AP Calculus BC7.4 Reasoning Using Slope Fields - FRQs - Exam Style Questions
No-CalcQuestion
The depth of seawater at a location can be modeled by the function \(H\) that satisfies the differential equation \[ \frac{dH}{dt}=\tfrac12\,(H-1)\cos\!\Big(\tfrac{t}{2}\Big), \] where \(H(t)\) is measured in feet and \(t\) is measured in hours after noon (\(t=0\)). It is known that \(H(0)=4\).
(a) A portion of the slope field for the differential equation is provided. Sketch the solution curve, \(y=H(t)\), through the point \((0,4)\).
(b) For \(0<t<5\), it can be shown that \(H(t)>1\). Find the value of \(t\), for \(0<t<5\), at which \(H\) has a critical point. Determine whether the critical point corresponds to a relative minimum, a relative maximum, or neither. Justify your answer.
(c) Use separation of variables to find \(y=H(t)\), the particular solution to the differential equation \(\displaystyle \frac{dH}{dt}=\tfrac12\,(H-1)\cos\!\big(\tfrac{t}{2}\big)\) with initial condition \(H(0)=4\).
Most-appropriate topic codes:
• TOPIC 7.3: Sketching Slope Fields (part a) | • TOPIC 7.4: Reasoning Using Slope Fields (part a) | • TOPIC 7.1: Modeling Situations with Differential Equations (critical point from \(dH/dt\)) (part b) | • TOPIC 7.7: Finding Particular Solutions Using Initial Conditions and Separation of Variables (part c)
▶️ Answer/Explanation
(a) Sketch through \((0,4)\)
Compute initial slope: \(\displaystyle \frac{dH}{dt}\big|_{t=0}=\tfrac12(4-1)\cos(0)=\tfrac32\).Therefore the solution increases initially from \((0,4)\).
Since \(\cos\!\big(\tfrac{t}{2}\big)=0\) at \(t=\pi\), slopes change sign there.
The curve rises to a highest point near \(t=\pi\) and then decreases, matching the slope field.
(b) Critical point and classification
Given \(H(t)>1\) on \(0<t<5\Rightarrow H-1>0\).
Hence the factor that can vanish is \(\cos\!\big(\tfrac{t}{2}\big)=0\).
Solve: \(\tfrac{t}{2}=\tfrac{\pi}{2}\Rightarrow t=\pi\ (\approx 3.142).\)
Sign test: for \(0<t<\pi\), \(\cos\!\big(\tfrac{t}{2}\big)>0\Rightarrow \dfrac{dH}{dt}>0\) (increasing).
For \(\pi<t<5\), \(\cos\!\big(\tfrac{t}{2}\big)<0\Rightarrow \dfrac{dH}{dt}<0\) (decreasing).
Therefore \(t=\pi\) is a \(\boxed{\text{relative maximum}}\) of \(H\).
(Second-derivative check: \(\displaystyle \frac{d^{2}H}{dt^{2}}\big|_{t=\pi}=-\tfrac14(H-1)<0\), confirming maximum.)
(c) Particular solution by separation of variables
Start with \(\displaystyle \frac{dH}{dt}=\tfrac12(H-1)\cos\!\big(\tfrac{t}{2}\big)\).Separate: \(\displaystyle \frac{dH}{H-1}=\tfrac12\cos\!\big(\tfrac{t}{2}\big)\,dt\).
Integrate both sides: \(\displaystyle \int\frac{dH}{H-1}=\int\tfrac12\cos\!\big(\tfrac{t}{2}\big)\,dt\).
Obtain: \(\displaystyle \ln|H-1|=\sin\!\big(\tfrac{t}{2}\big)+C\).
Use \(H(0)=4\): \(\ln 3=\sin(0)+C\Rightarrow C=\ln 3\).
Because \(H>1\) on the interval of interest, \(|H-1|=H-1\).
Thus \(\displaystyle \ln(H-1)=\sin\!\big(\tfrac{t}{2}\big)+\ln 3\).
Exponentiate: \(\displaystyle H-1=3\,e^{\sin(t/2)}\).
Final particular solution: \(\displaystyle \boxed{H(t)=1+3\,e^{\sin(t/2)}}\).