Answer :
To solve the problem of identifying when the water depth in the harbor reaches its maximum within the first 24 hours, we'll analyze the function:
[tex]\[ f(t) = 4.1 \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) + 19.7 \][/tex]
### Step-by-Step Solution:
1. Understand the Sine Function:
The sine function, [tex]\(\sin(x)\)[/tex], reaches its maximum value of 1 at angles [tex]\(x = \frac{\pi}{2} + 2\pi n\)[/tex], where [tex]\(n\)[/tex] is an integer.
2. Set the Expression to Maximize Sine:
To find when the sine part of the water depth function is at its maximum, set:
[tex]\[
\frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} + 2\pi n
\][/tex]
3. Solve for [tex]\(t\)[/tex]:
Rearrange the equation and solve for [tex]\(t\)[/tex]:
[tex]\[
\frac{\pi}{6} t = \frac{\pi}{2} + \frac{\pi}{3} + 2\pi n
\][/tex]
Simplify the right-hand side:
[tex]\[
\frac{\pi}{2} + \frac{\pi}{3} = \frac{3\pi}{6} + \frac{2\pi}{6} = \frac{5\pi}{6}
\][/tex]
Now substitute back:
[tex]\[
\frac{\pi}{6} t = \frac{5\pi}{6} + 2\pi n
\][/tex]
Multiply both sides by [tex]\(\frac{6}{\pi}\)[/tex] to isolate [tex]\(t\)[/tex]:
[tex]\[
t = 5 + 12n
\][/tex]
This gives general times [tex]\(t = 5 + 12n\)[/tex], but we need to consider times within the first 24 hours.
4. Calculate Specific Times:
For [tex]\(n = 0\)[/tex], we get [tex]\(t = 5\)[/tex].
For [tex]\(n = 1\)[/tex], we get [tex]\(t = 17\)[/tex].
For [tex]\(n = 2\)[/tex], [tex]\(t = 29\)[/tex] which is outside the 24-hour range, so we don't include it.
5. Consider Cycle and Frequency:
The sine wave also has periodic maxima. With a period of [tex]\(\frac{12}{\pi} \times 2\pi = 12\)[/tex] hours, add 6 hours (half the period) to find intermediate maxima:
- [tex]\(5 + 6 = 11\)[/tex]
- [tex]\(17 + 6 = 23\)[/tex]
6. List All Maximum Times:
Compiling all valid times within the first 24 hours, we get: 5, 11, 17, and 23 hours.
Thus, the water depth reaches its maximum at 5, 11, 17, and 23 hours within the first 24 hours.
[tex]\[ f(t) = 4.1 \sin \left(\frac{\pi}{6} t - \frac{\pi}{3}\right) + 19.7 \][/tex]
### Step-by-Step Solution:
1. Understand the Sine Function:
The sine function, [tex]\(\sin(x)\)[/tex], reaches its maximum value of 1 at angles [tex]\(x = \frac{\pi}{2} + 2\pi n\)[/tex], where [tex]\(n\)[/tex] is an integer.
2. Set the Expression to Maximize Sine:
To find when the sine part of the water depth function is at its maximum, set:
[tex]\[
\frac{\pi}{6} t - \frac{\pi}{3} = \frac{\pi}{2} + 2\pi n
\][/tex]
3. Solve for [tex]\(t\)[/tex]:
Rearrange the equation and solve for [tex]\(t\)[/tex]:
[tex]\[
\frac{\pi}{6} t = \frac{\pi}{2} + \frac{\pi}{3} + 2\pi n
\][/tex]
Simplify the right-hand side:
[tex]\[
\frac{\pi}{2} + \frac{\pi}{3} = \frac{3\pi}{6} + \frac{2\pi}{6} = \frac{5\pi}{6}
\][/tex]
Now substitute back:
[tex]\[
\frac{\pi}{6} t = \frac{5\pi}{6} + 2\pi n
\][/tex]
Multiply both sides by [tex]\(\frac{6}{\pi}\)[/tex] to isolate [tex]\(t\)[/tex]:
[tex]\[
t = 5 + 12n
\][/tex]
This gives general times [tex]\(t = 5 + 12n\)[/tex], but we need to consider times within the first 24 hours.
4. Calculate Specific Times:
For [tex]\(n = 0\)[/tex], we get [tex]\(t = 5\)[/tex].
For [tex]\(n = 1\)[/tex], we get [tex]\(t = 17\)[/tex].
For [tex]\(n = 2\)[/tex], [tex]\(t = 29\)[/tex] which is outside the 24-hour range, so we don't include it.
5. Consider Cycle and Frequency:
The sine wave also has periodic maxima. With a period of [tex]\(\frac{12}{\pi} \times 2\pi = 12\)[/tex] hours, add 6 hours (half the period) to find intermediate maxima:
- [tex]\(5 + 6 = 11\)[/tex]
- [tex]\(17 + 6 = 23\)[/tex]
6. List All Maximum Times:
Compiling all valid times within the first 24 hours, we get: 5, 11, 17, and 23 hours.
Thus, the water depth reaches its maximum at 5, 11, 17, and 23 hours within the first 24 hours.
