Trig TH/CM Question

I have to practice Ocean tides, oscillating springs, and ferris wheel problems for a test tomorrow

Desired result is a full example of one of the problems, or a link to a website that has examples, im also concerned about getting good communication marks so i need to know how to do it properly on this question. one full example on how to solve this question so i can use it as a model to solve others would be optimal please and thank you

Sample question:


The water depth in a harbour is 21 m at high tide ad 11 m at low tide. One cycle is completed aproximaately every 12 h.

a) find the equation for the water depth as a function of time , t hours , after low tide

b) draw a graph of the function for the 48 h after low tide which occurred at 14:00

d) find the depth of the water at
i)17:45
ii) 21:00

i would like full answers to the quiestions above if possible because time is of the essence

i am not asking you to do my homework i just need an exaample, as you can see that is 1/2 of that question and i still need to practice other 3 situations.. i only put up essential part of the question

PLease(Bow) Ty

Hello s.a.n.t.a.n.a

Here's the answer to (a). I have to go now. I'll post answers to (b) and (d) later unless someone else does first.

Grandad

Quote:

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Originally Posted by s.a.n.t.a.n.a

The water depth in a harbour is 21 m at high tide ad 11 m at low tide. One cycle is completed aproximaately every 12 h.

a) find the equation for the water depth as a function of time , t hours , after low tide

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(a) Suppose that the equation is $\displaystyle d=a+b\sin (\omega t+\alpha)$.

Since $\displaystyle \sin (\omega t+\alpha)$ varies between $\displaystyle +1$ and $\displaystyle -1$, the maximum and minimum values of $\displaystyle d$ are $\displaystyle a+b$ and $\displaystyle a-b$.

So: $\displaystyle a+b=21$ and $\displaystyle a-b=11$.

This gives $\displaystyle a=16$ and $\displaystyle b=5$.

Now $\displaystyle \sin (\omega t+\alpha)$ is periodic, with period $\displaystyle \frac{2\pi}{\omega}$. So:

$\displaystyle \frac{2\pi}{\omega}=12$

$\displaystyle \therefore \omega = \frac{\pi}{6}$

Also, low tide is at time $\displaystyle t=0$. So when $\displaystyle t=0$, $\displaystyle d=11$. So:

$\displaystyle 11=16+5\sin\alpha$

$\displaystyle \therefore \sin\alpha=-1$

$\displaystyle \therefore \alpha=\frac{3 \pi}{2}$

So the equation is $\displaystyle d=16+5\sin(\frac{\pi}{6}t+\frac{3\pi}{2})$

Hello again

Quote:

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Originally Posted by s.a.n.t.a.n.a

b) draw a graph of the function for the 48 h after low tide which occurred at 14:00

d) find the depth of the water at
i)17:45
ii) 21:00

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(b) Sketch graph attached.

(d)

(i) At 17:45, t = 17.75 - 14.0 = 3.75. Put this value into the formula for d. Depth = 17.91 m.

(ii) At 21:00, t = 7 and d = 20.33 m

Grandad

Thanx alot really helpd:)))