AP Calculus BC 7.9 Logistic Models with Differential Equations - Exam Style Questions - MCQs - New Syllabus
Question
If \(\,10\,\) people have heard the news at time \(\,t=0\,\), how many people have heard the news when the rate at which the news is spreading changes from increasing to decreasing?
The number of people \(\,P\,\) (at time \(\,t\,\) days) who have heard a piece of news at a business is modeled by the function \(\,P\). The rate at which the news is spreading is modeled by the logistic differential equation \[ \frac{dP}{dt}=144\,P\!\left(1-\frac{P}{4800}\right). \]
A) \(\,72\,\)
B) \(\,144\,\)
C) \(\,2400\,\)
D) \(\,4800\,\)
B) \(\,144\,\)
C) \(\,2400\,\)
D) \(\,4800\,\)
▶️ Answer/Explanation
Detailed solution
For a logistic model \( \displaystyle \frac{dP}{dt}=kP\!\left(1-\frac{P}{K}\right) \), the rate \( \dfrac{dP}{dt} \) is a concave-down quadratic in \(P\) with a maximum at \(P=\dfrac{K}{2}\).
Here \(K=4800\Rightarrow P=\dfrac{4800}{2}=2400\). That is the moment when the rate switches from increasing to decreasing.
✅ Correct: C) \(2400\)
Question
In a community of \(6000\) people, the number of people informed about an upcoming event after \(t\) weeks is modeled by a function \(P(t)\) that satisfies a logistic differential equation. When \(1000\) people are informed, the rate of growth is \(100\) people per week. If all people in the community eventually become informed, which differential equation best models the situation?
(A) \( \displaystyle \frac{dP}{dt}=0.12\,t\!\left(1-\frac{t}{1000}\right) \)
(B) \( \displaystyle \frac{dP}{dt}=0.12\,t\!\left(1-\frac{t}{6000}\right) \)
(C) \( \displaystyle \frac{dP}{dt}=0.12\,P\!\left(1-\frac{P}{5000}\right) \)
(D) \( \displaystyle \frac{dP}{dt}=0.12\,P\!\left(1-\frac{P}{6000}\right) \)
(B) \( \displaystyle \frac{dP}{dt}=0.12\,t\!\left(1-\frac{t}{6000}\right) \)
(C) \( \displaystyle \frac{dP}{dt}=0.12\,P\!\left(1-\frac{P}{5000}\right) \)
(D) \( \displaystyle \frac{dP}{dt}=0.12\,P\!\left(1-\frac{P}{6000}\right) \)
▶️ Answer/Explanation
Logistic form: \( \dfrac{dP}{dt}=r\,P\!\left(1-\dfrac{P}{K}\right) \), where \(K=6000\).
Use \( \dfrac{dP}{dt}=100 \) at \(P=1000\): \[ 100 = r\cdot 1000\!\left(1-\frac{1000}{6000}\right) = r\cdot 1000\cdot \frac{5}{6} \;\Rightarrow\; r=\frac{100}{833.\overline{3}}=0.12. \] Therefore \[ \boxed{\dfrac{dP}{dt}=0.12\,P\!\left(1-\dfrac{P}{6000}\right)}. \] ✅ Answer: (D)
Use \( \dfrac{dP}{dt}=100 \) at \(P=1000\): \[ 100 = r\cdot 1000\!\left(1-\frac{1000}{6000}\right) = r\cdot 1000\cdot \frac{5}{6} \;\Rightarrow\; r=\frac{100}{833.\overline{3}}=0.12. \] Therefore \[ \boxed{\dfrac{dP}{dt}=0.12\,P\!\left(1-\dfrac{P}{6000}\right)}. \] ✅ Answer: (D)