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Answer :
We are given the function
[tex]$$
f(t)=4.1 \sin \left(\frac{\pi}{6}t-\frac{\pi}{3}\right)+19.7
$$[/tex]
which models the water depth in feet after [tex]$t$[/tex] hours. The maximum water depth occurs when the sine part reaches its maximum of [tex]$1$[/tex]. That is, we need to find the values of [tex]$t$[/tex] for which
[tex]$$
\sin \left(\frac{\pi}{6}t-\frac{\pi}{3}\right)=1.
$$[/tex]
Since the sine function achieves its maximum value when its argument equals [tex]$\frac{\pi}{2}$[/tex] plus any integer multiple of [tex]$2\pi$[/tex], we set:
[tex]$$
\frac{\pi}{6}t-\frac{\pi}{3}=\frac{\pi}{2}+2\pi k,\quad k\in\mathbb{Z}.
$$[/tex]
Next, we solve for [tex]$t$[/tex]. First, add [tex]$\frac{\pi}{3}$[/tex] to both sides:
[tex]$$
\frac{\pi}{6}t=\frac{\pi}{2}+\frac{\pi}{3}+2\pi k.
$$[/tex]
Find a common denominator for the right-hand side:
[tex]$$
\frac{\pi}{2}+\frac{\pi}{3}=\frac{3\pi}{6}+\frac{2\pi}{6}=\frac{5\pi}{6}.
$$[/tex]
So the equation becomes:
[tex]$$
\frac{\pi}{6}t=\frac{5\pi}{6}+2\pi k.
$$[/tex]
Multiply both sides by [tex]$\frac{6}{\pi}$[/tex] to isolate [tex]$t$[/tex]:
[tex]$$
t=5+12k.
$$[/tex]
We now must find the values of [tex]$t$[/tex] within the first 24 hours, i.e., [tex]$0\le t\le24$[/tex]. Consider integer values of [tex]$k$[/tex]:
1. For [tex]$k=0$[/tex],
[tex]$$
t=5+12(0)=5.
$$[/tex]
2. For [tex]$k=1$[/tex],
[tex]$$
t=5+12(1)=17.
$$[/tex]
3. For [tex]$k=-1$[/tex],
[tex]$$
t=5+12(-1)=-7\quad \text{(which is not within the interval)}.
$$[/tex]
4. For [tex]$k=2$[/tex],
[tex]$$
t=5+12(2)=29\quad \text{(which is greater than 24)}.
$$[/tex]
Thus, the only times between [tex]$0$[/tex] and [tex]$24$[/tex] hours when [tex]$t=5$[/tex] and [tex]$t=17$[/tex] are the correct answers.
Therefore, during the first 24 hours, the water depth reaches a maximum at [tex]$t=5$[/tex] hours and [tex]$t=17$[/tex] hours.
[tex]$$
f(t)=4.1 \sin \left(\frac{\pi}{6}t-\frac{\pi}{3}\right)+19.7
$$[/tex]
which models the water depth in feet after [tex]$t$[/tex] hours. The maximum water depth occurs when the sine part reaches its maximum of [tex]$1$[/tex]. That is, we need to find the values of [tex]$t$[/tex] for which
[tex]$$
\sin \left(\frac{\pi}{6}t-\frac{\pi}{3}\right)=1.
$$[/tex]
Since the sine function achieves its maximum value when its argument equals [tex]$\frac{\pi}{2}$[/tex] plus any integer multiple of [tex]$2\pi$[/tex], we set:
[tex]$$
\frac{\pi}{6}t-\frac{\pi}{3}=\frac{\pi}{2}+2\pi k,\quad k\in\mathbb{Z}.
$$[/tex]
Next, we solve for [tex]$t$[/tex]. First, add [tex]$\frac{\pi}{3}$[/tex] to both sides:
[tex]$$
\frac{\pi}{6}t=\frac{\pi}{2}+\frac{\pi}{3}+2\pi k.
$$[/tex]
Find a common denominator for the right-hand side:
[tex]$$
\frac{\pi}{2}+\frac{\pi}{3}=\frac{3\pi}{6}+\frac{2\pi}{6}=\frac{5\pi}{6}.
$$[/tex]
So the equation becomes:
[tex]$$
\frac{\pi}{6}t=\frac{5\pi}{6}+2\pi k.
$$[/tex]
Multiply both sides by [tex]$\frac{6}{\pi}$[/tex] to isolate [tex]$t$[/tex]:
[tex]$$
t=5+12k.
$$[/tex]
We now must find the values of [tex]$t$[/tex] within the first 24 hours, i.e., [tex]$0\le t\le24$[/tex]. Consider integer values of [tex]$k$[/tex]:
1. For [tex]$k=0$[/tex],
[tex]$$
t=5+12(0)=5.
$$[/tex]
2. For [tex]$k=1$[/tex],
[tex]$$
t=5+12(1)=17.
$$[/tex]
3. For [tex]$k=-1$[/tex],
[tex]$$
t=5+12(-1)=-7\quad \text{(which is not within the interval)}.
$$[/tex]
4. For [tex]$k=2$[/tex],
[tex]$$
t=5+12(2)=29\quad \text{(which is greater than 24)}.
$$[/tex]
Thus, the only times between [tex]$0$[/tex] and [tex]$24$[/tex] hours when [tex]$t=5$[/tex] and [tex]$t=17$[/tex] are the correct answers.
Therefore, during the first 24 hours, the water depth reaches a maximum at [tex]$t=5$[/tex] hours and [tex]$t=17$[/tex] hours.
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