Answer :
To determine the times when the water depth reaches a maximum during the first 24 hours, we need to analyze the function [tex]\( f(t) = 4.1 \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) + 19.7 \)[/tex].
The key to solving this is knowing that the sine function, [tex]\( \sin(x) \)[/tex], reaches its maximum value of 1. Therefore, we focus on when [tex]\( \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) = 1 \)[/tex].
1. Understand the Condition for Maximum:
The sine function reaches 1 at angles that are [tex]\( \frac{\pi}{2} \)[/tex] plus multiples of [tex]\( 2\pi \)[/tex]. So, we set up the equation:
[tex]\[
\frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} + 2n\pi
\][/tex]
where [tex]\( n \)[/tex] is an integer.
2. Solve for [tex]\( t \)[/tex]:
Let's solve the equation:
- Add [tex]\( \frac{\pi}{3} \)[/tex] to both sides:
[tex]\[
\frac{\pi}{6} t = \frac{\pi}{2} + \frac{\pi}{3} + 2n\pi
\][/tex]
- Combine terms on the right side:
- Convert [tex]\( \frac{\pi}{2} \)[/tex] and [tex]\( \frac{\pi}{3} \)[/tex] to a common denominator:
[tex]\[
\frac{\pi}{2} = \frac{3\pi}{6}, \quad \frac{\pi}{3} = \frac{2\pi}{6}
\][/tex]
- So:
[tex]\[
\frac{\pi}{6} t = \frac{3\pi}{6} + \frac{2\pi}{6} + 2n\pi = \frac{5\pi}{6} + 2n\pi
\][/tex]
- Solve for [tex]\( t \)[/tex]:
[tex]\[
t = \frac{6}{\pi} \left(\frac{5\pi}{6} + 2n\pi\right) = 5 + 12n
\][/tex]
3. Find the Relevant Values of [tex]\( t \)[/tex]:
We are interested in the values of [tex]\( t \)[/tex] that are within the first 24 hours (i.e., [tex]\( 0 \leq t < 24 \)[/tex]).
- Start with [tex]\( n = 0 \)[/tex]:
[tex]\[
t = 5 + 12(0) = 5
\][/tex]
- Next, try [tex]\( n = 1 \)[/tex]:
[tex]\[
t = 5 + 12(1) = 17
\][/tex]
- Since [tex]\( 5 \)[/tex] and [tex]\( 17 \)[/tex] are both within the interval [0, 24], these are the times when the water depth reaches a maximum.
Thus, the water depth reaches a maximum at 5 and 17 hours during the first 24 hours.
The key to solving this is knowing that the sine function, [tex]\( \sin(x) \)[/tex], reaches its maximum value of 1. Therefore, we focus on when [tex]\( \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) = 1 \)[/tex].
1. Understand the Condition for Maximum:
The sine function reaches 1 at angles that are [tex]\( \frac{\pi}{2} \)[/tex] plus multiples of [tex]\( 2\pi \)[/tex]. So, we set up the equation:
[tex]\[
\frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} + 2n\pi
\][/tex]
where [tex]\( n \)[/tex] is an integer.
2. Solve for [tex]\( t \)[/tex]:
Let's solve the equation:
- Add [tex]\( \frac{\pi}{3} \)[/tex] to both sides:
[tex]\[
\frac{\pi}{6} t = \frac{\pi}{2} + \frac{\pi}{3} + 2n\pi
\][/tex]
- Combine terms on the right side:
- Convert [tex]\( \frac{\pi}{2} \)[/tex] and [tex]\( \frac{\pi}{3} \)[/tex] to a common denominator:
[tex]\[
\frac{\pi}{2} = \frac{3\pi}{6}, \quad \frac{\pi}{3} = \frac{2\pi}{6}
\][/tex]
- So:
[tex]\[
\frac{\pi}{6} t = \frac{3\pi}{6} + \frac{2\pi}{6} + 2n\pi = \frac{5\pi}{6} + 2n\pi
\][/tex]
- Solve for [tex]\( t \)[/tex]:
[tex]\[
t = \frac{6}{\pi} \left(\frac{5\pi}{6} + 2n\pi\right) = 5 + 12n
\][/tex]
3. Find the Relevant Values of [tex]\( t \)[/tex]:
We are interested in the values of [tex]\( t \)[/tex] that are within the first 24 hours (i.e., [tex]\( 0 \leq t < 24 \)[/tex]).
- Start with [tex]\( n = 0 \)[/tex]:
[tex]\[
t = 5 + 12(0) = 5
\][/tex]
- Next, try [tex]\( n = 1 \)[/tex]:
[tex]\[
t = 5 + 12(1) = 17
\][/tex]
- Since [tex]\( 5 \)[/tex] and [tex]\( 17 \)[/tex] are both within the interval [0, 24], these are the times when the water depth reaches a maximum.
Thus, the water depth reaches a maximum at 5 and 17 hours during the first 24 hours.
