course Mth 173 ˡJ gy|ƒέassignment #009
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10:39:00 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-corrector method is based on taking a wide prediction of the action of a differential equation. This prediction is then corrected by solving the differential equation in smaller parts to check the accuracy. We do this by using the rate dT/dt with an initial T. We then predict the change by using the rate over the time (t) to find the final T, and creating a dT/dt for this period. An average value of the initial and final dT/dt values is then found. We then continue this process of choosing a period t and finding a new T and dT/dt at the end of the period until we reach the t we want.
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10:42:46 ** 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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10:43:15 Problem 1.5.13. amplitude, period of 5 + cos(3x)
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RESPONSE --> Amplitude: 1 Period: 2pi / 3
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10:43:31 *&*& 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. I missed the vertical shift.
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10:44:41 Explain how you determine the amplitude and period of a given sine or cosine function.
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RESPONSE --> The amplitude can be determined by finding the coefficient of the trigonometric function. The period can be found by dividing 2pi by the b value in y = sin(bx) + c.
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10:44:55 *&*& 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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10:47:45 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 (a) = 3 Vertical shift (c) = 3 Period = 8 = 2'pi / b b = 2'pi / 8 = 'pi / 4 y = 3sin('pi/4 x) + c
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10:48:11 What is a possible formula for the graph?
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RESPONSE --> Found in prior answer: y = 3sin('pi/4 x) + 3
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10:48:40 ** 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 --> Ok.
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10:53:30 problem 1.5.28. Solve 1 = 8 cos(2x+1) - 3 for x.
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RESPONSE --> First add 3 then divide both sides by 8: (1-3) / 8 = cos(2x + 1) Take the inverse cosine: cos^-1((1-3)/8) = 2x + 1 Shuffle: (cos^-1((1-3)/8) - 1) / 2 = x
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10:55:59 ** 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 --> Hmm, I need to study this more. Why 'pi / 3 + 2n 'pi?
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14:53:58 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 arccos function is the cos function flipped through the origin. To make sure that the arccos function does not have two outputs for one input, the function must be restricted to the domain of [-1, 1], creating a range of ['pi, 0]. The function is decreasing at a decreasing rate over the interval [-1, 0] (making this part concave up), and decreasing at an increasing rate over the interval [0, 1] (making this part concave down).
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15:12:56 ** 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 --> Ok.
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15:15:23 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Not much. Solving the cosine function like that was new, but other than that I think it was pretty much review.
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15:15:27 None.
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RESPONSE --> Ok.
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