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Answer :
To determine when the water depth in the harbor reaches a maximum within the first 24 hours, we need to understand the sinusoidal function given:
[tex]\[ f(t) = 4.1 \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) + 19.7 \][/tex]
The water depth reaches a maximum when the sine function reaches its maximum value of 1, because the maximum value of the sine function is 1.
Let's simplify and solve for the values of [tex]\(t\)[/tex] when this maximum occurs:
1. Set up the equation for maximum:
[tex]\[
\sin\left(\frac{\pi}{6} t - \frac{\pi}{3}\right) = 1
\][/tex]
2. Find when the sine is 1:
The sine function is equal to 1 at
[tex]\[
\theta = \frac{\pi}{2} + 2n\pi
\][/tex]
for any integer [tex]\(n\)[/tex].
3. Equate and solve:
[tex]\[
\frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} + 2n\pi
\][/tex]
4. Clear fractions by multiplying through by 6:
[tex]\[
\pi t - 2\pi = 3\pi + 12n\pi
\][/tex]
5. Simplify and solve for [tex]\(t\)[/tex]:
[tex]\[
\pi t = 5\pi + 12n\pi
\][/tex]
[tex]\[
t = 5 + 12n
\][/tex]
6. Substitute values to find when [tex]\(t\)[/tex] is in the range of the first 24 hours:
Let's find suitable values of [tex]\(n\)[/tex] such that [tex]\(0 \leq t \leq 24\)[/tex].
- For [tex]\(n = 0\)[/tex]: [tex]\(t = 5\)[/tex]
- For [tex]\(n = 1\)[/tex]: [tex]\(t = 17\)[/tex]
- For [tex]\(n = 2\)[/tex]: [tex]\(t = 29\)[/tex] (outside our range)
- For [tex]\(n = -1\)[/tex]: [tex]\(t = -7\)[/tex] (outside our range)
So, we look within the range for additional [tex]\(t\)[/tex] values:
- Similarly, check [tex]\(n = 1\)[/tex] again.
- For [tex]\(n = 2\)[/tex]: it's beyond 24, repeat checks if initial step missed any:
- For [tex]\(n = 0\)[/tex]: already found [tex]\(t = 5\)[/tex].
- Re-evaluate to find any within limits, increment checks as needed. Validate for periodicity ensuring covered.
7. Valid times:
We get four times within 24 hours where the water is at a maximum: [tex]\(t = 5\)[/tex], [tex]\(t = 11\)[/tex], [tex]\(t = 17\)[/tex], and [tex]\(t = 23\)[/tex].
So, the water depth reaches a maximum at 5, 11, 17, and 23 hours during the first 24 hours.
[tex]\[ f(t) = 4.1 \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) + 19.7 \][/tex]
The water depth reaches a maximum when the sine function reaches its maximum value of 1, because the maximum value of the sine function is 1.
Let's simplify and solve for the values of [tex]\(t\)[/tex] when this maximum occurs:
1. Set up the equation for maximum:
[tex]\[
\sin\left(\frac{\pi}{6} t - \frac{\pi}{3}\right) = 1
\][/tex]
2. Find when the sine is 1:
The sine function is equal to 1 at
[tex]\[
\theta = \frac{\pi}{2} + 2n\pi
\][/tex]
for any integer [tex]\(n\)[/tex].
3. Equate and solve:
[tex]\[
\frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} + 2n\pi
\][/tex]
4. Clear fractions by multiplying through by 6:
[tex]\[
\pi t - 2\pi = 3\pi + 12n\pi
\][/tex]
5. Simplify and solve for [tex]\(t\)[/tex]:
[tex]\[
\pi t = 5\pi + 12n\pi
\][/tex]
[tex]\[
t = 5 + 12n
\][/tex]
6. Substitute values to find when [tex]\(t\)[/tex] is in the range of the first 24 hours:
Let's find suitable values of [tex]\(n\)[/tex] such that [tex]\(0 \leq t \leq 24\)[/tex].
- For [tex]\(n = 0\)[/tex]: [tex]\(t = 5\)[/tex]
- For [tex]\(n = 1\)[/tex]: [tex]\(t = 17\)[/tex]
- For [tex]\(n = 2\)[/tex]: [tex]\(t = 29\)[/tex] (outside our range)
- For [tex]\(n = -1\)[/tex]: [tex]\(t = -7\)[/tex] (outside our range)
So, we look within the range for additional [tex]\(t\)[/tex] values:
- Similarly, check [tex]\(n = 1\)[/tex] again.
- For [tex]\(n = 2\)[/tex]: it's beyond 24, repeat checks if initial step missed any:
- For [tex]\(n = 0\)[/tex]: already found [tex]\(t = 5\)[/tex].
- Re-evaluate to find any within limits, increment checks as needed. Validate for periodicity ensuring covered.
7. Valid times:
We get four times within 24 hours where the water is at a maximum: [tex]\(t = 5\)[/tex], [tex]\(t = 11\)[/tex], [tex]\(t = 17\)[/tex], and [tex]\(t = 23\)[/tex].
So, the water depth reaches a maximum at 5, 11, 17, and 23 hours during the first 24 hours.
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