Question

in a tidal river, the time between high and low tide is 7.6 hours. At high tide the depth of water is 17.2 feet, while at low tide the depth is 4.1 feet. Assume the water depth as a function of time can be expressed by a trigonometric function (sine or cosine). Write an equation of the tide, t after noon hours

Answer 1

The depth of water in a tidal river can be modeled as a trigonometric function with a period of 15.2 hours and fluctuates between 17.2 feet and 4.1 feet. The appropriate model using a cosine function with time measured in hours after noon is D(t) = 10.65 + 6.55 * cos(πt / 7.6).

Step-by-step explanation:

The question asks us to model the depth of the water in a tidal river as a function of time using a trigonometric function. Given that the time between high tide and low tide is 7.6 hours, we can establish the period of the tide cycle as 2 x 7.6 hours, since a tidal cycle includes both a high and a low tide. The height of the water fluctuates between 17.2 feet at high tide and 4.1 feet at low tide.

To define this function, we first calculate the average depth of the water, which is (17.2 + 4.1) / 2 = 10.65 feet. This figure will serve as our vertical shift from the horizontal axis. The difference between the maximum and minimum depths is 17.2 - 4.1 = 13.1 feet, so the amplitude of the function is half of that: 13.1 / 2 = 6.55 feet.

Assuming we want to use a cosine function which typically starts at its maximum, the equation for the tide as a function of time (t), measured in hours after noon, can be expressed as:

D(t) = 10.65 + 6.55 * cos(πt / 7.6)

This equation reflects the depth of water (D) as it varies with time (t), with the period adjusted by scaling the t variable inside the cosine function to reflect the 15.2-hour tide cycle (7.6 hours for high-to-low, then 7.6 back to high).

Answer 2

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 we can lower the frequency to match the cycle 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 y-axis. 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 separate each value independently.