A depth-of-seawater quantity satisfies a given separable first-order differential equation with a stated initial condition. Students sketch the solution curve on a slope field, locate a critical point of the solution and classify it as a relative max, min, or neither, and solve the equation by separation of variables for the particular solution.
A depth-of-seawater quantity satisfies a given separable first-order differential equation with a stated initial condition. Students sketch the solution curve on a slope field, locate a critical point of the solution and classify it as a relative max, min, or neither, and solve the equation by separation of variables for the particular solution.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
On the given slope field for $\frac{dH}{dt}=\frac{1}{2}(H-1)\cos\!\left(\frac{t}{2}\right)$, sketch the solution curve $y=H(t)$ through $(0,4)$. At $t=0$ the slope is $\frac{1}{2}(4-1)\cos 0=\frac{3}{2}>0$, so the curve initially rises from $(0,4)$. Because $\cos\!\left(\frac{t}{2}\right)>0$ for $0<t<\pi$ and $<0$ for $\pi<t<5$ (with $H>1$ throughout), the curve rises until $t=\pi$, reaches a high point there, then falls. Draw the curve passing through $(0,4)$, following the slope marks, increasing on $(0,\pi)$ and decreasing on $(\pi,5)$, and extending to about $t=4.5$ as indicated on the field.
1 point (solution curve): the sketch must pass through (0,4), be consistent with the slope marks, and extend reasonably across the provided field (to about t=4.5). Common loss: a curve that does not pass through (0,4) or that conflicts with the slope marks (e.g. fails to turn around near t=pi).
A critical point of $H$ occurs where $\frac{dH}{dt}=0$. Since $H(t)>1$ on $0<t<5$, the factor $\frac{1}{2}(H-1)>0$, so $$\frac{dH}{dt}=0\iff \cos\!\left(\frac{t}{2}\right)=0.$$ On $0<t<5$, $\cos\!\left(\frac{t}{2}\right)=0$ requires $\frac{t}{2}=\frac{\pi}{2}$, i.e. $t=\pi$. (The next zero, $t=3\pi\approx 9.4$, is outside the interval.) So the only critical point is $t=\pi$. Apply the first-derivative test. For $0<t<\pi$, $\frac{t}{2}\in\left(0,\frac{\pi}{2}\right)$ so $\cos\!\left(\frac{t}{2}\right)>0$ and $\frac{dH}{dt}>0$. For $\pi<t<5$, $\frac{t}{2}\in\left(\frac{\pi}{2},2.5\right)$ so $\cos\!\left(\frac{t}{2}\right)<0$ and $\frac{dH}{dt}<0$. Because $\frac{dH}{dt}$ changes from positive to negative at $t=\pi$, the critical point at $t=\pi$ is a **relative maximum** of the depth $H$.
3 points: 1 for considering the sign of dH/dt and reducing the critical-point condition to cos(t/2)=0 (using H-1>0); 1 for identifying the critical point t=pi on 0<t<5; 1 for the relative-maximum answer with a first-derivative-test justification (dH/dt changes + to -). A second-derivative test that shows d^2H/dt^2<0 at t=pi is an accepted alternative justification. Stating 'relative max' without sign evidence does not earn the justification point.
Solve $\frac{dH}{dt}=\frac{1}{2}(H-1)\cos\!\left(\frac{t}{2}\right)$ with $H(0)=4$ by separation of variables. Separate and integrate: $$\frac{dH}{H-1}=\frac{1}{2}\cos\!\left(\frac{t}{2}\right)dt.$$ The left side antidifferentiates to $\ln|H-1|$. For the right side, $\int\frac{1}{2}\cos\!\left(\frac{t}{2}\right)dt=\sin\!\left(\frac{t}{2}\right)$ (since $\frac{d}{dt}\sin\frac{t}{2}=\frac{1}{2}\cos\frac{t}{2}$). Thus $$\ln|H-1|=\sin\!\left(\frac{t}{2}\right)+C.$$ Apply $H(0)=4$: $\ln(4-1)=\sin 0+C$, so $C=\ln 3$. Since $H>1$, $|H-1|=H-1$, giving $$\ln(H-1)=\sin\!\left(\frac{t}{2}\right)+\ln 3.$$ Exponentiate: $$H-1=3\,e^{\sin(t/2)} \;\Rightarrow\; \boxed{H(t)=1+3e^{\sin(t/2)}}.$$ Check: $H(0)=1+3e^0=4$ matches the initial condition.
5 points, awarded sequentially: P1 separates variables (dH/(H-1) = (1/2)cos(t/2)dt); P2 antiderivative of the H-side, ln|H-1|; P3 antiderivative of the t-side, sin(t/2); P4 constant of integration plus the initial condition H(0)=4 to get C=ln 3; P5 solves explicitly for H to reach H(t)=1+3e^{sin(t/2)}. No separation earns 0/5; omitting the constant of integration caps the score (the constant/IC and solve points are then unavailable).
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