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anonymous
 5 years ago
In a tidal river, the time between high and low tide is 7.6 hours. At high tide the depth of water is 17.7 feet, while at low tide the depth is 5.1 feet. Assume the water depth as a function of time can be expressed by a trigonometric function (sine or cosine).Write an equation for the depth f(t) of the tide (in feet) t hours after 12:00 noon.
anonymous
 5 years ago
In a tidal river, the time between high and low tide is 7.6 hours. At high tide the depth of water is 17.7 feet, while at low tide the depth is 5.1 feet. Assume the water depth as a function of time can be expressed by a trigonometric function (sine or cosine).Write an equation for the depth f(t) of the tide (in feet) t hours after 12:00 noon.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0man what a question. Alright, so the difference between tidal heights is 12.6 ft. right? for every integer cos(x) is multiplied, say a*cos(x) the amplitude doubles. So then, (1/2)=(n/12.6) will gives us a value for a that will match the differences in tidal height. a=6.3. Now let's say that 2pi will give us a full cycle from high to low back to high again. If 12:00 noon is our high, it takes 15.2hrs to complete a cycle back to high again. 2pi = 6.28 roughly, and we know that is less than our midpoint in a whole cycle, 7.6 hrs. Also, for every value k>1 in cos(kx), this increases the frequency of waves in a given period. So it seems that we need a value of k less than 1. So wee can lower the frequency to match the cyle we are looking for, which is to come back to 17.7 ft by 15.2hrs. So to find our value k we divide 2pi by 15.2. This is 0.41. We do it that way based on the ratio for the standard period to what we need, which is a number that can get us to 15.2. So far we have f(t) = 6.3cos(0.41x). Now the problem we have now is that we are not at our height for max, nor our min, even though our wave matches the difference between them, we are between 6.3 and 6.3 so far. so for every value of c in cos(x)+c, this adds to the height cos starts on the yaxis. Since we are at 6.3 already, all we need to do is subtract 6.3 from 17.6 to get c. So C is 11.3 and our final equation is f(t)=6.3cos(0.41x)+11.3. This gives our tides in 15.2 hr periods, starting at 12 noon = x=0, with extremes at intervals of 7.6 hours. Let me know if you have any more questions about it, it's kind of rough to explain. We transformed by f(t)=a*cos(kx)+c altogether, but I thought it might help to seperate each value independently.
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