







Don't forget that we have our design and measurment test tomarrow morning and our sequence test tomarrow afternoon!
next scribe is ~*~ Just_in ~*~
Expanding the walls of our classroom. This is an interactive learning ecology for students and parents in my Applied Math 40S class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.
There are 2 different ways to find the answers for these questions.
Then enter...
Then hit 2nd, Mode to exit to your main screen. Then enter u(110) (to enter u, press 2nd, 7 ) Press enter to get the 110th term which is 332.
Now, go back to your main screen and enter u(21), then hit enter to get the 21st term which is 3145728.
Go back to your main screen and enter u(27), then hit enter to get the 27th term which is 196418.
On this question we are now looking for the "term".
Then hit 2nd, Graph to get to the table. Scroll down till you end up with 1535.
So the term of 1535 in the sequence is 10.
This is how we start our graph. It stated that the normal depth of the water if 9 metres therefore it is the average value of function(sinusoidal axis). It is where our graph will start. We then add 10 metres to the sinusoidal axis since the amplitude is 10. It tell us how far to go from the sinusoidal axis. This will give us the maximum(19 metres). We then find the minimum by subtracting 10 from 9(-1).
We can now construct the graph.
The graph first goes down since it is stated in the information that the water first goes down from its normal depth. This wave has an amplitude of 10(as stated earlier) therefore the wave goes down its minimum depth which is -1. It also says that it then goes back to its normal depth at the same equal time. Therefore we divide the whole period(which is 15 ) into 4 parts because it first goes down then up, then up again and finally down.
We can now find "DABC"
A= -10 ( amplitude. it is negative because the first movement of the wave is down)
B=2(pi)/15 ( determines the period , we place 15 because a period is every 15 mins)
C=0 (since the wave started at 0 minutes and at the sinusoidal axis, there is no horizontal shift)
D= 9 ( this tells us that the whole graph will move 9 metres up, it is the sinusoidal axis)
Now we can put these informations into an equation.
The amplitude
2 Pi(3.14) / 15 = o.42
The vertical shift\
A) Assuming that the depth of the wave varies sinusoidally with time as the tsunami passes, predict the depth of the water at the following times after the tsunami first reaches the pier.
i) 2 minutes ii) 4 minutes iii) 12 minutes
For this problem, we could just insert these times into the equation in place of x to determine the depth of the water. For example, this is the equation for 2 minutes:

We can see the lowest depth of the water by looking at our graph. It is -1 since it is the lowest the wave can go. In the real world this is kind of confusing. How can the lowest depth be -1 , this would mean it digs under see bottom. Therefore we state it as , for lenght of time the wave is at its lowest, we can see the bottom.
C) The wavelenght of the water is the distance a crest of the wave travels in one period. It is also equal to the distance between two adjacent crests. If a tsunami travels at 800 kilometers per hour, what is the wavelenght?
For this question. it is stated that the the wave is traveling 800 km/h. In our graph, one period of the wave is 15 minutes. We found the answer for this question by cross multiply.
THE NEXT SCRIBE IS CHRIS A.!!!!
The first question asked "How does it take for the spaceship to go around the earth."- To answer this you are trying to find the period or (B)

- The next question was "What are the distances at the Perigree and the Apogree?"
To find the distance for the Perigree you take D-A or 550-450 = 100
To find the distance for the Apogree you take D+A or 550+450 = 1000
Question 2 is explained in the slides posted from today.
Period 2
We did our per-test.
The test is on Wednesday people and we start our final unit tomorrow.
The next scribe is Jowell.

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