Wei is standing in wavy water and notices the depth of the waves varies in a periodic way that can be modeled by a trigonometric function. He starts a stopwatch to time the waves. $\newline$ After $1$ . $1$ seconds, and then again every $3$ seconds, the water just touches his knees. Between peaks, the water recedes to his ankles. Wei's ankles are $12 \mathrm{~cm}$ off the ocean floor, and his knees are $55 \mathrm{~cm}$ off the ocean floor. $\newline$ Find the formula of the trigonometric function that models the depth $D$ of the water $t$ seconds after Wei starts the stopwatch. Define the function using radians. $\newline$ $D(t)=\square$

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Q. Wei is standing in wavy water and notices the depth of the waves varies in a periodic way that can be modeled by a trigonometric function. He starts a stopwatch to time the waves.

\newline

After

1

1

seconds, and then again every

3

seconds, the water just touches his knees. Between peaks, the water recedes to his ankles. Wei's ankles are

12 \mathrm{~cm}

off the ocean floor, and his knees are

55 \mathrm{~cm}

off the ocean floor.

\newline

Find the formula of the trigonometric function that models the depth

D

of the water

t

seconds after Wei starts the stopwatch. Define the function using radians.

\newline

D(t)=\square

Determine Amplitude: Determine the amplitude of the wave. $\newline$ The amplitude is half the distance between the maximum and minimum values of the wave. The maximum depth is at Wei's knees ( $55\,\text{cm}$ ) and the minimum depth is at his ankles ( $12\,\text{cm}$ ). $\newline$ Amplitude ( $A$ ) = (Maximum depth - Minimum depth) / $2$ $\newline$ $A = (55\,\text{cm} - 12\,\text{cm}) / 2$ $\newline$ $A = 43\,\text{cm} / 2$ $\newline$ $A = 21.5\,\text{cm}$
Determine Vertical Shift: Determine the vertical shift of the wave. $\newline$ The vertical shift is the average of the maximum and minimum values of the wave. $\newline$ Vertical shift $D_0$ = $\frac{\text{Maximum depth} + \text{Minimum depth}}{2}$ $\newline$ $D_0 = \frac{55\,\text{cm} + 12\,\text{cm}}{2}$ $\newline$ $D_0 = \frac{67\,\text{cm}}{2}$ $\newline$ $D_0 = 33.5\,\text{cm}$
Determine Period: Determine the period of the wave. $\newline$ The period $T$ is the time it takes for the wave to complete one full cycle. Wei notices the water touches his knees every $3$ seconds, so the period is $3$ seconds.
Convert to Radians: Convert the period from seconds to radians. $\newline$ The period in radians is given by $2\pi$ radians for a full cycle. Since the period is $3$ seconds, we need to find how many radians correspond to $3$ seconds. $\newline$ $T$ (in radians) $= \frac{2\pi}{\text{Period (in seconds)}}$ $\newline$ $T$ (in radians) $= \frac{2\pi}{3}$
Determine Phase Shift: Determine the phase shift of the wave. $\newline$ The phase shift is the horizontal shift of the wave. Since the wave touches Wei's knees at $t = 1.1$ seconds, and this is the first time it happens after he starts the stopwatch, the phase shift ( $\varphi$ ) will be negative and correspond to this time. $\newline$ Phase shift ( $\varphi$ ) = $-1.1$ seconds
Write Trig Function: Write the formula for the trigonometric function. $\newline$ We will use the cosine function because it starts at a maximum point, which corresponds to the water touching Wei's knees when he starts the stopwatch. The general form of the trigonometric function is: $\newline$ $D(t) = A \cdot \cos(B(t - \varphi)) + D_0$ $\newline$ Where: $\newline$ $A$ is the amplitude, $\newline$ $B$ is the frequency ( $B = \frac{2\pi}{T}$ ), $\newline$ $\varphi$ is the phase shift, $\newline$ $D_0$ is the vertical shift. $\newline$ Now we substitute the values we found: $\newline$ $D(t) = 21.5 \cdot \cos\left(\frac{2\pi}{3}(t - (-1.1))\right) + 33.5$ $\newline$ $D(t) = 21.5 \cdot \cos\left(\frac{2\pi}{3}(t + 1.1)\right) + 33.5$