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52844811
A patient takes a \(20\,\text{mg}\) tablet once each day at the same time. During each day, the body eliminates \(30\%\) of the medication currently present. a) Make a table of the amount in the body immediately after each dose for days \(1\) through \(5\). Before the first dose, the amount is \(0\,\text{mg}\). b) On what day does the amount first exceed \(50\,\text{mg}\)? c) Find the theoretical long-term amount in the body immediately after a dose.

Hints

- At each step, first keep \(70\%\) of the previous amount, then add the new dose. - Compare the table values with \(50\,\text{mg}\). - At equilibrium, the amount after one full step is unchanged.

Solution

1. If \(x_n\) is the amount immediately after the dose on day \(n\), then \(x_n=0.7x_{n-1}+20\), with \(x_0=0\). 2. \(x_1=20\), \(x_2=0.7\cdot20+20=34\), \(x_3=0.7\cdot34+20=43.8\), \(x_4=0.7\cdot43.8+20=50.66\), and \(x_5=0.7\cdot50.66+20=55.462\), all in milligrams. 3. The amount first exceeds \(50\,\text{mg}\) on day \(4\). 4. At equilibrium, \(L=0.7L+20\). Thus \(0.3L=20\), so \(L=\frac{200}{3}\approx66.67\,\text{mg}\).

Answer

a) <table> <tr> <th>Day</th> <th>Amount after dose (mg)</th> </tr> <tr><td>\(1\)</td><td>\(20\)</td></tr> <tr><td>\(2\)</td><td>\(34\)</td></tr> <tr><td>\(3\)</td><td>\(43.8\)</td></tr> <tr><td>\(4\)</td><td>\(50.66\)</td></tr> <tr><td>\(5\)</td><td>\(55.462\)</td></tr> </table> b) Day \(4\) c) \(\frac{200}{3}\,\text{mg}\approx66.67\,\text{mg}\)
52845011
Immediately after the final dose of an earlier treatment, a patient has \(40\,\text{mg}\) of a medication in the body at day \(0\). The patient now takes \(5\,\text{mg}\) once daily. Before each new dose, the body eliminates \(20\%\) of the medication present. a) Find the amount immediately after the dose on days \(1\) and \(2\). b) Show that the amount decreases over time despite the daily dose, and find the value to which it stabilizes. c) How would the limiting amount change if the body eliminated \(25\%\) each day instead of \(20\%\)?

Hints

- Compare the daily amount eliminated with the daily amount added. - At equilibrium, the amount remains unchanged after one step. - Increasing the elimination rate changes the equilibrium equation.

Solution

1. The recurrence is \(A_{n+1}=0.8A_n+5\), with \(A_0=40\). 2. \(A_1=0.8\cdot40+5=37\,\text{mg}\), and \(A_2=0.8\cdot37+5=34.6\,\text{mg}\). 3. The daily change is \(A_{n+1}-A_n=5-0.2A_n\), which is negative whenever \(A_n>25\). Also, \(A_{n+1}-25=0.8(A_n-25)\), so the terms stay above \(25\) and approach it. The limiting amount is \(25\,\text{mg}\). 4. With a \(25\%\) daily elimination, the limit satisfies \(L=0.75L+5\), giving \(L=20\,\text{mg}\).

Answer

a) Day \(1\): \(37\,\text{mg}\); day \(2\): \(34.6\,\text{mg}\) b) The amount decreases toward \(25\,\text{mg}\). c) The limiting amount would decrease to \(20\,\text{mg}\).
52845711
A newly planted forest can support at most \(4000\) trees. Initially, \(800\) young trees are planted. Each year, the number of trees increases by \(25\%\) of the remaining capacity, the difference between \(4000\) and the current number of trees. a) Write a recurrence for \(B_{n+1}\) in terms of \(B_n\). b) Find the model values for the first five years after planting. Round to the nearest whole tree. c) Explain why the model represents bounded growth and describe how the annual increases behave over time.

Hints

- Express the remaining capacity as a difference. - Add the annual increase to the current number of trees. - Consider how the gap to capacity changes over time. - Compare the annual increases from one year to the next.

Solution

1. The annual increase is \(0.25(4000-B_n)\), so \(B_{n+1}=B_n+0.25(4000-B_n)=0.75B_n+1000\), with \(B_0=800\). 2. Without rounding intermediate values, \(B_1=1600\), \(B_2=2200\), \(B_3=2650\), \(B_4=2987.5\approx2988\), and \(B_5=3240.625\approx3241\). 3. The population is bounded above by \(4000\). As the gap to capacity becomes smaller, the annual increase also becomes smaller and approaches \(0\).

Answer

a) \(B_{n+1}=B_n+0.25(4000-B_n)\), or \(B_{n+1}=0.75B_n+1000\) b) \(B_1=1600\), \(B_2=2200\), \(B_3=2650\), \(B_4\approx2988\), \(B_5\approx3241\) c) The capacity \(4000\) is an upper bound, and the annual increases approach \(0\).
52844911
An artificial pond begins empty. Each day, \(15\%\) of the water evaporates, and then \(60\,\text{L}\) of fresh water is added. a) Find the amount of water immediately after the addition on days \(2\), \(3\), and \(4\). b) Find the theoretical upper limit of the water amount. c) On what day does the amount first exceed \(250\,\text{L}\)?

Hints

- Write a recurrence relating one day's amount to the previous day's amount. - At the limiting amount, the daily loss equals the daily addition. - Combine the percentage loss and fixed addition in one equation. - Use an explicit formula or test consecutive days near the threshold.

Solution

1. Let \(W_n\) be the amount immediately after the addition on day \(n\). Then \(W_{n+1}=0.85W_n+60\), with \(W_0=0\). 2. \(W_1=60\), \(W_2=0.85\cdot60+60=111\), \(W_3=0.85\cdot111+60=154.35\), and \(W_4=0.85\cdot154.35+60=191.1975\approx191.20\), in liters. 3. At equilibrium, \(L=0.85L+60\). Therefore, \(0.15L=60\), so \(L=400\,\text{L}\). 4. The explicit form is \(W_n=400(1-0.85^n)\). It gives \(W_6\approx249.14\,\text{L}\) and \(W_7\approx271.77\,\text{L}\), so the amount first exceeds \(250\,\text{L}\) on day \(7\).

Answer

a) Day \(2\): \(111\,\text{L}\); day \(3\): \(154.35\,\text{L}\); day \(4\): about \(191.20\,\text{L}\) b) \(400\,\text{L}\) c) Day \(7\)
53351411
A town has \(20{,}000\) residents. A new car-sharing service begins with \(2000\) registered users. Each month, the company expects to register \(20\%\) of the residents who are not yet users. a) Find the number of users after each of the first three months and complete the table. <table> <tr><th>Month \(x\)</th><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr> <tr><th>Users \(N(x)\)</th><td>\(2000\)</td><td>...</td><td>...</td><td>...</td></tr> </table> b) Write an explicit function for the number of users after \(x\) months. c) About how many users does the model predict after one year? In the graph, curve \(n\) shows the user model and the dashed line \(s\) shows the saturation level. Use the graph to explain why the model never exceeds \(20{,}000\) users.
Figure for problem 533514

Hints

- Each month, first find how many residents are still nonusers. - Add \(20\%\) of the nonusers to the current user count. - Model the remaining gap to the saturation value as an exponential decay. - Interpret the upper horizontal asymptote.

Solution

1. Each month, add \(20\%\) of the remaining nonusers. Month \(1\): \(2000+0.20\cdot(20{,}000-2000)=5600\). Month \(2\): \(5600+0.20\cdot(20{,}000-5600)=8480\). Month \(3\): \(8480+0.20\cdot(20{,}000-8480)=10{,}784\). 2. The number of nonusers begins at \(18{,}000\) and is multiplied by \(0.8\) each month. Therefore, \(N(x)=20{,}000-18{,}000(0.8)^x\). 3. \(N(12)=20{,}000-18{,}000\cdot(0.8)^{12}\approx18{,}763\). 4. Since \(18{,}000(0.8)^x>0\) for every finite \(x\), \(N(x)<20{,}000\). The graph approaches the horizontal asymptote \(y=20{,}000\) from below.

Answer

a) Month \(1\): \(5600\); month \(2\): \(8480\); month \(3\): \(10{,}784\) b) \(N(x)=20{,}000-18{,}000(0.8)^x\) c) About \(18{,}763\) users; the graph approaches \(20{,}000\) from below and never reaches or exceeds it.

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