Math / Algebra

QuestionThe figure shows the depth of water at the end of a boat dock. The depth is 6 feet at low tide and 14 feet at high tide. On a certain day, low tide occurs at 6 AM and high tide at noon. If $y$ represents the depth of the water $x$ hours after midnight, use a cosine function of the form $y=A \cos B x+D$ to model the water's depth.
Write the equation for this problem in the form $y=A \cos B x+D$ . $\mathrm{y}=\square$ (Type an expression using $x$ as the variable. Type an exact answer, using as needed. Use integers or fractions for any numbers in the expression.)

Studdy Solution

STEP 1

What is this asking? We need to find a cosine function that describes the depth of the water at any time, knowing the highest and lowest depths and when they occur. Watch out! Remember that the cosine function starts at its maximum value, so we might need to adjust our function to match the given times for high and low tides.

STEP 2

1. Find the midline.
2. Find the amplitude.
3. Find the period.
4. Find the phase shift.
5. Write the equation.

STEP 3

The midline is the average of the highest and lowest depths.
Think of it like the "center" of the tide's motion!

STEP 4

$D = \frac{\textbf{14 feet} + \textbf{6 feet}}{2} = \frac{20 \text{ feet}}{2} = \textbf{10 feet}$ So, our midline is at $\textbf{10 feet}$ .

STEP 5

The amplitude is half the difference between the highest and lowest points.
It tells us how much the tide goes up and down from the midline.

STEP 6

$A = \frac{\textbf{14 feet} - \textbf{6 feet}}{2} = \frac{8 \text{ feet}}{2} = \textbf{4 feet}$ Our amplitude is $\textbf{4 feet}$ .

STEP 7

The period is the time it takes for one full cycle of the tide (from high to low and back to high).
We know high tide is at noon and low tide is at 6 AM, so half a cycle is 6 hours.

STEP 8

A full cycle must be double that, which is $2 \cdot 6 \text{ hours} = \textbf{12 hours}$ .

STEP 9

Now, we can find $B$ using the formula $\text{Period} = \frac{2\pi}{B}$ .

STEP 10

$\textbf{12} = \frac{2\pi}{B}$ $B = \frac{2\pi}{\textbf{12}} = \frac{\pi}{6}$ So, $B$ is $\frac{\pi}{6}$ .

STEP 11

A regular cosine function starts at its maximum.
In our problem, high tide (the maximum depth) is at noon, which is $\textbf{12}$ hours after midnight.
This means our cosine function is shifted $\textbf{12}$ hours to the right.

STEP 12

To shift the graph of $\cos(x)$ to the right by $c$ units, we replace $x$ with $x - c$ .
In our case, $c = \textbf{12}$ , so we have $\cos(x - \textbf{12})$ .
Since $B$ is multiplied by $x$ , we have $\cos(B(x - \textbf{12})) = \cos(Bx - 12B)$ .

STEP 13

Plugging in our value for $B$ , we get $\frac{\pi}{6}(x - 12) = \frac{\pi}{6}x - 2\pi$ .

STEP 14

Putting it all together, our equation is: $y = A \cos(Bx - 12B) + D$ $y = 4 \cos\left(\frac{\pi}{6}x - 2\pi\right) + 10$ Since $\cos(x - 2\pi) = \cos(x)$ , we can simplify to: $y = 4 \cos\left(\frac{\pi}{6}x\right) + 10$

STEP 15

The equation for the water's depth is $y = 4 \cos\left(\frac{\pi}{6}x\right) + 10$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the midline.
2. Find the amplitude.
3. Find the period.
4. Find the phase shift.
5. Write the equation.

STEP 3

The midline is the average of the highest and lowest depths.
Think of it like the "center" of the tide's motion!

STEP 4

$D = \frac{\textbf{14 feet} + \textbf{6 feet}}{2} = \frac{20 \text{ feet}}{2} = \textbf{10 feet}$ So, our midline is at $\textbf{10 feet}$ .

STEP 5

The amplitude is half the difference between the highest and lowest points.
It tells us how much the tide goes up and down from the midline.

STEP 6

$A = \frac{\textbf{14 feet} - \textbf{6 feet}}{2} = \frac{8 \text{ feet}}{2} = \textbf{4 feet}$ Our amplitude is $\textbf{4 feet}$ .

STEP 7

The period is the time it takes for one full cycle of the tide (from high to low and back to high).
We know high tide is at noon and low tide is at 6 AM, so half a cycle is 6 hours.

STEP 8

A full cycle must be double that, which is $2 \cdot 6 \text{ hours} = \textbf{12 hours}$ .

STEP 9

Now, we can find $B$ using the formula $\text{Period} = \frac{2\pi}{B}$ .

STEP 10

$\textbf{12} = \frac{2\pi}{B}$ $B = \frac{2\pi}{\textbf{12}} = \frac{\pi}{6}$ So, $B$ is $\frac{\pi}{6}$ .

STEP 11

A regular cosine function starts at its maximum.
In our problem, high tide (the maximum depth) is at noon, which is $\textbf{12}$ hours after midnight.
This means our cosine function is shifted $\textbf{12}$ hours to the right.

STEP 12

To shift the graph of $\cos(x)$ to the right by $c$ units, we replace $x$ with $x - c$ .
In our case, $c = \textbf{12}$ , so we have $\cos(x - \textbf{12})$ .
Since $B$ is multiplied by $x$ , we have $\cos(B(x - \textbf{12})) = \cos(Bx - 12B)$ .

STEP 13

Plugging in our value for $B$ , we get $\frac{\pi}{6}(x - 12) = \frac{\pi}{6}x - 2\pi$ .

STEP 14

Putting it all together, our equation is: $y = A \cos(Bx - 12B) + D$ $y = 4 \cos\left(\frac{\pi}{6}x - 2\pi\right) + 10$ Since $\cos(x - 2\pi) = \cos(x)$ , we can simplify to: $y = 4 \cos\left(\frac{\pi}{6}x\right) + 10$

STEP 15

The equation for the water's depth is $y = 4 \cos\left(\frac{\pi}{6}x\right) + 10$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the midline.2. Find the amplitude.3. Find the period.4. Find the phase shift.5. Write the equation.

STEP 3

The midline is the average of the **highest** and **lowest** depths.Think of it like the "center" of the tide's motion!

STEP 4

D=14 feet+6 feet2=20 feet2=10 feet D = \frac{\textbf{14 feet} + \textbf{6 feet}}{2} = \frac{20 \text{ feet}}{2} = \textbf{10 feet} D=214 feet+6 feet​=220 feet​=10 feet So, our midline is at 10 feet\textbf{10 feet}10 feet.

STEP 5

The amplitude is *half* the difference between the **highest** and **lowest** points.It tells us how much the tide goes up and down from the midline.

STEP 6

A=14 feet−6 feet2=8 feet2=4 feet A = \frac{\textbf{14 feet} - \textbf{6 feet}}{2} = \frac{8 \text{ feet}}{2} = \textbf{4 feet} A=214 feet−6 feet​=28 feet​=4 feet Our amplitude is 4 feet\textbf{4 feet}4 feet.

STEP 7

The period is the time it takes for one full cycle of the tide (from high to low and back to high).We know high tide is at noon and low tide is at 6 AM, so half a cycle is 6 hours.

STEP 8

A full cycle must be **double** that, which is 2⋅6 hours=12 hours2 \cdot 6 \text{ hours} = \textbf{12 hours}2⋅6 hours=12 hours.

STEP 9

Now, we can find BBB using the formula Period=2πB \text{Period} = \frac{2\pi}{B} Period=B2π​.

STEP 10

12=2πB \textbf{12} = \frac{2\pi}{B} 12=B2π​ B=2π12=π6 B = \frac{2\pi}{\textbf{12}} = \frac{\pi}{6} B=122π​=6π​ So, BBB is π6\frac{\pi}{6}6π​.

STEP 11

A regular cosine function starts at its **maximum**.In our problem, high tide (the maximum depth) is at noon, which is 12\textbf{12}12 hours after midnight.This means our cosine function is shifted 12\textbf{12}12 hours to the right.

STEP 12

STEP 13

Plugging in our value for BBB, we get π6(x−12)=π6x−2π\frac{\pi}{6}(x - 12) = \frac{\pi}{6}x - 2\pi6π​(x−12)=6π​x−2π.

STEP 14

STEP 15

The equation for the water's depth is y=4cos⁡(π6x)+10y = 4 \cos\left(\frac{\pi}{6}x\right) + 10y=4cos(6π​x)+10.

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1. Find the midline.
2. Find the amplitude.
3. Find the period.
4. Find the phase shift.
5. Write the equation.

The midline is the average of the highest and lowest depths.
Think of it like the "center" of the tide's motion!

$D = \frac{\textbf{14 feet} + \textbf{6 feet}}{2} = \frac{20 \text{ feet}}{2} = \textbf{10 feet}$ So, our midline is at $\textbf{10 feet}$ .

The amplitude is half the difference between the highest and lowest points.
It tells us how much the tide goes up and down from the midline.

$A = \frac{\textbf{14 feet} - \textbf{6 feet}}{2} = \frac{8 \text{ feet}}{2} = \textbf{4 feet}$ Our amplitude is $\textbf{4 feet}$ .

The period is the time it takes for one full cycle of the tide (from high to low and back to high).
We know high tide is at noon and low tide is at 6 AM, so half a cycle is 6 hours.

A full cycle must be double that, which is $2 \cdot 6 \text{ hours} = \textbf{12 hours}$ .

Now, we can find $B$ using the formula $\text{Period} = \frac{2\pi}{B}$ .

$\textbf{12} = \frac{2\pi}{B}$ $B = \frac{2\pi}{\textbf{12}} = \frac{\pi}{6}$ So, $B$ is $\frac{\pi}{6}$ .

A regular cosine function starts at its maximum.
In our problem, high tide (the maximum depth) is at noon, which is $\textbf{12}$ hours after midnight.
This means our cosine function is shifted $\textbf{12}$ hours to the right.

Plugging in our value for $B$ , we get $\frac{\pi}{6}(x - 12) = \frac{\pi}{6}x - 2\pi$ .

The equation for the water's depth is $y = 4 \cos\left(\frac{\pi}{6}x\right) + 10$ .