Trig TH/CM Question

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December 15th 2008, 05:24 PM
s.a.n.t.a.n.a

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
December 16th 2008, 12:20 AM
Grandad

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:

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

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

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

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

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

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

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

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

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

$11=16+5\sin\alpha$

$\therefore \sin\alpha=-1$

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

So the equation is $d=16+5\sin(\frac{\pi}{6}t+\frac{3\pi}{2})$
December 16th 2008, 05:03 AM
Grandad

1 Attachment(s)

Harmonic graph

Hello again

Quote:

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

(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
December 16th 2008, 04:56 PM
s.a.n.t.a.n.a

Thanx alot really helpd:)))