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Answer :
To find when the water depth in the harbor reaches a maximum during the first 24 hours, we need to analyze the given function:
[tex]\[ f(t) = 4.1 \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) + 19.7 \][/tex]
The sine function, [tex]\(\sin(x)\)[/tex], reaches its maximum value of 1 at certain points. For the function [tex]\(f(t)\)[/tex], this happens when:
[tex]\[ \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) = 1 \][/tex]
The sine function equals 1 at:
[tex]\[ x = \frac{\pi}{2} + 2n\pi \][/tex]
(where [tex]\(n\)[/tex] is an integer)
Now, set the inside of the sine function equal to these points:
[tex]\[ \frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} + 2n\pi \][/tex]
To solve for [tex]\(t\)[/tex], rearrange the above equation:
1. Add [tex]\(\frac{\pi}{3}\)[/tex] to both sides:
[tex]\[ \frac{\pi}{6} t = \frac{\pi}{2} + \frac{\pi}{3} + 2n\pi \][/tex]
2. Find a common denominator for the terms on the right:
[tex]\[ \frac{\pi}{2} = \frac{3\pi}{6} \][/tex]
[tex]\[ \frac{\pi}{3} = \frac{2\pi}{6} \][/tex]
So,
[tex]\[ \frac{3\pi}{6} + \frac{2\pi}{6} = \frac{5\pi}{6} \][/tex]
Thus,
[tex]\[ \frac{\pi}{6} t = \frac{5\pi}{6} + 2n\pi \][/tex]
3. Multiply the entire equation by 6 to solve for [tex]\(t\)[/tex]:
[tex]\[ t = 5 + 12n \][/tex]
Now, we determine the specific values of [tex]\(t\)[/tex] within the first 24 hours, where [tex]\(0 \leq t < 24\)[/tex]:
- For [tex]\(n = 0\)[/tex]:
[tex]\[ t = 5 + 12 \times 0 = 5 \][/tex]
- For [tex]\(n = 1\)[/tex]:
[tex]\[ t = 5 + 12 \times 1 = 17 \][/tex]
Since 17 is within the 24-hour period, these are the times when the water depth reaches a maximum.
Thus, the water depth in the harbor reaches its maximum at 5 hours and 17 hours during the first 24-hour period.
[tex]\[ f(t) = 4.1 \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) + 19.7 \][/tex]
The sine function, [tex]\(\sin(x)\)[/tex], reaches its maximum value of 1 at certain points. For the function [tex]\(f(t)\)[/tex], this happens when:
[tex]\[ \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) = 1 \][/tex]
The sine function equals 1 at:
[tex]\[ x = \frac{\pi}{2} + 2n\pi \][/tex]
(where [tex]\(n\)[/tex] is an integer)
Now, set the inside of the sine function equal to these points:
[tex]\[ \frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} + 2n\pi \][/tex]
To solve for [tex]\(t\)[/tex], rearrange the above equation:
1. Add [tex]\(\frac{\pi}{3}\)[/tex] to both sides:
[tex]\[ \frac{\pi}{6} t = \frac{\pi}{2} + \frac{\pi}{3} + 2n\pi \][/tex]
2. Find a common denominator for the terms on the right:
[tex]\[ \frac{\pi}{2} = \frac{3\pi}{6} \][/tex]
[tex]\[ \frac{\pi}{3} = \frac{2\pi}{6} \][/tex]
So,
[tex]\[ \frac{3\pi}{6} + \frac{2\pi}{6} = \frac{5\pi}{6} \][/tex]
Thus,
[tex]\[ \frac{\pi}{6} t = \frac{5\pi}{6} + 2n\pi \][/tex]
3. Multiply the entire equation by 6 to solve for [tex]\(t\)[/tex]:
[tex]\[ t = 5 + 12n \][/tex]
Now, we determine the specific values of [tex]\(t\)[/tex] within the first 24 hours, where [tex]\(0 \leq t < 24\)[/tex]:
- For [tex]\(n = 0\)[/tex]:
[tex]\[ t = 5 + 12 \times 0 = 5 \][/tex]
- For [tex]\(n = 1\)[/tex]:
[tex]\[ t = 5 + 12 \times 1 = 17 \][/tex]
Since 17 is within the 24-hour period, these are the times when the water depth reaches a maximum.
Thus, the water depth in the harbor reaches its maximum at 5 hours and 17 hours during the first 24-hour period.
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