qiery 9

qiery 9

This should have been posted, but isn't. You didn't send an access number, so its' possible I transcribed it wrong when I looked it up. However that should have caused an error when I posted it, so chances are I messed up in some other way. Gotta go back and look at what might not have been posted but should have.

In any case see my notes and let me know if you have questions. Respond to that thing about the inverse functions.

I don't think the calculus is going to give you as much trouble once you get the precalculus straight. If it's been 7 years, it would be expected that you'd need a bit of a refresher. Which is what Chapter 1 is for, of course.

assignment #009 â{a\úlXipx~ Calculus I 06-24-2005

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19:30:07 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 --> did it

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19:30:24 ** 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 --> ok

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19:39:07 Problem 1.5.13. amplitude, period of 5 + cos(3x)

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RESPONSE --> so pos. 5=shifted 5 up 1x cos= amplitude of complete cycle is 1 3x= period of complete cycle is 2pi/3

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19:39:42 *&*& 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 --> ok.

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20:09:52 Explain how you determine the amplitude and period of a given sine or cosine function.

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RESPONSE --> the amp. is like the radius of the circle. so 2cos x amp.=2. 360*=2pi. sin and cos are =2pi : peroid is 2pi/x

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20:11:21 *&*& 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 --> oh thats right. amplitude is deviation from the axis.

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20:24:08 query Problem 1.5.24. trig fn graph given, defined by 3 pts (0,3), (2,6), (4,3), (6,0), (8,3)

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RESPONSE -->

y=0 min, 6max. ave.=3. 6-3=3 amp.=3, also shifted 3 from axis. period is 0 thru 8. so per.is 8. 8=2pi/x 3+3cos(2pi/8)

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20:24:37 What is a possible formula for the graph?

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RESPONSE --> the equation?

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20:26:18 ** The complete cycle goes from y = 3 to a max of y = 6, back to y = 3 then to a min of y = 0, then back to y = 3. The cycle has to go thru both the max and the min. So the period is 8.

The mean y value is 3, and the difference between the mean y value and the max y value is the amplitude 3.

The function starts at it mean when x = 0 and moves toward its maximum, which is the behavior of the sine function (the cosine, by contrast, starts at its max and moves toward its mean value).

So we have the function 3 sin( 2 `pi / 8 * x), which would have mean value 0, shifted vertically 3 units so the mean value is 3. The function is therefore

y = 3 + 3 sin( `pi / 4 * x). **

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RESPONSE --> so how do i know if it's sin or cos?

The sine starts its cycle its horizontal axis of symmetry, right between its y values, while the cosine starts at its maximum positive displacement from the axis.

The given function starts at y value 3, halfway between the extreme y values of 6 and 0.

So it's the sine.

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20:38:10 problem 1.5.28. Solve 1 = 8 cos(2x+1) - 3 for x.

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RESPONSE --> 1 = 8 cos(2x+1) - 3 4=8 cos(2x+1) 32=cos(2x+1) ??

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20:38:48 ** 2x + 1 = `pi/3, or `pi / 3 + 2 `pi, or `pi / 3 + 4 `pi, ... or `pi / 3 + 2n `pi, ... with n = an integer.

So x = (`pi / 3 - 1) / 2.

Or x = (`pi / 3 + 2 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + `pi.

Or x = (`pi / 3 + 4 `pi - 1) / 2 = (`pi / 3 - 1) / 2 + 2 `pi. Or x = (`pi / 3 + 2 n `pi - 1) / 2 = (`pi / 3 - 1) / 2 + n `pi ... x = (`pi / 3 - 1) / 2 + n `pi, n an integer, is the general solution.

Since 0 (`pi / 3 - 1) / 2 < `pi he first two solutions are between 0 and 2 `pi, and are the only solutions between 0 and 2 `pi.

We generally want at least the solutions between 0 and 2 `pi. **

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RESPONSE --> oh.

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20:40:25 problem 1.5.42 (was 1.9.34) arccos fndescribe the graph of the arccos or cos^-1 function, including a statement of its domain and range, increasing/decreasing behavior, asymptotes if any, and concavityexplain why the domain and range are as you describe

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RESPONSE --> ok. arccos? fndescribe? i had trig about 7 years ago, so i need help

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20:41:48 ** To get the graph of arccos(x) we take the inverse of the graph of cos(x), restricting ourselves to 0 < = x < = `pi, where y = cos(x) takes each possible y value exactly once. The graph of this portion of y = cos(x) starts at (0, 1), decreases at an increasing rate (i.e., concave down) to (`pi/2, 0) the decreases at a decreasing rate (i.e., concave up) to (`pi, -1).

The inverse function reverses the x and y values. Since y values of the cosine function range from -1 to 1, the x values of the inverse function will run from -1 to 1. The graph of the inverse function therefore runs from x = -1 to x = 1.

The graph of the inverse function starts at (-1, `pi) and runs to (0, `pi/2), reversing the values of y = cos(x) between (`pi/2, 0) and (`pi, -1). Since the graph of y = cos(x) is decreasing at a decreasing rate (i.e., concave up) between these points, the graph of the inverse function will be decreasing at a decreasing rate (i.e., concave up) over the corresponding region.

The graph of the inverse function continues from (0, `pi/2) to (1, 0), where it decreases at an increasing rate.

The values of the inverse cosine function range from 0 to `pi; the domain of this function is [-1, 1].**

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RESPONSE --> help?

To start with think of a table for the cosine function from x = 0 to x = +pi. We use these values because the value of the cosine goes from 1 to -1, i.e., through its full range, between these x values.

x cos(x)

0 1 pi/4 .71 pi/2 0 3pi/4 -.71 pi -1.

If we reverse the columns of this table we get a partial table for the inverse cosine function. To avoid confusion with x in the above table we'll use z for the variable:

z arcCos(z)

1 0 .71 pi/4 0 pi/2 -.71 3pi/4 -1 pi

Graph both the cos(x) vs. x and arcCos(z) vs. z functions according to these tables.

Then make an extended table for cos(x) using x = 0, pi/6, pi/3, pi/2, 2pi/3, 5pi/6 and pi.

Finally reverse this table to get a table for arcCos(z) vs. z.

These tables should allow you to refine your graphs.

Finally follow the same scheme for sin(x) vs. x for x = -pi/2 to pi/2, reversing columns to get arcsin(z) vs. z.

Send me a description of what you get and I'll tell you more.

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20:42:57 Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> i did't remember trig to be this complicated...anything i can use as a refresher?

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20:43:31 None.

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RESPONSE --> umm ok. that's just plain freaky .....

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