Another sine/cosine question i don't get
in vancouver on certain day, high tide is 20m at 2am, the next low tide is 8m at 6am, what an equation that represents height (h)of water at any time (t)since midnight)
what is the height of water at 8:45AM?
What is height of water at 8:30PM?
i tried attempting this but i think i only got the amplitude
I went 20-8/2 and got 6 as the amplitude
Im using the cos function so I have 6cos
and then for the phase shift i only got (t-2) And I don't get how to find the rest. Im using the original graphs of the sine and cosine to solve, i think that makes it easier.
you will need the period
high tide = 2:0 am
low tide = 6:00 am
time between is 4 hrs, which is half the period
(to get the full period you would go from high to low (4hrs) and back to high again(another 4 hrs)
period = 8
2π/k = 8
8k = 2π
k = 2π/8 = π/4
so you have y = 6 cos (π/4)t + h , the height
remember the min of the equation we have so far is -6, but we want that to be 8 , so we have to raise the curve 14 m
gives us:
y = 6 cos (π/4)t + 14
need : y = 6cos (π/4)(t + d) + 14
we know that at t=2 , y = 20
20 = 6 cos (π/4)(2-d) + 14
cos (π/4)(2-d) = (20-14)/6 = 1
but we know cos 0 = 1
so (π/4)(2-d) = 0
2-d=0
d=2 , but I just noticed you already had that
y = 6 cos (π/4)(t-2) + 14
check for other data:
at 6:00 am , h = 8 ---- t=6 should give us y = 8
y = 6 cos (π/4)(6-2) + 14
= 6 cos (π) + 14
= 6(-1) = 14 = 8
looks like we nailed it
To represent the height of water at any time since midnight, you can use a combination of sine and cosine functions. Let's break down the problem step by step.
1. Amplitude:
The amplitude is the difference between the high tide and low tide divided by 2. In this case, the high tide is 20m and the low tide is 8m. So the amplitude is (20 - 8) / 2 = 12 / 2 = 6m.
2. Period:
The period of the tide cycle is the time it takes for one complete cycle, which is 12 hours (from high tide to high tide or low tide to low tide).
3. Phase Shift:
The given information is that high tide occurs at 2am. Since we want to represent the height of water at any time since midnight, we need to shift the function horizontally to the right by 2 hours so that it aligns with the given time. The phase shift is (t - 2) since we are shifting the function t units to the right.
4. Equation:
Using the information above, the equation to represent the height of water at any time (t) since midnight is h(t) = 6 cos((2π/12)(t - 2)).
Now, let's calculate the height of the water at 8:45AM and 8:30PM.
For 8:45AM:
Convert 8:45AM to a decimal time value. Since there are 60 minutes in an hour, 45 minutes is 45/60 = 0.75.
Add the decimal time value to 8 hours (since it's in the morning). 8 + 0.75 = 8.75.
Now substitute this value into the equation:
h(8.75) = 6 cos((2π/12)(8.75 - 2))
For 8:30PM:
Convert 8:30PM to a decimal time value. Since it's in the evening, add 12 hours to it. 8 + 12 = 20.
Add the decimal time value to 20 hours. 20 + 0.5 = 20.5.
Now substitute this value into the equation:
h(20.5) = 6 cos((2π/12)(20.5 - 2))
By evaluating these equations, you will find the height of the water at 8:45AM and 8:30PM.