0) represents the initial number of words per minute at the start of the course.
3. The problem states that after 10 weeks in the course, the average student can take 50 words of dictation per minute. We can use this information to find the value of k.
Substituting t = 10 and Q(t) = 50 in the given equation, we have:
dQ/dt = k(85 - Q)
k(85 - 50) = 50
k = 50 / (85 - 50)
4. Now that we have the value of k, we can find the number of words per minute the average student can take after completing the course. The problem specifies a 20-week course, so we need to find Q(20).
Using the given equation dQ/dt = k(85 - Q), we can separate variables and integrate both sides to solve for Q(t):
∫dQ / (85 - Q) = ∫k dt
Applying the appropriate integration techniques, we find:
ln|85 - Q| = kt + C
Now, let's use the initial condition Q(0) = 0 to find the value of the constant C:
ln|85 - 0| = k * 0 + C
ln(85) = C
Therefore, the equation becomes:
ln|85 - Q| = kt + ln(85)
We can rearrange this equation to solve for Q:
|85 - Q| = e^(kt + ln(85))
Considering the absolute value, we get two possible solutions:
85 - Q = e^(kt + ln(85)) or Q - 85 = e^(kt + ln(85))
5. Now we substitute t = 20 into the equation and find Q(20). Using the form Q - 85 = e^(kt + ln(85)), we have:
Q - 85 = e^(k * 20 + ln(85))
Rearranging and solving for Q, we get:
Q = e^(k * 20 + ln(85)) + 85
6. Finally, plug in the value of k that we found earlier to calculate Q:
Q = e^((50/(85-50)) * 20 + ln(85)) + 85
Evaluate this expression to find the average number of words per minute the student can take after completing the course.