course mth173 ??????Y?Ry????J???assignment #009
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11:21:50 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 --> This what we did in the last QA but essentially you take the time ti and tf and find the slope of both by averaging the slopes of both then find the 'dy by multiplying the 'dt by average slope and subtracting the 'dy from the original y value.
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11:24:18 ** 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 --> okay
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11:36:28 Problem 1.5.13. amplitude, period of 5 + cos(3x)
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RESPONSE --> amplitude= 4.8 period = 2
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11:40:13 *&*& 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 --> so breaking down the equation we get 5cos(3x) the 5 coresponds to the amplitude because it multiplies the intensity of the function. the 3x coresonds to the period where the cos function is standing where the 2pi should be over 3. So the period of a cosine function is 2pi over the number in the cosine function itself.
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11:45:36 Explain how you determine the amplitude and period of a given sine or cosine function.
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RESPONSE --> The amplitude is given by the multiplier of the function for example 5cos(3x) the amplitude is 5, or on a graph it is from the xaxis to the highest point the wave. The period is found by the number itself in the function such as cos(3x) since cos is (2pi/) the period is 2pi/3 or on a graph the period is the distance it takes to complete one wave.
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11:45:51 *&*& 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 --> okay
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11:51:49 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 --> amplitude: 3 Period: 8pi 3cos(.25x)
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11:52:08 What is a possible formula for the graph?
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RESPONSE --> 3cos(.25x)
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11:59:46 ** 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 --> I found the amplitude by using the method you have, and the period was 8. However i used the cos function when sin should have been used, I said 3cos(.25x) I got the multiplier right but I didn't account for the 3unit shift, to make it 3+3... when is sin used and when is cos used??
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12:09:27 problem 1.5.28. Solve 1 = 8 cos(2x+1) - 3 for x.
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RESPONSE --> 1= 8cos(2x+1)-3 4=8cos(2x+1) .5= cos(2x+1) cos^-1(.5)= 2x+1 (cos^-1(.5)-1) /2 = x .0236 = x
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12:30:38 ** 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 --> I am pretty sure I understand the pi rad concept, a pi radian is 1 radius wrapped around the curcimference of a circle and I understand sin and cos when dealing with triangles, and how to find the amplitude and period. However I don't understand how to break down the sine/cosine functions into the fractions as shown in the answer also I don't understand how the sine and cosine functions differ in that aspect.
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12:39:58 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 --> The cos^-1 funtion is the inverse of the cos function it works just like the 'sqrt and the ^2 function work, it undoes cos. y=cos(x) {(0, 1),(2, -.42)(4, -.65), (6, .96)} y = cox^-1(x) {(1, 0), (-.42, 2), (-.65, 4), (.96, 6)} it forms a mirror image of the cos wave
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12:46:18 ** 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 --> as long as the 0
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12:48:18 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I'm finding the cosine/sine ideas a little hard to grasp I think that I'm missing the 1 little idea that is very important and that is what is making it confusing to me, however with a little explaining I think that I will get the concepts with no problem.
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12:48:23 None.
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RESPONSE --> okay
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