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# Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water depth at low tide is about 2.0 m and at high tide it is about 12.0 m. The natural period of oscillation is about 12 hours and on June 30, 2009, high tide occurred at 6:45 am. Find a function involving the cosine function that models the water depth $D(t)$ (in meters) as a functionof time $t$ (in hours after midnight) on that day.

## The water depth $D(t)$ can be modeled by a cosine function with amplitude $\frac{12-2}{2}=5 \mathrm{m},$ average magnitude $\frac{12+2}{2}=7 \mathrm{m}$and period 12 hours. High tide occurred at time 6: 45 AM $(t=6.75 \mathrm{h})$, so the curve begins a cycle at time $t=6.75 \mathrm{h}$ (shift6.75 units to the right). Thus, $D(t)=5 \cos \left[\frac{2 \pi}{12}(t-6.75)\right]+7=5 \cos \left[\frac{\pi}{6}(t-6.75)\right]+7,$ where $D$ is in meters and $t$ is thenumber of hours after midnight.

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can this problem were given some information about the tide at the Bay of Fundy, and we're trying to write an equation that will model the depth of the water at any time T in hours after midnight. Okay. They gave us a big hint. They say to use the cosine, so we know that our equation is gonna look like this. A co sign. Be T minus sea put plus deep. Okay, so a here is gonna be the amplitude of the wave D is going to be the vertical shift. How far off of zero does it start? B is gonna fix the period so that it re occurs at the right time and then see is gonna be the time shift. All right. So first we know that at low tide, the water is 2 ft, and then high tide. The water is 12. 0, sorry. Meters, 12 m. Okay, So here's what it's gonna look like. And this is at 6. 45 right here. Okay, so you can see that the amplitude here remember to find the amplitude, you gotta find half of the distance from the top to the bottom. So the distance from the top to the bottom is 10. So half of this five. So we know the amplitude is five. Okay, so this is seven right here. So that's gonna be the vertical shift. Okay, so we've shifted up. Seven. The amplitude is gonna be five top time tion hours after midnight, so she is gonna be 6 45 six hours, 45 minutes. So C is gonna be 6.75 45 minutes is 75% of an hour. Okay, so we get a convert that Okay? So now the only other thing we have to worry about is B. And it says that it takes 12 hours for it to repeat his period again. So we need, um, to pie divided by B t equal a 12. So two pi over b equals 12 or B equals two pi over 12. So then we have d of tea is five times the cosine two pi over 12. T minus 6.75 That's ours. Plus de, which was 7 ft. Okay, so now our picture is shifted over. Okay. By this verdict, this horizontal shift right here t minus 6.75 So when you put 6.75 in there, then that's gonna be where you get 12. So you get zero here cause I have zero is 15 plus seven. That's 12. Okay, then this one will not be at six. Even though the period is 12, it'll be 6 45 plus 6, 12, 45. Okay, so I'll give you six there. That will give you the coastline of pie. That'll be negative. Five plus seven. Okay, It all works. It's great. There you go.

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