Answer :
To determine when the water depth in the harbor reaches its maximum during the first 24 hours, we can analyze the function:
[tex]\[ f(t) = 4.1 \sin\left(\frac{\pi}{6} t - \frac{\pi}{3}\right) + 19.7 \][/tex]
This function models the water depth in feet after [tex]\( t \)[/tex] hours. The sine function, [tex]\(\sin(\theta)\)[/tex], reaches its maximum value of 1. So, the maximum value of [tex]\( f(t) \)[/tex] occurs when:
[tex]\[ \sin\left(\frac{\pi}{6} t - \frac{\pi}{3}\right) = 1 \][/tex]
To find the values of [tex]\( t \)[/tex] that satisfy this equation within the first 24 hours, follow these steps:
1. Find the angle for which the sine is 1:
The sine of an angle is 1 at [tex]\(\frac{\pi}{2}\)[/tex] plus any integer multiple of [tex]\(2\pi\)[/tex]:
[tex]\[ \frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} + 2k\pi \][/tex]
where [tex]\( k \)[/tex] is any integer.
2. Solve for [tex]\( t \)[/tex]:
Start with the principal value when [tex]\( k = 0 \)[/tex]:
[tex]\[ \frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} \][/tex]
Solve for [tex]\( t \)[/tex]:
Add [tex]\(\frac{\pi}{3}\)[/tex] to both sides:
[tex]\[ \frac{\pi}{6} t = \frac{\pi}{2} + \frac{\pi}{3} \][/tex]
Convert the right side to a common denominator:
[tex]\[ \frac{\pi}{6} t = \frac{3\pi}{6} + \frac{2\pi}{6} = \frac{5\pi}{6} \][/tex]
Multiply both sides by 6:
[tex]\[ \pi t = 5\pi \][/tex]
Divide both sides by [tex]\(\pi\)[/tex]:
[tex]\[ t = 5 \][/tex]
3. Determine additional times within the first 24 hours:
Since the period of the sine function [tex]\(\sin\left(\frac{\pi}{6} t - \frac{\pi}{3}\right)\)[/tex] is [tex]\( \frac{2\pi}{\frac{\pi}{6}} = 12 \)[/tex] hours, the water level reaches its maximum every 12 hours. Therefore, by adding the period or half-periods, we find more times:
- First maximum: [tex]\( t_1 = 5 \)[/tex] hours
- Second maximum: [tex]\( t_1 + 6 = 11 \)[/tex] hours
- Third maximum: [tex]\( t_1 + 12 = 17 \)[/tex] hours
- Fourth maximum: [tex]\( t_2 + 12 = 23 \)[/tex] hours
Therefore, during the first 24 hours, the water depth reaches a maximum at 5, 11, 17, and 23 hours.
[tex]\[ f(t) = 4.1 \sin\left(\frac{\pi}{6} t - \frac{\pi}{3}\right) + 19.7 \][/tex]
This function models the water depth in feet after [tex]\( t \)[/tex] hours. The sine function, [tex]\(\sin(\theta)\)[/tex], reaches its maximum value of 1. So, the maximum value of [tex]\( f(t) \)[/tex] occurs when:
[tex]\[ \sin\left(\frac{\pi}{6} t - \frac{\pi}{3}\right) = 1 \][/tex]
To find the values of [tex]\( t \)[/tex] that satisfy this equation within the first 24 hours, follow these steps:
1. Find the angle for which the sine is 1:
The sine of an angle is 1 at [tex]\(\frac{\pi}{2}\)[/tex] plus any integer multiple of [tex]\(2\pi\)[/tex]:
[tex]\[ \frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} + 2k\pi \][/tex]
where [tex]\( k \)[/tex] is any integer.
2. Solve for [tex]\( t \)[/tex]:
Start with the principal value when [tex]\( k = 0 \)[/tex]:
[tex]\[ \frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} \][/tex]
Solve for [tex]\( t \)[/tex]:
Add [tex]\(\frac{\pi}{3}\)[/tex] to both sides:
[tex]\[ \frac{\pi}{6} t = \frac{\pi}{2} + \frac{\pi}{3} \][/tex]
Convert the right side to a common denominator:
[tex]\[ \frac{\pi}{6} t = \frac{3\pi}{6} + \frac{2\pi}{6} = \frac{5\pi}{6} \][/tex]
Multiply both sides by 6:
[tex]\[ \pi t = 5\pi \][/tex]
Divide both sides by [tex]\(\pi\)[/tex]:
[tex]\[ t = 5 \][/tex]
3. Determine additional times within the first 24 hours:
Since the period of the sine function [tex]\(\sin\left(\frac{\pi}{6} t - \frac{\pi}{3}\right)\)[/tex] is [tex]\( \frac{2\pi}{\frac{\pi}{6}} = 12 \)[/tex] hours, the water level reaches its maximum every 12 hours. Therefore, by adding the period or half-periods, we find more times:
- First maximum: [tex]\( t_1 = 5 \)[/tex] hours
- Second maximum: [tex]\( t_1 + 6 = 11 \)[/tex] hours
- Third maximum: [tex]\( t_1 + 12 = 17 \)[/tex] hours
- Fourth maximum: [tex]\( t_2 + 12 = 23 \)[/tex] hours
Therefore, during the first 24 hours, the water depth reaches a maximum at 5, 11, 17, and 23 hours.
