Question

example 2 modeling tides
by studying the tides in a seaport the following sinusoidal function was found to
model the depth of the water, d(t), in meters, at the docking position.
d(t) = 3 sin 0.16π(t − 3) + 10.2 where t is the time in hours
a) sketch the function below in order to set your window, and use your graphing
calculator.
b) a large freighter needs a depth of at least 9 m to dock safely. for how many
hours per tide cycle can the ship dock safely?

Explanation:

Step1: Set up the inequality

We need to find when \( d(t) \geq 9 \). So we set up the inequality:
\( 3\sin(0.16\pi(t - 3))+ 10.2\geq9 \)
Subtract \( 10.2 \) from both sides:
\( 3\sin(0.16\pi(t - 3))\geq9 - 10.2=- 1.2 \)
Divide both sides by \( 3 \):
\( \sin(0.16\pi(t - 3))\geq\frac{-1.2}{3}=- 0.4 \)

Step2: Solve the sine inequality

We know that the general solution for \( \sin(x)\geq a \) (where \( - 1\leq a\leq1 \)) can be found using the inverse sine function. Let \( x = 0.16\pi(t - 3) \). Then:
\( \arcsin(-0.4)\leq0.16\pi(t - 3)\leq\pi-\arcsin(-0.4) \)
We know that \( \arcsin(-0.4)\approx - 0.4115 \) (in radians). So:
\( - 0.4115\leq0.16\pi(t - 3)\leq\pi + 0.4115\approx3.5531 \)

Step3: Solve for \( t \)

First, divide all parts of the inequality by \( 0.16\pi \):
\( \frac{-0.4115}{0.16\pi}\leq t - 3\leq\frac{3.5531}{0.16\pi} \)
Calculate \( \frac{-0.4115}{0.16\pi}\approx\frac{-0.4115}{0.5027}\approx - 0.8186 \) and \( \frac{3.5531}{0.16\pi}\approx\frac{3.5531}{0.5027}\approx7.068 \)
Then add \( 3 \) to all parts:
\( - 0.8186+3\leq t\leq7.068 + 3 \)
\( 2.1814\leq t\leq10.068 \)

But we also need to consider the period of the sine function. The period \( T \) of \( y = A\sin(B(x - C))+D \) is \( T=\frac{2\pi}{B} \). Here \( B = 0.16\pi \), so \( T=\frac{2\pi}{0.16\pi}=\frac{2}{0.16}=12.5 \) hours.

The general solution for the inequality \( \sin(0.16\pi(t - 3))\geq - 0.4 \) over one period (since the tide is a periodic phenomenon) will be \( t\in[3 + \frac{\arcsin(-0.4)}{0.16\pi},3+\frac{\pi-\arcsin(-0.4)}{0.16\pi}] \) and we can also consider the other interval in the period. But since we are looking for the time per tide cycle (one period), we can calculate the length of the interval \( [2.1814,10.068] \) (after adjusting for the period, but in this case, the first interval we found within one period). The length of the interval is \( 10.068 - 2.1814\approx7.8866 \) hours. Wait, but let's do it more accurately.

Let's go back to the inequality \( \sin(0.16\pi(t - 3))\geq - 0.4 \)

The solutions of \( \sin\theta=-0.4 \) are \( \theta=\arcsin(-0.4)+2k\pi \) or \( \theta=\pi-\arcsin(-0.4)+2k\pi \), \( k\in\mathbb{Z} \)

For \( k = 0 \):
\( \theta_1=\arcsin(-0.4)\approx - 0.4115 \)
\( \theta_2=\pi-\arcsin(-0.4)\approx3.1416 + 0.4115 = 3.5531 \)

Now, \( \theta = 0.16\pi(t - 3) \)

So for \( \theta_1 \):
\( 0.16\pi(t - 3)=-0.4115 \)
\( t - 3=\frac{-0.4115}{0.16\pi}\approx\frac{-0.4115}{0.5027}\approx - 0.8186 \)
\( t\approx3 - 0.8186 = 2.1814 \)

For \( \theta_2 \):
\( 0.16\pi(t - 3)=3.5531 \)
\( t - 3=\frac{3.5531}{0.16\pi}\approx\frac{3.5531}{0.5027}\approx7.068 \)
\( t\approx3 + 7.068 = 10.068 \)

The length of the interval \( [2.1814,10.068] \) is \( 10.068 - 2.1814 = 7.8866\approx7.89 \) hours. But let's check with the period. The period is \( T = 12.5 \) hours.

Alternatively, we can use the formula for the length of time when \( \sin(x)\geq a \). The length of the interval where \( \sin(x)\geq a \) in \( [0,2\pi] \) is \( 2(\pi-\arcsin(a)) \) when \( a\geq0 \) and \( 2(\pi-\arcsin(-a)) \) when \( a < 0 \)? Wait, no. Wait, when \( a=-0.4 \) (negative), the solutions in \( [0,2\pi] \) are from \( \arcsin(-0.4) \) to \( \pi-\arcsin(-0.4) \)? No, wait, the sine function is greater than or equal to \( - 0.4 \) between \( \arcsin(-0.4) \) and \( \pi-\arcsin(-0.4) \) in \( [0,2\pi] \)? Wait, no, let's think about the unit circle. The sine is negative in the third and fourth quadrants. \( \sin\theta=-0.4 \) has solutions in the third ( \( \pi+\arcsin(0.4) \)) and fourth ( \( 2\pi-\arcsin(0.4) \)) quadrants? Wait, I made a mistake earlier.

Wait, \( \sin\theta=-0.4 \)

\( \theta=\arcsin(-0.4)+2k\pi \) or \( \theta=\pi-\arcsin(-0.4)+2k\pi \)? No, that's wrong. The correct solutions are \( \theta=\arcsin(-0.4)+2k\pi \) or \( \theta=\pi-\arcsin(-0.4)+2k\pi \) is incorrect. The correct formula is: if \( \sin\theta = y \), then \( \theta=\arcsin(y)+2k\pi \) or \( \theta=\pi-\arcsin(y)+2k\pi \), \( k\in\mathbb{Z} \). But when \( y \) is negative, \( \arcsin(y) \) is negative. So for \( y = - 0.4 \), \( \arcsin(-0.4)\approx - 0.4115 \), so the solutions are \( \theta=-0.4115 + 2k\pi \) or \( \theta=\pi+0.4115 + 2k\pi \) (because \( \pi-\arcsin(-0.4)=\pi + 0.4115 \))

Ah, here is the mistake. So the correct solutions for \( \sin\theta=-0.4 \) are \( \theta=\arcsin(-0.4)+2k\pi \) or \( \theta=\pi-\arcsin(-0.4)+2k\pi=\pi+\vert\arcsin(-0.4)\vert+2k\pi \)

So for \( k = 0 \):
\( \theta_1=-0.4115 \)
\( \theta_2=\pi + 0.4115\approx3.5531 \)

For \( k = 1 \):
\( \theta_3=-0.4115 + 2\pi\approx - 0.4115+6.2832 = 5.8717 \)
\( \theta_4=\pi + 0.4115+2\pi=3\pi + 0.4115\approx9.8357 \)

Now, our \( \theta = 0.16\pi(t - 3) \), and we are looking for \( t \) in one period ( \( T = 12.5 \) hours), so \( \theta\in[0,2\pi] \) (since when \( t = 0 \), \( \theta=-0.48\pi\approx - 1.508 \); when \( t = 12.5 \), \( \theta=0.16\pi(12.5 - 3)=0.16\pi\times9.5 = 1.52\pi\approx4.775 \)). Wait, maybe it's better to consider \( t \) from \( 0 \) to \( 12.5 \) (one period).

We need to find \( t \) such that \( \sin(0.16\pi(t - 3))\geq - 0.4 \)

Let's find the values of \( t \) where \( \sin(0.16\pi(t - 3))=-0.4 \)

First solution:
\( 0.16\pi(t - 3)=\arcsin(-0.4)+2\pi \) (because \( \arcsin(-0.4) \) is negative, so we add \( 2\pi \) to get it in \( [0,2\pi] \))
\( 0.16\pi(t - 3)=-0.4115 + 6.2832 = 5.8717 \)
\( t - 3=\frac{5.8717}{0.16\pi}\approx\frac{5.8717}{0.5027}\approx11.68 \) (but \( t = 11.68+3 = 14.68 \), which is more than one period \( 12.5 \), so we take \( k = 0 \) for the other solution)

Second solution:
\( 0.16\pi(t - 3)=\pi-\arcsin(-0.4)=\pi + 0.4115 = 3.5531 \)
\( t - 3=\frac{3.5531}{0.16\pi}\approx7.068 \)
\( t\approx10.068 \)

First solution in the period:
\( 0.16\pi(t - 3)=\arcsin(-0.4)+2\pi \) is too big, so we take \( 0.16\pi(t - 3)=\arcsin(-0.4) \)
\( t - 3=\frac{-0.4115}{0.16\pi}\approx - 0.8186 \)
\( t\approx2.1814 \)

Wait, but when \( t = 2.1814 \), \( \theta=0.16\pi(2.1814 - 3)=0.16\pi(-0.8186)\approx - 0.4115 \), which is \( \arcsin(-0.4) \)

And when \( t = 10.068 \), \( \theta=0.16\pi(10.068 - 3)=0.16\pi(7.068)\approx3.5531 \), which is \( \pi-\arcsin(-0.4) \)

Now, the sine function is greater than or equal to \( - 0.4 \) between \( \theta=\arcsin(-0.4) \) and \( \theta=\pi-\arcsin(-0.4) \) (because in the unit circle, between the fourth quadrant angle and the second quadrant angle, the sine is greater than or equal to \( - 0.4 \))

So the length of the interval in \( \theta \) is \( (\pi-\arcsin(-0.4))-\arcsin(-0.4)=\pi - 2\arcsin(-0.4)=\pi + 2\arcsin(0.4) \) (since \( \arcsin(-x)=-\arcsin(x) \))

Calculate \( \arcsin(0.4)\approx0.4115 \), so \( \pi + 2\times0.4115\approx3.1416+0.823\approx3.9646 \) radians

Now, the length of time is \( \frac{\text{length of } \theta \text{ interval}}{B} \), where \( B = 0.16\pi \)

So time length \( =\frac{3.9646}{0.16\pi}\approx\frac{3.9646}{0.5027}\approx7.886 \) hours, approximately \( 7.89 \) hours.

But let's check with the function \( d(t)=3\sin(0.16\pi(t - 3))+10.2 \)

We can also use the formula for the period \( T = 12.5 \) hours. The amplitude is \( 3 \), midline is \( 10.2 \). The minimum depth is \( 10.2 - 3 = 7.2 \) meters, maximum is \( 10.2 + 3 = 13.2 \) meters. We need depth \( \geq9 \) meters.

So \( 3\sin(0.16\pi(t - 3))+10.2\geq9 \)
\( \sin(0.16\pi(t - 3))\geq\frac{9 - 10.2}{3}=-0.4 \)

The general solution for \( \sin(x)\geq - 0.4 \) is \( 2k\pi+\arcsin(-0.4)\leq x\leq2k\pi+\pi-\arcsin(-0.4) \), \( k\in\mathbb{Z} \)

Substitute \( x = 0.16\pi(t - 3) \):

\( 2k\pi+\arcsin(-0.4)\leq0.16\pi(t - 3)\leq2k\pi+\pi-\arcsin(-0.4) \)

Divide by \( 0.16\pi \):

\( \frac{2k\pi}{0.16\pi}+\frac{\arcsin(-0.4)}{0.16\pi}\leq t - 3\leq\frac{2k\pi}{0.16\pi}+\frac{\pi-\arcsin(-0.4)}{0.16\pi} \)

Simplify:

\( \frac{2k}{0.16}+\frac{\arcsin(-0.4)}{0.16\pi}\leq t - 3\leq\frac{2k}{0.16}+\frac{\pi-\arcsin(-0.4)}{0.16\pi} \)

\( 12.5k+\frac{-0.4115}{0.5027}\leq t - 3\leq12.5k+\frac{3.5531}{0.5027} \)

\( 12.5k - 0.8186\leq t - 3\leq12.5k + 7.068 \)

\( 12.5k + 2.1814\leq t\leq12.5k + 10.068 \)

For \( k = 0 \), the interval is \( [2.1814,10.068] \), length \( 10.068 - 2.1814 = 7.8866\approx7.89 \) hours.

So the ship can dock safely for approximately \( 7.89 \) hours per tide cycle.

Answer:

Approximately \( 7.89 \) hours (or more precisely, using exact calculations, we can get the length as \( \frac{\pi - 2\arcsin(-0.4)}{0.16\pi}=\frac{\pi + 2\arcsin(0.4)}{0.16\pi}\approx\frac{3.1416+0.823}{0.5027}\approx\frac{3.9646}{0.5027}\approx7.89 \) hours)