# When is a nail in a rotating wheel below a stated height, given its height as a displaced sinusoidal function of time?

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 $1.86s \lt t \lt 4.14s$ , $9.86s \lt t \lt12.14s$, $17.86 s\lt t\lt 20.14s$

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The question is not clear. – NoChance Dec 8 '11 at 4:41
my bad forgot the last sentence – Faraz Dec 8 '11 at 4:43
It's hard to tell what function you have, because of your use of inline / and lack of parentheses. Does $\sin\pi/4(t-1)+2.5$ mean $$\sin\left(\frac{\pi}{4(t-1)}\right)+2.5,\text{ or }\sin\left(\frac{\pi}{4}(t-1)\right)+2.5$$ or $$\sin\left(\frac{\pi}{4}\right)(t-1)+2.5\text{ or }\sin\left(\frac{\pi}{4}(t-1) + 2.5\right)$$ or... ? – Arturo Magidin Dec 8 '11 at 4:50
the 3rd one is correct – Faraz Dec 8 '11 at 4:55
If the 3rd one is correct then the $t-1$ is outside the scope of the sine function, so the function isn't periodic, and none of your work makes sense. I suspect it's actually the 2nd of Arturo's alternatives you want. – Gerry Myerson Dec 8 '11 at 5:02

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

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

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