IB-Trigonometry,please help!

Leah5467

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Feb 28, 2019
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Hello! This is the question i couldn't solve:
maths.png
It has to use the formula of asin(b(t-c)+d
I found that a is 7,and that d is 0. But...i don't have any clues how to do the rest. Please help! Thank you!
 

Harry_the_cat

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a=7 correct (half the distance between high and low tide)
The average time difference between high tides is the period of the function. So \(\displaystyle \frac{2\pi}{b} = 12.4\). So b=??

The value of d depends on how you define H.
If you let H= "height above low tide" then d=7. In this case H=0 at low tide.
If you let H = "height above "mid-tide" " then d=0. Here H=-7 at low tide.

Also. the value of c depends on where you are measuring time from.
If you measure time from mid-tide on a rising tide c=0.
If you measure time from high tide then c = \(\displaystyle \frac{1}{4}*12.4 = 3.1\)

Personally I would choose the bold options above. The equation would be: H = ???

The graph would have the equilibrium line at y=7 (ie d). It would begin at the maximum point (0, 14), cut the equilibrium line at(3.1, 7) have a minimum on the x-axis at (6.2, 0) up to the equilibrium line again at (9.3, 7) up to a max again at (12.4, 14).

Other answers would be possible depending on how H and t are defined ie where they are measured from.
 
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Leah5467

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ahhh i see! Thank you for the detailed explanation! I don't understand why c is 1/4*12.4? Why is it 1/4? B is 0.51. H i think i would pick 7.
 
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Harry_the_cat

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Draw the graph starting at high tide. You'll see that the original sin graph has to move 1/4 of a period to the left.
 

Jomo

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This is the 2nd time recently I saw y= asin(b(t-c))+d [= asin(bt-bc)+d = asin(bt-C)+d, where C=bc]
Isn't y = asin(bt-c)+d easier to work with?
One period of the sin graph starts when bt-c=0 or t = c/b
That period ends when bt-c =2pi or t = (2pi+c)/b
And the period is therefore 2pi/b
 

Dr.Peterson

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This is the 2nd time recently I saw y= asin(b(t-c))+d [= asin(bt-bc)+d = asin(bt-C)+d, where C=bc]
Isn't y = asin(bt-c)+d easier to work with?
One period of the sin graph starts when bt-c=0 or t = c/b
That period ends when bt-c =2pi or t = (2pi+c)/b
And the period is therefore 2pi/b
Both are useful, and are easy to use once you learn the relevant formulas; but I personally prefer the factored form a sin(b(t - c)) + d, because the shift is explicitly given as c, rather than having to be extracted by dividing. I've taught from books that do it either way, and live with it.

In general, the form f(a(x - b)) represents a "stretch" by 1/a followed by a "shift" of b; f(ax - b) represents a shift by b followed by a stretch by 1/a which modifies the shift to b/a, so that the shift you are doing is not the shift you see in the graph.

Actually, now that I think about it, if I were writing the book, I might use f((x - b)/a), so that x is multiplied by a and then increased by b. Then I could point out that u = (x - b)/a is the inverse of x = au + b, and a lot of things fall into place (specifically, why things work backward). But I don't want to get too far out of step with what all the textbooks say. Then, in the case of the sine, we'd have y= a sin((t - c)/b) + d, where the period is b times 2 pi and the horizontal shift is c, just as the amplitude is a and the vertical shift is d.
 

Leah5467

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Hello,when i came back to this question,i discovered that i don't understand why the graph should be drawn from high tides. What is the reason for it?
 

Dr.Peterson

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I don't think there is a "should" involved here! The question leaves it for you to define what t represents - that is, where the origin is.

Using sines, the easiest thing to do is to start from the point between high and low (mid-tide), since then no phase shift is needed. For practical purposes, you would want to start at midnight; but we don't have that information. In light of that, I would tend to start from low or high tide, which are more natural times to observe; but then I would naturally use a cosine model.

So who are you taking as the authority on what you "should" do?
 

Leah5467

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Thank you!
 
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