Question

the tide reached a max heights of 10.8 feet at midnight, and at 9am, it reached the min of 5.8 feet. what was the height of the tide at 7am? (relationship is sinusidal)

My answer was 5.58 feet.. but it is supposed to be 6.4 feet?

Answer 1

@sourwing ?

Answer 2

What is the period of your sine wave?
What is the center of your sine wave?

Answer 3

i found the period to be 2pi/9, and I don't understand what you mean by center? like the displacement?

Answer 4

@tkhunny

Answer 5

\(y = sin(x)\) -- Average value, or center, is y = 0

\(y = sin(x)+1\) -- Average value, or center, is y = 1

\(y = sin(x)-5\) -- Average value, or center, is y = -5

It is half way between the minimum and maximum values.

Answer 6

Half a cycle is 9 hours. The period is 18 hours. \(\dfrac{2\pi}{What} = 18\)

Answer 7

so the period is pi/9? I still don't understand the average value thing? I was taught to find the amplitude, vertical displacement, and phase shift? my equation was \[2.5\sin(\pi/9x)+8.3\]

Answer 8

@tkhunny

Answer 9

How did you get 8.3?

Answer 10

10.8+5.8=result/2

Answer 11

Very good. Does it work? See of you can reproduce the two known values.

Answer 12

amplitude: 10.8-5.8=result/2
and now I don't know how to get to 7am? or am I supposed to plug it into my equation: 7=2.5sin(pi/9)+8.3? @tkhunny

Answer 13

First, you need the right function. Don't bail and start trying to find the answer when you haven't yet managed to find the right function.

2.5sin(π/9(0))+8.3 Is that 10.8?
2.5sin(π/9(9))+8.3 Is that 5.8?

Answer 14

oh. they are not equal... that's strange when I graphed it the graph came out ok D;

Answer 15

You knew what you wanted it to look like. This is a common error. You must get your opinion out of it and PROVE IT!

y = 2.5sin((π/9)x)+8.3

This has a maximum at x = 9/2. How shall we fix that? We need the maximum to be at x = 0!

Answer 16

I thought the max was at 10.8 ft?

Answer 17

The maximum value is y = 10.8 This maximum is attained at x = 0 (midnight)

Answer 18

oohh ok

Answer 19

Keep your eyes on the x and y and make sure you know which one you're talking about.

Answer 20

Oh the phase shift!

y = 2.5sin((π/9)(x+4.5))+8.3?? I saw that the graph was 4.5 to the right, so I moved ti backed to the left where x is 0, and y is 10.8

Answer 21

Now THAT is the right function. It is time to evaluate at x = 7!!

Answer 22

ok so 7=2.5sin((pi/9)(x+4.5))+8.3?

i just have one more question: how do you find the phase shift without a calculator? I found it through looking at a graph @tkhunny

Answer 23

Several ways. It's particularly easy with the calculus. Can we use that?

Answer 24

nope I'm only in pre-calc 12 ):

Answer 25

or maybe you can still do it that way and I can see if I get it?

Answer 26

Okay, then... Let's think about only the very first period of the sine function.

Where is the maximum value of y = sin(x)?? For what value of x does sin(x) attain it's maximum value?

Answer 27

y would be at 1?

Answer 28

Right. Where is x when that happens?

Answer 29

x is at zero

Answer 30

No, that's the maximum value of the cosine. We need the maximum value for the sine.

Answer 31

sine represents 'y' portion of a graph, so the max value of y is at 1?

Answer 32

You said that already. Are we talking about x or y? Read very carefully!

You have NOT said what value of x gives the maximum value of y for \(y = \sin(x)\). You tried x = 0, but that was incorrect.

Answer 33

It is confusing to say "the maximum value of y is at 1". It is a little more clear to say "the maximum value of y is 1". It reads just a little differently.

Answer 34

honestly, I don't know? I thought the max value of x would be 0, since when x is zero, y is 1

Answer 35

Why do you repeat that? It is not correct.

cos(0) = 1
sin(0) = 0

sin(What) = 1? <== This is the one you are not answering correctly.

Answer 36

OH. sinepi/2 = 1

Answer 37

That's it!!!!

For y = sin(x), the maximum is at x = pi/2
When we messed with the period, to where did we move this maximum? Do you remember \(\dfrac{\pi}{9?}\). Try it out. \(\dfrac{\pi/2}{\pi/9} = \dfrac{9}{2}\), and there we have it!

Answer 38

ohhh I did not understand your question at all lol. Thank you so much for clarifying!so the max is supposed to be 9/2 or?

Answer 39

Note: Had we used cosine, instead of sine, this would have been a bit easier, since the cosine already has a maximum at x = 0 and messing with the period doesn't move it about. Of course, it the problem statement DEMANDS a sine function, we would then need to convert it.

Answer 40

You keep saying it in a very confusing way. You MAY mean the right thing, I just can't quite tell. I suspect you are confusing yourself. Try to be more exact and precise in your language and your notation. These things will save you!

\(y = sin(x)\)

The maximum is y = 1 at x = pi/2

\(y = sin((\pi/9)x)\)

The maximum is y = 1 at x = 9/2

\(y = sin((\pi/9)(x+9/2))\)

The maximum is y = 1 at x = 0

\(y = 2.5*sin((\pi/9)(x+9/2))\)

The maximum is y = 2.5 at x = 0

\(y = 2.5*sin((\pi/9)(x+9/2)) + 8.3\)

The maximum is y = 10.8 at x = 0

Answer 41

oh I see. you're right, my teacher always reminds me to correct my mathematical language. thank you so much for a detailed explanation! this definitely helps for my exam tomorrow!

Answer 42

Super. Go get 'em!!