assn 9pt1

course Mth 173

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Calculus I

07-02-2007

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11:07:55

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

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You choose a number and plug it into the equation and then you get back a number that should be in the graph.

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11:10:07

** 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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The method is much more complex than the one you outlined, and it is on the test.

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Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions (to which I will respond).

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11:10:52

Problem 1.5.13. amplitude, period of 5 + cos(3x)

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The amplitude is 1

The period is 2pi divided by 3

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11:11:07

*&*& 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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right

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11:13:09

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

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The amplitude is how far up it goes before it reaches its beginning point. You can determine this by only looking at the equation. It is the beginning of the equation, for example, 5sin(3t) the amplitude is 5. the period is what comes after the sin or cos function. i.e. in the above equation it would be 3

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11:13:25

*&*& 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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right

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11:14:21

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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? not sure of the question here...

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11:15:38

What is a possible formula for the graph?

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2cos(2pi/5)

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11:16:03

** 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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&#

This also requires a self-critique.

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11:21:49

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

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I begin by adding 3 to both sides: 4=8 cos(2x+1)

Next I divide by 8: 1/2=cos (2x+1)

Then i switch cos to cos^ -1 and move to the other side of the function: cos^ -1(1/2)= 2x +1

Last i subtract 1 and divide by 2: {cos^ -1 (1/2)-1}/2

This leaves me with x; which is approximately .02359

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11:22:08

** 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

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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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The exact form of the solution is preferred, when an exact solution can be found (e.g., when the inverse function can be found using a basic 30-60 or 45-45 right triangle).

The text really doesn't go far into multiple solutions, but you should be aware of the possibility of multiple solutions.

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11:22:23

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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11:22:27

** 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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&#

You need a detailed self-critique here.

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11:23:16

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

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I am still having some trouble when i see a graph to determining the period, i ended up working all the problems in that section and still had trouble with it. I am not quit sure that i have grasped it yet.

This is something that should have been covered throughly in your prerequisite courses.

You can also check out my Precalculus II page at http://vhmthphy.vhcc.edu/ > precalculus II. That course pretty much starts out with this idea. The q_a_ and Class Notes might be helpful.

&#You are always welcome to ask self-contained questions about anything. By self-contained questions I mean a question that includes a brief statement of the problem or topic you are asking about (in order to give everyone the best responses I can, I can't take time to look problems up in the text, which I don't carry with me in any case), and a statement of precisely what you do and do not understand about the situation. If it's a problem, you should include a list of things you have tried in attempting to solve (or to understand) the problem. Depending on the problem this might include a description of any diagrams, listings of concepts and topics you think might be helpful, and other relevant information. This can be relatively brief, but the more you can tell me, the more you will learn in the process, and the more specifically I can address my response. &#

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assn 9pt1

The method is much more complex than the one you outlined, and it is on the test.

&#

&#

This also requires a self-critique.

The exact form of the solution is preferred, when an exact solution can be found (e.g., when the inverse function can be found using a basic 30-60 or 45-45 right triangle). The text really doesn't go far into multiple solutions, but you should be aware of the possibility of multiple solutions.

&#

You need a detailed self-critique here.

This is something that should have been covered throughly in your prerequisite courses. You can also check out my Precalculus II page at http://vhmthphy.vhcc.edu/ > precalculus II. That course pretty much starts out with this idea. The q_a_ and Class Notes might be helpful.

&#

See my notes and let me know if you have questions. &#

The exact form of the solution is preferred, when an exact solution can be found (e.g., when the inverse function can be found using a basic 30-60 or 45-45 right triangle).

The text really doesn't go far into multiple solutions, but you should be aware of the possibility of multiple solutions.

This is something that should have been covered throughly in your prerequisite courses.

You can also check out my Precalculus II page at http://vhmthphy.vhcc.edu/ > precalculus II. That course pretty much starts out with this idea. The q_a_ and Class Notes might be helpful.