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Question 3

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The depth of seawater at a spot is H(t)H(t)H(t) feet, ttt hours after noon. It satisfies

dHdt=12 (H−1)cos⁡ ⁣(t2),H(0)=4.\frac{dH}{dt} = \frac{1}{2}\,(H - 1)\cos\!\left(\frac{t}{2}\right), \qquad H(0) = 4.dtdH​=21​(H−1)cos(2t​),H(0)=4.
1234562468tH(0, 4)
  1. (a)\text{(a)}(a)

    Part of the slope field is drawn. Sketch the solution curve y=H(t)y = H(t)y=H(t) through (0,4)(0, 4)(0,4).

  2. (b)\text{(b)}(b)

    For 0<t<50 < t < 50<t<5 it can be shown that H(t)>1H(t) > 1H(t)>1. Find the ttt in 0<t<50 < t < 50<t<5 where HHH has a critical point, and decide whether it is a relative minimum, a relative maximum, or neither. Justify your answer.

  3. (c)\text{(c)}(c)

    Use separation of variables to find the particular solution y=H(t)y = H(t)y=H(t) with H(0)=4H(0) = 4H(0)=4.

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This question is about the depth of seawater rising and falling over time

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