0
$\begingroup$

I have this question from my textbook, however I keep getting "unidentified", which is not the answer at the back. I was wondering what I'm doing wrong.

The question is: Given that in a water wheel the height (in meters) of a nail above the surface of the water is, as a function of time (in seconds), $h(t) = -4\sin(\frac{\pi}{4}(t-1))+2.5$, during what periods of time is the nail below the water in the first 24s of the wheel rotating?

What I tried was:

first i got the period $\begin{align*} \text{period} &= 2\pi \times 4/\pi = 8\\ &-4\sin\left(\frac{\pi}{4}\right)(t-1) + 2.5\\ -2.5/-4 &= \sin(\pi/4)t - \sin(\pi)\\ (\pi/4)t &= \sin^{-1}(-2.5/4 + \sin(\pi/4))\\ &= \text{unidentified..} \end{align*}$

normally after that I usually get a answer find the quadrant its in find the actuate angle then two possible angles(then more from adding the period), however i got unidentified

the answer in the textbook is $1.86s \lt t \lt 4.14s$ , $9.86s \lt t \lt12.14s$, $17.86 s\lt t\lt 20.14s$

  • 0
    @Faraz: I've edited your question so that it corresponds to what I think your textbook must have intended, i.e., $h(t) = -4\sin\left(\frac{\pi}{4}(t-1)\right)+2.5$. (See my answer below.)2011-12-09

1 Answers 1

1

graph of <span class=h(t)">

The problem requires finding the values of $t$ in the interval $(0,24)$, such that $h(t) = -4\sin(\frac{\pi}{4}(t-1))+2.5 < 0.$ Because $h$ is periodic with period $\frac{2\pi}{\frac{\pi}{4}} = 8$, the solution consists of three disjoint subintervals, which can be written as
$(a,b)$ $(a+8, b+8)$ $(a+16, b+16)$ The problem is then to find the values $a$ and $b$.

Now $a$ is just the least positive $t$ such that $h(t) = 0$, i.e., $\sin(\frac{\pi}{4}(t-1)) = \frac{5}{8}$:

$a = 1 + \frac{4}{\pi}\sin^{-1}(\frac{5}{8}) \approx 1.86.$

To find $b$, notice that $h(t)$ is a sinusoid whose minima occur for $t$ such that $\frac{\pi}{4}(t-1) \in \{\frac{\pi}{2} + 2\pi k: k \in \mathbb{Z} \}$, i.e., $t \in \{3 + 8k: k \in \mathbb{Z} \}$; furthermore, the least positive such $t$ (which is $3+0=3$) is just the midpoint of the subinterval $(a,b)$. Thus, $b = 3 + (3 - a) = 6 - a$.

Therefore, the required subintervals are

$(a,6-a) \approx (1.86,4.14)$ $(a+8, 14-a) \approx (9.86, 12.14)$ $(a+16, 20-a) \approx (17.86, 20.14).$