course Mth 173 衭}wѲassignment #009
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15:28:06 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 --> The predictor part of the chart: You have a value for t You have a value for T You have the value for dT/dt, which is the change in T/ change in t for the next row of values. You have the dt, which is the change in t from the current row to the next. Then you have the predictor section which is information obtained by putting the values into the formula. Next we have the corrector section, which is the two values of dT/dt averaged and then the actual change in T for the next corrector.
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15:28:11 ** 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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15:31:58 Problem 1.5.13. amplitude, period of 5 + cos(3x)
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RESPONSE --> The 3x means that it has made a complete cycle from 0 to 2 pi / 3. The period is 2 pi/3. Then the ""cos"" * 1 = 1. Then it is took up 5 points.
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15:32:01 *&*& 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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15:34:01 Explain how you determine the amplitude and period of a given sine or cosine function.
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RESPONSE --> The amplitude is what the ""cos"" is multiplied by. The period is 2 pi divided by the number that is with x.
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15:34:14 *&*& 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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15:42:37 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, 0, 3, 3, 6 The mean is 3. mean - max = amplitude 3 - 6 = 3 x = 0, 2, 4, 6, 8 min - max = period 0 - 8 = 8 y =3(mean) + 3(amplitude) sin(2 pi/ 8(period) x) Simplified = 3 + 3 sin (pi/4x)
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15:43:16 What is a possible formula for the graph?
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RESPONSE --> y = 3 + 3 sin (pi/4x)
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15:43:20 ** 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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w w}ݵr assignment #009 ލ܈y[ HĒj Calculus I 02-15-2007
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08:08:54 problem 1.5.28. Solve 1 = 8 cos(2x+1) - 3 for x.
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RESPONSE --> 8 cos 2x + 1 - 3 = 1 8 cos 2x -3 = 0 8 cos 2x = 3 cos 2x = 3/8 cos^-1(cos(2x)) = cos^-1 (3/8) 2x = cos ^ -1 (3/8) x = 1/2 cos^-1 (3/8)
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08:16:47 ** 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 don't understand what I done wrong.
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08:17:49 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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08:17:50 ** 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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08:18:25 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I was sure that I understood how to solve the cos problem. I don't understand what I done wrong.