There is a tidal estuary in Darwin. In the middle of the channel, the depth of water can be measured using a trig function.

Each cycle of high tides occur every 12.5 hours. The maximum depth of the high tide measures 2m and the minimum measures 0.6 meters.

1.) Write the trig function equation that models the depth of water in the Darwin estuary. High tide begins at 10am.

** I figured that the equation is: 0.7cos(2TT/12.5)+1.3 (But please tell me if I'm wrong.)

2.) Using your model, find the depth of water in the estuary at 12pm (noon.) (High tide begins at 10am.)

Question

There is a tidal estuary in Darwin.  In the middle of the channel, the depth of water can be measured using a trig function.

Each cycle of high tides occur every 12.5 hours.  The maximum depth of the high tide measures 2m and the minimum measures 0.6 meters.

1.) Write the trig function equation that models the depth of water in the Darwin estuary.  High tide begins at 10am.

** I figured that the equation is: 0.7cos(2TT/12.5)+1.3 (But please tell me if I'm wrong.)

2.) Using your model, find the depth of water in the estuary at 12pm (noon.)  (High tide begins at 10am.)

dumbcow · Answer

equation looks good

according to your equation, hide tide is at t=0 or 10am
therefor 12pm corresponds to t=2
plug in 2 for t to get water depth

anonymous · Answer

Hi dumbcow, thank you very much for taking the time to respond.

Did you mean for me to plug in 0.7cos(2TT/12.5) + 1.3 = 2 ?

If so I've tried that but cos^-1 (1) ends up being 0.  Which ultimately means the tide at 12pm is 0 meters.  (When the minimum depth is 0.6 meters.) 

0.7cos(2TT/12.5) + 1.3 = 2
cos (2TT/12.5) = 1
2TT/12.5 = cos^-1 1 = 0
0 * 12.5/2TT = 0

It doesn't make any sense.  xD!

dumbcow · Answer

no your function is a function of time .... f(t)
$$f(t) = 0.7\cos(\frac{2\pi}{12.5}t) + 1.3$$
the depth 2 hours after high tide is
$$f(2) = 0.7\cos(\frac{2\pi}{12.5}*2) +1.3$$

anonymous · Answer

Sorry I was never taught to complete the equation that way in class.

We were given: y = 0.7cos(2TTt/12.5) + 1.3

And as a result we substituted 'y' with things like temperature and time.

Is there a possibility that you could give me the correct number to plug in for 'y'  to solve?

Thank you!

dumbcow · Answer

?? 
but "y" is what you are solving for.
y refers to water depth at certain time "t"
you are given the time (noon) or t=2 ... then you solve for "y"

dumbcow · Answer

actually there is no solving just plugging in 2 for t and computing the value

anonymous · Answer

Sorry I still don't quite get what you mean, but I drew up an example and I hope this can help clarify things

dumbcow · Answer

ok to draw a parallel , your example asked at what times she was 20 ft off the ground, that is same as asking "At what time is the water depth 1.5 m?"  
However, for this problem it does not say that, it says "What is the water depth at 12 pm?"
In your example, you set the equation equal to 20 and solved for "x" right.
well what if you had to solve for y....you would substitute in a value for x right
In other words, you are given the "x" value and are asked what the "y" value is

$$y = 0.7\cos(\frac{2\pi}{12.5}x) +1.3$$
x=2, substitute it into equation
$$y = 0.7\cos(\frac{2\pi}{12.5}*2) +1.3$$
$$y = 0.7\cos(\frac{4\pi}{12.5}) +1.3$$
$$y=0.7*.5358 +1.3$$
$$y = 1.675$$

anonymous · Answer

Ahh, I see.  That method is still new to me as we never ran it through in class.  But with the layout using the equation button, it really made sense to me visually seeing it.

Thank you!