Question

the depth of seawater at a location can be modeled by the function h that satisfies the differential equation\\(\frac{dh}{dt}=\frac{1}{2}(h - 1)\cos(\frac{t}{2})\\)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 a relative minimum nor a relative maximum of the depth of seawater at the location. justify your answer.
(c) use separation of variables to find y = h(t), the particular solution to the differential equation\\(\frac{dh}{dt}=\frac{1}{2}(h - 1)\cos(\frac{t}{2})\\)with initial condition h(0)=4.

Explanation:

Part (b)

Step1: Recall the condition for critical points

Critical points occur where \(\frac{dH}{dt}=0\). Given \(\frac{dH}{dt}=\frac{1}{2}(H - 1)\cos(\frac{t}{2})\), and since \(H(t)>1\) for \(0 < t<5\), we set \(\cos(\frac{t}{2}) = 0\).

Step2: Solve \(\cos(\frac{t}{2})=0\)

We know that \(\cos(x)=0\) when \(x=(2n + 1)\frac{\pi}{2}\), \(n\in\mathbb{Z}\). So, \(\frac{t}{2}=(2n + 1)\frac{\pi}{2}\), which gives \(t=(2n + 1)\pi\). For \(0 < t<5\), when \(n = 0\), \(t=\pi\approx3.14\).

Step3: Use the first - derivative test

We consider intervals around \(t = \pi\). Let's choose a test - point \(t_1=\frac{\pi}{2}\) (in the interval \(0For \(t=\frac{\pi}{2}\), \(\frac{dH}{dt}=\frac{1}{2}(H - 1)\cos(\frac{\frac{\pi}{2}}{2})=\frac{1}{2}(H - 1)\cos(\frac{\pi}{4})>0\) (since \(H>1\) and \(\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}>0\)).
For \(t=\frac{3\pi}{2}\), \(\frac{dH}{dt}=\frac{1}{2}(H - 1)\cos(\frac{\frac{3\pi}{2}}{2})=\frac{1}{2}(H - 1)\cos(\frac{3\pi}{4})<0\) (since \(\cos(\frac{3\pi}{4})=-\frac{\sqrt{2}}{2}<0\)).
Since \(\frac{dH}{dt}\) changes sign from positive to negative at \(t = \pi\), by the first - derivative test, \(H\) has a relative maximum at \(t=\pi\).

Step1: Separate the variables

Starting with \(\frac{dH}{dt}=\frac{1}{2}(H - 1)\cos(\frac{t}{2})\), we rewrite it as \(\frac{dH}{H - 1}=\frac{1}{2}\cos(\frac{t}{2})dt\).

Step2: Integrate both sides

Integrate \(\int\frac{dH}{H - 1}\) and \(\int\frac{1}{2}\cos(\frac{t}{2})dt\).
For \(\int\frac{dH}{H - 1}\), let \(u = H-1\), then \(du=dH\), and \(\int\frac{dH}{H - 1}=\ln|H - 1|+C_1\).
For \(\int\frac{1}{2}\cos(\frac{t}{2})dt\), let \(v=\frac{t}{2}\), then \(dt = 2dv\), and \(\int\frac{1}{2}\cos(\frac{t}{2})dt=\int\cos(v)dv=\sin(v)+C_2=\sin(\frac{t}{2})+C_2\).
So, \(\ln|H - 1|=\sin(\frac{t}{2})+C\).

Step3: Use the initial condition \(H(0)=4\)

Substitute \(t = 0\) and \(H=4\) into \(\ln|H - 1|=\sin(\frac{t}{2})+C\).
\(\ln|4 - 1|=\sin(0)+C\), \(\ln(3)=C\).

Step4: Solve for \(H\)

We have \(\ln(H - 1)=\sin(\frac{t}{2})+\ln(3)\) (since \(H>1\)). Using the property of logarithms \(\ln a-\ln b=\ln\frac{a}{b}\), we can rewrite it as \(\ln(\frac{H - 1}{3})=\sin(\frac{t}{2})\).
Exponentiating both sides gives \(\frac{H - 1}{3}=e^{\sin(\frac{t}{2})}\), and then \(H(t)=3e^{\sin(\frac{t}{2})}+1\).

Answer:

The value of \(t\) is \(\pi\). The critical point corresponds to a relative maximum.

Part (c)