To solve this problem, let's break it down step by step. We want to find a function [tex]\( d(t) \)[/tex] that models the depth of the water in feet as a function of time [tex]\( t \)[/tex] in minutes past high tide.
1. Given Data:
- Maximum depth (high tide): [tex]\( 8 \)[/tex] feet
- Minimum depth (low tide): [tex]\( 4 \)[/tex] feet
- The time it takes for the tide to go from high tide to low tide and back to high tide again is typically around 12.4 hours (744 minutes).
2. Key Information:
- The midpoint (average depth) between the maximum and minimum depth:
[tex]\[
\text{Average depth} = \frac{8 + 4}{2} = 6 \text{ feet}
\][/tex]
- The amplitude of the sine function, which represents the variation from the average depth:
[tex]\[
\text{Amplitude} = \frac{8 - 4}{2} = 2 \text{ feet}
\][/tex]
- The period of the sine function, which is the time it takes to complete one full cycle (high tide to low tide and back to high tide):
[tex]\[
\text{Period} = 744 \text{ minutes}
\][/tex]
- Since we're working with a sine function, the period can be written as:
[tex]\[
\frac{2\pi}{k} = 744 \implies k = \frac{2\pi}{744} = \frac{\pi}{372}
\][/tex]
3. Forming the Function:
- The general sine function with the above details can be written as:
[tex]\[
d(t) = A \sin(Bt) + C
\][/tex]
where [tex]\( A \)[/tex] is the amplitude, [tex]\( B \)[/tex] determines the period, and [tex]\( C \)[/tex] is the average depth.
- Plugging in our values:
[tex]\[
d(t) = 2 \sin\left(\frac{\pi t}{372}\right) + 6
\][/tex]
This matches one of the given options exactly:
[tex]\[
d(t) = 2 \sin\left(\frac{\pi t}{372}\right) + 6
\][/tex]
Thus, the correct function that models the depth [tex]\( d(t) \)[/tex] in feet as a function of time [tex]\( t \)[/tex] in minutes past high tide is:
[tex]\[
d(t) = 2 \sin\left(\frac{\pi t}{372}\right) + 6
\][/tex]
