course Mth 173 I just submitted one with not all the information needed. This is the correct one ?w?????????????assignment #009
......!!!!!!!!...................................
20:42:35 `qNote that there are 9 questions in this assignment. `q001. The process we used in the preceding series of exercises to approximate the graph of y corresponding to the graph of y ' can be significantly improved. Recall that we used the initial slope to calculate the change in the y graph over each interval. For example on the first interval we used the initial slope -6 for the entire interval, even though we already knew that the slope would be -5 by the time we reached the t = 10 point at the end of that interval. We would have been more accurate in our approximation if we had used the average -5.5 of these two slopes. Using the average of the two slopes, what point would we end up at when t = 10?
......!!!!!!!!...................................
RESPONSE --> If the average value slope = -5.5 is used between t = 0 and t = 10, we find that the rise is -55 and we go from (0,100) to (10,45) confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:42:42 If the average value slope = -5.5 is used between t = 0 and t = 10, we find that the rise is -55 and we go from (0,100) to (10,45).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:43:38 `q002. We find that using the average of the two slopes we reach the point (10, 45). At this point we have y ' = -5. By the time we reach the t = 20 point, what will be the slope? What average slope should we therefore use on the interval from t = 10 to t = 20? Using this average slope what point will we end up with when t = 20?
......!!!!!!!!...................................
RESPONSE --> The slope at t = 20 is y ' = -4. Averaging this with the slope y ' = -5 at t = 10 average slope = (-5 + -4) / 2 = -4.5 between t = 10 and t = 20. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:44:06 The slope at t = 20 is y ' = -4. Averaging this with the slope y ' = -5 at t = 10 we estimate an average slope of (-5 + -4) / 2 = -4.5 between t = 10 and t = 20. This would imply a rise of -45, taking the graph from (10,45) to (20, 0).
......!!!!!!!!...................................
RESPONSE --> this takes the graph from (10,45) to (20,0) self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:44:50 `q003. We found that we reach the point (20, 0), where the slope is -4. Following the process of the preceding two steps, what point will we reach at t = 30?
......!!!!!!!!...................................
RESPONSE --> The slope at t = 20 is y ' = -3. Averaging this with the slope y ' = -4 at t = 20 average slope = (-4 + -3) / 2 = -4.5 between t =20 and t = 30. a rise of -35, taking the graph from (20,0) to (30, -35) confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:44:58 The slope at t = 20 is y ' = -3. Averaging this with the slope y ' = -4 at t = 20 we estimate an average slope of (-4 + -3) / 2 = -4.5 between t =20 and t = 30. This would imply a rise of -35, taking the graph from (20,0) to (30, -35).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:46:04 `q004. Continue this process up through t = 70. How does this graph differ from the preceding graph, which was made without averaging slopes?
......!!!!!!!!...................................
RESPONSE --> The average slopes over the next four intervals is -2.5, -1.5, -.5 and +.5. These slopes rises of -25, -15, -5 and +5, taking the graph from (30,-35) to (40, -60) then to (50, -75) then to (60, -80) and finally to (70, -75). confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:46:10 The average slopes over the next four intervals would be -2.5, -1.5, -.5 and +.5. These slopes would imply rises of -25, -15, -5 and +5, taking the graph from (30,-35) to (40, -60) then to (50, -75) then to (60, -80) and finally to (70, -75). Note that the graph decreases at a decreasing rate up to the point (60, -80) before turning around and increasing to the ponit (70, -75).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:46:55 `q005. Now suppose that y = -.2 t^2 + 5 t + 100 represents depth vs. clock time. What is the corresponding rate function y ' ?
......!!!!!!!!...................................
RESPONSE --> The rate function at y = a t^2 + b t + c y ' = 2 a t + b. In this example a = -.2, b = 5 and c = 100 so y ' = 2(-.2) t + 5, or y ' = -.4 t + 5 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:47:14 The rate function corresponding to y = a t^2 + b t + c is y ' = 2 a t + b. In this example a = -.2, b = 5 and c = 100 so y ' = 2(-.2) t + 5, or y ' = -.4 t + 5.
......!!!!!!!!...................................
RESPONSE --> right self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:48:19 `q006. From the depth function and the rate function we can find the coordinates of the t = 30 point of the graph, as well as the rate of change of the function at that point. The rate of change of the function is identical to the slope of the graph at that point, so we can find the slope of the graph. What are the coordinates of the t = 30 point and what is the corresponding slope of the graph? Sketch a graph of a short line segment through that point with that slope.
......!!!!!!!!...................................
RESPONSE --> At t = 30 we have y = -.2 * 30^2 + 5 * 30 + 100 = 70, y ' = -.4 * 30 + 5 = -7. The graph consist of the point (30, 70) anda line segment with slope -7 confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:48:26 At t = 30 we have y = -.2 * 30^2 + 5 * 30 + 100 = 70, and y ' = -.4 * 30 + 5 = -7. The graph will therefore consist of the point (30, 70) and a short line segment with slope -7.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:49:18 `q007. What is the equation of the straight line through the t = 30 point of the depth function of the preceding example, having a slope matching the t = 30 rate of change of that function?
......!!!!!!!!...................................
RESPONSE --> y - 70 = -7 ( x - 30), found by the point-slope form of a straight line. This equation is y = -7 x + 280. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:49:24 A straight line through (30, 70) with slope -7 has equation y - 70 = -7 ( x - 30), found by the point-slope form of a straight line. This equation is easily rearranged to the form y = -7 x + 280.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:50:08 `q008. Evaluate the depth function y at t = 30, 31, 32. Evaluate the y coordinate of the straight line of the preceding question at t = 30, 31, 32. What are your results?
......!!!!!!!!...................................
RESPONSE --> Plugging in t = 30, 31, 32 y = -.2 t^2 + 5 t + 100 we get y = 70, 62.8, 55.2 respectively. Plugging t = 30, 31, 32 into the straight-line equation y = -7 t + 280 we get y = 70, 63, 56 respectively. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:50:13 Plugging t = 30, 31, 32 into the original function y = -.2 t^2 + 5 t + 100 we get y = 70, 62.8, 55.2 respectively. Plugging t = 30, 31, 32 into the straight-line equation y = -7 t + 280 we get y = 70, 63, 56 respectively.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
20:51:25 `q009. By how much does the straight-line function differ from the actual depth function at each of the three points? How would you describe the pattern of these differences?
......!!!!!!!!...................................
RESPONSE --> the functions at t = 30 are the same At t = 31 the values are 62.8 and 63; the function is .2 units below the straight line. At t = 32 the values are 55.2 and 56; the function is now .8 units below the straight line. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:51:31 At t = 30 the two functions are identical. At t = 31 the values are 62.8 and 63; the function is .2 units below the straight line. At t = 32 the values are 55.2 and 56; the function is now .8 units below the straight line. The pattern of differences is therefore 0, -.2, -.8. The function falls further and further below the straight line as we move away from the point (30, 70).
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
}???????q???? assignment #009 ??K???????? Calculus I 09-28-2008
......!!!!!!!!...................................
20:52:50 09-28-2008 20:52:50 Query class notes #10 Describe in your own words how the predictor-corrector method of solving a differential equation works
......!!!!!!!!...................................
NOTES -------> calculate a dT/dt for this T. The two values of dT / dt then averaged This is then used to calculate a new change in T. This change is added to the original T.
.......................................................!!!!!!!!...................................
20:52:52 Query class notes #10 Describe in your own words how the predictor-corrector method of solving a differential equation works
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
20:52:59 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:54:07 Problem 1.5.13. amplitude, period of 5 + cos(3x)
......!!!!!!!!...................................
RESPONSE --> 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. period of this function =2 pi /3. The cosine function is multiplied by 1 so the amplitude is 1. The function is vertically shifted 5 units
.................................................
......!!!!!!!!...................................
20:54:11 *&*& 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. *&*&
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:54:57 Explain how you determine the amplitude and period of a given sine or cosine function.
......!!!!!!!!...................................
RESPONSE --> 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 / x
.................................................
......!!!!!!!!...................................
20:55:01 *&*& 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. *&*&
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:57:37 query Problem 1.5.28. trig fn graph given, defined by 3 pts (0,3), (2,6), (4,3), (6,0), (8,3)
......!!!!!!!!...................................
RESPONSE --> The cycle goes from y = 3 to y = 6, back to y = 3 then to a min of y = 0, then back to y = 3. the period = 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 3 sin( 2 `pi / 8 * x), mean = 0, shifted vertically 3 units mean = 3. The function = y = 3 + 3 sin( `pi / 4 * x).
.................................................
......!!!!!!!!...................................
20:58:10 What is a possible formula for the graph?
......!!!!!!!!...................................
RESPONSE --> y = 3 + 3 sin( `pi / 4 * x).
.................................................
......!!!!!!!!...................................
20:58:15 ** 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). **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:59:34 problem 1.5.30. Solve 1 = 8 cos(2x+1) - 3 for x.
......!!!!!!!!...................................
RESPONSE --> 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. x = (`pi / 3 - 1) / 2 + n `pi, n an integer, is the solution. Since 0 (`pi / 3 - 1) / 2 < `pi t he first two solutions are between 0 and 2 `pi and are the only solutions between 0 and 2 `pi.
.................................................
......!!!!!!!!...................................
20:59:48 ** 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. **
......!!!!!!!!...................................
RESPONSE --> basically what I said
.................................................
......!!!!!!!!...................................
21:00:24 problem 1.5.47 (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
......!!!!!!!!...................................
RESPONSE --> confused
.................................................
......!!!!!!!!...................................
21:01:29 ** 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].**
......!!!!!!!!...................................
RESPONSE --> ok I understand now 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
.................................................
......!!!!!!!!...................................
21:01:38 Query Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> no comments
.................................................
......!!!!!!!!...................................
21:01:45 None.
......!!!!!!!!...................................
RESPONSE --> k
.................................................