query 9

query 9

Good answer on the predictor-corrector thing. It's the precalculus review that seems to be giving you trouble, but once you're through that I think you'll be fine with the calculus.

The first chapter of the text is a review of precalculus and isn't covered in the notes; you should have covered the circular definition of the trig functions and the entire subject of amplitude, period, phase shift, etc. in that course. Not your fault if you didn't, of course, and I'll help you get it straight. See my notes.

]{ҿxvzTȞ assignment #009 â{a\úlXipx~ Calculus I 05-26-2005

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16:21:51 Query class notes #10 Describe in your own words how the predictor-corrector method of solving a differential equation works

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RESPONSE --> if you use the derived rate of the function @ the start of your interval for the whole interval, it will not be completely accurate, because the rate of the functions actual change changes itself. but the rate of change that you use to calculate the next point doesn't change because it is figured for a specific point. so the p.c. method solves this by taking into account the differing rate of change. it does this by averaging the different rates/slopes at the first and last point of the interval. this gives a more accurate result, because it takes the slope change into account.

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16:22:43 ** We start with the rate dT/dt calculated at the beginning of the t interval for some given initial T. We then use this dT/dt to calculate the amount of change in T over the interval. Then T is calculated for the end of the interval.

We then calculate a dT/dt for this T.

The two values of dT / dt then averaged to obtain a corrected value.

This is then used to calculate a new change in T. This change is added to the original T.

The process is then continued for another interval, then another, until we reach the desired t value. **

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RESPONSE --> oh. ok

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19:16:22 Problem 1.5.13. amplitude, period of 5 + cos(3x)

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RESPONSE --> no where in the notes linked for this lesson can i find what the amplitude is. the period is 2pi/3?

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19:18:58 *&*& The function goes through a complete cycle as 3x changes from 0 to 2 pi. This corresponds to a change in x from 0 to 2 pi / 3. So the period of this function is 2 pi /3.

The cosine function is multiplied by 1 so the amplitude is 1.

The function is then vertically shifted 5 units. *&*&

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RESPONSE --> uuhhh. ok.

This first chapter is a review of precalculus and isn't covered in the notes. The notes address calculus and the ideas that lead up to it, such as rates, slopes, areas and accumulated quantities.

Remember from precalculus:

y = A sin(omega t) + c is a sine function with period 2 pi / omega, vertically stretched by factor A so that instead of peaking/'valleying' at y = 1 and y = -1 it peaks/'valleys' at y = A and y = -A, then shifted vertically c units so that the 'midline' is at y = c rather than y = 0.

The same description applies if you replace 'sin' by 'cos' and 'sine function' by 'cosine function'.

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19:19:30 Explain how you determine the amplitude and period of a given sine or cosine function.

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RESPONSE --> help???

The text addresses these questions, which should be familiar from precalculus.

Review the text and let me know what isn't clear to you. You can use my synopsis above as a point of reference.

If you didn't cover the circular definition of sine and cosine functions in the second semester of precalculus then someone missed giving you a very standard topic. In any case I'll help you through it if you ask questions based on what you see in the text.

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19:20:06 *&*& GOOD ANSWER FROM STUDENT:

Amplitude is the multiplication factor of the cosine function, such as the absolute value of a in y = a cos(x).

the period is 2`pi divided by the coefficient of x. *&*&

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RESPONSE --> ok.

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