On a typical day at an ocean port, the water has a maximum depth of 20 m at 8:00 a.m. The minimum depth of 8 m occurs 6.2 hours later. Assume that the relation between the depth of the water and time is a sinusoidal function.
Let's draw a graph!
a) What is the period of the function?
From the information we have been given...
* We can set the 8am as t = 0 hours.
* A maximum value is when t = 0 hours and when d = 20.
* A minimum value is when t = 6.2 and d = 8.
We can see from the graph that the period is 12.4 hours.
b) Write an equation for the depth of the water at any time, t hours.
cosine equation's parameters...
A = 6
B = (2pi) / 12.4 = pi/6.2
C = 0
D = 14
To get A, amplitude, calculate the distance from the sinusoidal axis to a maximum value or minimum value.
sinusoidal axis = (20+8)/2 = 14
amplitude = 14-8 = 6
B = pi/period = pi/6.2
C, the phase shift, is 0.
D is the sinusoidal axis, 14.
D(t) = 6cos [(pi/6.2)t] + 14
c) Determine the depth of the water at 10:00 a.m.
10:00 am = 2 hrs from when t = 0 or 8:00 am. Plug in the 2 as t into the equation to get the answer.
D(2) = 6 cos [(pi/6.4)2] + 14 = 17.1738 metres
d) Determine one time when the water is 10 m deep.
The wave is 10 metres deep, so the qestion is asking for what the time is when D = 10. Plug in 10 as D, then solve for t.
10 = 6cos[(pi/6.2)t] +14
-4 = 6cos[(pi/6.2)t]
-4/6 = cos[(pi/6.2)t]
arc cos(-4/6) = (pi/6.2)t
2.3005 = (pi/6.2)t
2.3005/(pi/6.2) = t
2.3005 x 6.2/pi = t
14.2632/pi = t
t = 4.5401
Convert the 4.5401 into "actual time" because we use hours:minutes:seconds to show time, so...
4.5401 hrs + 8am = 12.5401 hrs
Obviously, its not efficient to say .5401 hrs so we convert that to minutes.
0.5401 x 60 = 32.406 min.
12:32:24pm
We can round that to 12:30pm.
12 comments:
Well i think that you just have done a good job on showing your work. The only prblem that i have is that you didn't indicate your sinusoidal axis other wise good job Nothing Less Than The Best....Group / Lifted Research Group
Well you don't have to indicate the sinusoidal axis. I haven't checked the answer but the process seem correct. You don't have to round the time, I don't think. Good Job :)
Good job guys. It appears to be correct to me.
the solutions and process are looking pretty swell. good like a piece pie. everything is in lower case because i'm lazy to press the shift button. but just in case....in the test i'm going to show my sinusoidal axis with a dotted line, because i still think it's important to show that.
Well done guys, everything looks okay to me. I liked how you explained your work and it was easy to understand. Again, good job! d(-_-)b
I haven't checked your numbers, but I don't see anything wrong with your solution except a technical error:
On your graph at the beginning you labeled the y axis as D, whereas it should be D(t), because D is a function of t.
This also carries over to your solution for question d. D(t) = 10, not D = 10. You seemed to be switching between D and D(t), which is odd to me, but whatever. Thats the only thing I could see wrong with your solution.
Good job guys.
Yeah the process seems correct. Your question is really similar to our groups question. I liked that idea of making 8:00am = 0 seemed to make a lot more sense that way.
Great work everyone! I really like that your solution is easy to follow because of the added space. After checking your solutions, I therefore conclude that it is perfect. Now, I just need to check two more questions from the other groups.
-m@rk
"Nothing Less Than The Best..Group" haha I like your name. Anyways, that was totally besides the point. After finishing the problem myself, I made some errors of my own on Part D. But everything else was really well done. Looks good guys!
Looks good to me, good job guys. Hahah, thanks for showing/explaining how to turn your answer into 'proper time', because I got confused with that. ^^;;
Good job guys! Who ever posted this clearly explained the process and the solution. You explained what to do and what you were going to do well. Easy to follow and understandable. (y)
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