course Mth 163 }زâ·ï–ä}ËÅS¯ÍÇÈîšyäŽ÷¤£ðassignment #010
......!!!!!!!!...................................
20:47:21 query the family of linear functions, Problem 2. Describe the graphs of y = A f(x) for A = -.3 and A = 1.3 and compare; explain the comparison.
......!!!!!!!!...................................
RESPONSE --> The graph of y = Af(x) where f(x) is -.3 has a line moving in the negative direction through the (0,0) coordinate, while the line for f(x) = 1.3 is a steeper line moving through the (0,0) coordinate in a positive direction. confidence assessment: 2
.................................................
......!!!!!!!!...................................
20:47:45 ** For the basic linear function f(x) = x the A = -.3 graph is obtained by vertically stretching the y = x function by factor -.3, resulting in a straight line thru the origin with slope -.3, basic points (0,0) and (1, -.3), and the A = 1.3 graph is obtained by vertically stretching the y = x function by factor 1.3, resulting in is a straight line thru the origin with slope 1.3, basic points (0,0) and (1, 1.3). **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:01:39 describe the graphs of y = f(x) + c for c = .3 and c = -2.7 and compare; explain the comparison.
......!!!!!!!!...................................
RESPONSE --> When c = .3 we find a line that passes through the y-axis at coordinate (0, .3) and is moving in the positive direction. When c = -2.7 we find a line passing through the y-axis at coordinate (0, -2.7) which means it is 3 units down from the other, but is increasing in a positive manner. confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:01:52 ** The graphs will have slopes identical to that of the original function, but their y intercepts will vary from -2.7 to .3. **
......!!!!!!!!...................................
RESPONSE --> 3 self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:05:44 query problem 4. linear function y = f(x) = -1.77 x - 3.87 What are your symbolic expressions, using x1 and x2, for the corresponding y coordinates y1 and y2.
......!!!!!!!!...................................
RESPONSE --> x1 = -1 x2 = 4 y1 = -2.1 y2 = -10.95 confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:06:53 ** y1 = f(x1) = -1.77 x1 - 3.87 y2 = f(x2) = -1.77 x2 - 3.87. `dy = y2 - y1 = -1.77 x2 - 3.87 - ( -1.77 x1 - 3.87) = -1.77 x2 + 1.77 x1 - 3.87 + 3.87 = -1.77 ( x2 = x1). Thus slope = `dy / `dx = -1.77 (x2 - x1) / (x2 - x1) = -1.77. This is the slope of the straight line, showing that these symbolic calculations are consistent. **
......!!!!!!!!...................................
RESPONSE --> Not sure self critique assessment: 1
Deconstruct the given solution and explain in detail what you do and do not understand about every part.. &#
.................................................
......!!!!!!!!...................................
21:14:16 query problem 5. graphs of families for y = mx + b. Describe your graph of the family: m = 2, b varies from -3 to 3 by step 1.
......!!!!!!!!...................................
RESPONSE --> When finding point of y and x when using the equation y = 2x + -3, (-2, -1, 0 1, 2, 3) you get lines that are parallel to one another passing through the y-axis at different point which ar (0, -3), (0, -2) etc...to (0, 3). confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:14:31 ** The graphs will all have slope 2 and will pass thru the y axis between y = -3 and y = 3. The family will consist of all such graphs. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:21:34 query problem 6. three basic points graph of y = .5 x + 1 what are your three basic points?
......!!!!!!!!...................................
RESPONSE --> y-intercept = (0,1) x-intercept = (-2, 0) 1 unit to the right = (2, 2)
.................................................
......!!!!!!!!...................................
21:21:58 ** This is of the form y = mx + b with b= 1. So the y intercept is (0, 1). The point 1 unit to the right is (1, 1.5). The x-intercept occurs when y = 0, which implies .5 x + 1 = 0 with solution x = -2, so the x-intercept is (-2, 0). **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:23:06 query problem 6. three basic points graph of y = .5 x + 1 What are your three basic points?
......!!!!!!!!...................................
RESPONSE --> y-intercept = (0, 1) x- Intercept = (-2, 0) 1 unit to the right = (1, 1.5) confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:23:15 ** The y intercept occurs when x = 0, at which point we have y = .5 * 0 + 1 = 1. So one basic point is (0, 1). The point 1 unit to the right of the y axis occurs at x = 1, where we get y = .5 * 1 + 1 = 1.5 to give us the second basic point (1, 1.5) }The third point, which is not really necessary, is the x intercept, which occurs when y = 0. This gives us the equation 0 = .5 x + 1, with solution x = -2. So the third basic point is (-2, 0). **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:29:33 query problem 7. simple pendulum force vs. displacement What are your two points and what line do you get from the two points?
......!!!!!!!!...................................
RESPONSE --> The two points I got was (1.1, .21) and (4.1, 1.04). The line I got was y = .277x + b confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:30:06 STUDENT RESPONSE: The two points are (1.1, .21) and (2.0, .54). These points give us the two simultaneous equations .21- m(1.1) + b .54= m(2.0) +b. If we solve for m and b we will get our y = mx + b form. INSTRUCTOR COMMENT: I believe those are data points. I doubt if the best-fit line goes exactly through two data points. In the future you should use points on your sketched line, not data points. However, we'll see how the rest of your solution goes based on these points. **
......!!!!!!!!...................................
RESPONSE --> ok confidence assessment: 3
.................................................
......!!!!!!!!...................................
21:34:13 STUDENT RESPONSE: b= .21 m=.19 INSTRUCTOR COMMENT: ** b would be the y intercept, which is not .21 since y = .21 when x = 1.1 and the slope is nonzero. If you solve the two equations above for m and b you obtain m = .367 and b = -.193. This gives you equation y = mx + b or y = .367 x - .193. **
......!!!!!!!!...................................
RESPONSE --> My slope was equal to .277 My y-intercept = -.0947 self critique assessment: 2
.................................................
......!!!!!!!!...................................
21:36:31 ** Your linear regression model would be obtained using a graphing calculator or DERIVE. As a distance student you are not required to use these tools but you should be aware that they exist and you may need to use them in other courses. **
......!!!!!!!!...................................
RESPONSE --> OK self critique assessment: 1
.................................................
......!!!!!!!!...................................
21:47:58 What force would be required to hold the pendulum 47 centimeters from its equilibrium position? what equation did you solve to obtain this result?
......!!!!!!!!...................................
RESPONSE --> y = .277(47) * -.0947 y = 13.019 * -.0947 y = 1.233 confidence assessment: 1
.................................................
......!!!!!!!!...................................
21:48:33 ** If your model is y = .367 x - .193, with y = force and x= number of cm from equilibrium, then we have x = 47 and we get force = y = .367 * 47 - .193 = 17 approx. The force would be 17 force units. **
......!!!!!!!!...................................
RESPONSE --> If your model is y = .367 x - .193, with y = force and x= number of cm from equilibrium, then we have x = 47 and we get force = y = .367 * 47 - .193 = 17 approx. The force would be 17 force units. ** Got it. self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:50:55 STUDENT RESPONSE: I used the equation f= .10*47+.21 and got the answer 15.41 which would be to much force to push or pull INSTRUCTOR COMMENT: ** The problem is that you can't hold a pendulum further at a distance greater than its length from its equilibrium point--the string isn't long enough. **
......!!!!!!!!...................................
RESPONSE --> ** The problem is that you can't hold a pendulum further at a distance greater than its length from its equilibrium point--the string isn't long enough. ** self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:56:23 How far could you hold the pendulum from its equilibrium position using a string with a breaking strength of 25 pounds? what equation did you solve to obtain this result?
......!!!!!!!!...................................
RESPONSE --> 25 = .367x - .193 25.193 = .367x 25.193 / .367 = x = 68.65cm confidence assessment: 2
.................................................
......!!!!!!!!...................................
21:56:34 ** Using the model y = .367 * x - .193 with y = force = 25 lbs we get the equation 25 = .367 x - .193, which we solve to obtain x = 69 (approx.). Note that this displacement is also unrealistic for this pendulum. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:57:32 ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium we can use any two (x, y) points to get the rate of change. In all cases we will get rate of change = change in y / change in x = .367. The change in y is the change in the force, while the change in x is the change in position. The rate of change therefore tells us how much the force changes per unit of change in position (e.g., the force increases by 15 pounds for every inch of displacement). **
......!!!!!!!!...................................
RESPONSE --> The change in y is the change in the force, while the change in x is the change in position. The rate of change therefore tells us how much the force changes per unit of change in position (e.g., the force increases by 15 pounds for every inch of displacement). Average rate of change is equal to .367 self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:57:55 ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium the average slope is .367. **
......!!!!!!!!...................................
RESPONSE --> Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium the average slope is .367. self critique assessment: 3
.................................................
......!!!!!!!!...................................
21:59:56 As you gradually pull the pendulum from a point 30 centimeters from its equilibrium position to a point 80 centimeters from its equilibrium position, what average force must you exert?
......!!!!!!!!...................................
RESPONSE --> y = .367(80-30cm) - .193 y = 18.35 - .193 y = 18.157 confidence assessment: 2
.................................................
......!!!!!!!!...................................
22:00:51 ** if it was possible to pull the pendulum back this far and if the model applies you will get Force at 30 cm: y = .367 * 30 - .193 = 10.8 approx. and Force at 80 cm: y = .367 * 80 - .193 = 29 approx. so that ave force between 30 cm and 80 cm is therefore (10.8 + 29) / 2 = 20 approx.. **
......!!!!!!!!...................................
RESPONSE --> i did it a little different and cam up with approx. 18, but I now know it is wrong. I understand the correct way. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:03:24 ** STUDENT RESPONSE: I obtained model one by drawing a line through the data points and picking two points on the line and finding the slope between them. I then substituted this value for m and used one of my data points on my line for the x and y value and solved for b. the line I got was range(t) = -.95t + 112.38. y = -16/15x + 98 INSTRUCTOR COMMENT: This looks like a good model. According to the instructions it should however be expressed as range(time) = -16/15 * time + 98. **
......!!!!!!!!...................................
RESPONSE --> range(time) = -16/15 * time + 98 self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:04:06 ** the average rate of change is change in range / change in clock time. The average rate of change indicates the average rate at which range in cm is changing with respect to clock time in sec, i.e., the average number of cm / sec at which the range changes. Thus the average rate tells us how fast, on the average, the range changes. **
......!!!!!!!!...................................
RESPONSE --> the average rate of change is change in range / change in clock time. The average rate of change indicates the average rate at which range in cm is changing with respect to clock time in sec, i.e., the average number of cm / sec at which the range changes. Thus the average rate tells us how fast, on the average, the range changes. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:04:38 ** it's the average rate at which the range of the flow changes--the average rate at which the position of the end of the stream changes. It's the speed with which the point where the stream reaches the ground moves across the ground. **
......!!!!!!!!...................................
RESPONSE --> it's the average rate at which the range of the flow changes--the average rate at which the position of the end of the stream changes. It's the speed with which the point where the stream reaches the ground moves across the ground. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:09:02 ** Total wealth has to be expressed in terms of t. A graph of total wealth vs. t would have y intercept 3956, since that is the t = 0 value, and slope 8, since slope represents change in total wealth / change in t, i.e., the number of dollars per hour. A graph with y-intercept b and slope m has equation y = m t + b. Thus we have totalWealth(t) = 8 * t + 3956 . **
......!!!!!!!!...................................
RESPONSE --> Total wealth has to be expressed in terms of t. A graph of total wealth vs. t would have y intercept 3956, since that is the t = 0 value, and slope 8, since slope represents change in total wealth / change in t, i.e., the number of dollars per hour. A graph with y-intercept b and slope m has equation y = m t + b. Thus we have totalWealth(t) = 8 * t + 3956 . self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:10:23 At what clock time will your total wealth reach $4000? what equation did you solve to obtain this result?
......!!!!!!!!...................................
RESPONSE --> 4000 = 8 * t + 3956 44 = 8t 44/8 = t = 5.5 confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:10:34 STUDENT RESPONSE: To find the clock time when my total wealth will reach 4000 I solved the equation totalWealth(t) = 4000. The value I got when I solved for t was t = 5.5 hours. 4.4 hours needed to reach 4000 4000 = 10x + 3956 INSTRUCTOR COMMENT: Almost right. You should solve 4000 = 8 x + 3956, obtaining 5.5 hours. This is equivalent to solving totalWealth(t) = 4000 = 8 t + 3956, which is the more meaningful form of the relationship. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:11:00 GOOD STUDENT RESPONSE: The slope of the graph shows the steady rate at which money is earned on an hourly basis. It shows a steady increase in wealth.
......!!!!!!!!...................................
RESPONSE --> The slope of the graph shows the steady rate at which money is earned on an hourly basis. It shows a steady increase in wealth. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:14:21 If you make a graph of y = numberSold vs. x = price you have graph points (30, 200) and (28, 300). You need the equation of the straight line through these points. You plug these coordinates into the form y = m x + b and solve for m and b. Or you can use another method. Whatever method you use you get y = -50 x + 1700. Then to put this into the notation of the problem you write numberSold(price) instead of y and price instead of x. You end up with the equation numberSold(price) = -50 * price + 1700. **
......!!!!!!!!...................................
RESPONSE --> If you make a graph of y = numberSold vs. x = price you have graph points (30, 200) and (28, 300). You need the equation of the straight line through these points. You plug these coordinates into the form y = m x + b and solve for m and b. Or you can use another method. Whatever method you use you get y = -50 x + 1700. Then to put this into the notation of the problem you write numberSold(price) instead of y and price instead of x. You end up with the equation numberSold(price) = -50 * price + 1700. confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:14:50 ** If the variables are y and x, you know y so you can solve for x. For the function numberSold(price) = -50 * price + 1700 you substitute 220 for numbersold(price) and solve for price. You get the equation 220 = -50 * price + 1700 which you can solve to get price = 30, approx. **
......!!!!!!!!...................................
RESPONSE --> For the function numberSold(price) = -50 * price + 1700 you substitute 220 for numbersold(price) and solve for price. You get the equation 220 = -50 * price + 1700 which you can solve to get price = 30, approx. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:16:35 STUDENT RESPONSE: If each widget costs the store $25, then they should expect to earn a profit of 300 dollars from a selling price of $28, 250 dollars from a price of $29 and 200 dollars from a price of $30. To find this I solved the equations numberSold(28); numberSold(29), and numberSold(30). Solving for y after putting the price values in for p. They will sell 300, 250 and 200 widgets, respectively (found by solving the given equation). To get the total profit you have to multiply the number of widgets by the profit per widget. At $28 the profit per widgit is $3 and the total profit is $3 * 300 = $900; at $30 the profit per widgit is $5 and 200 are sold for profit $1000; at $29 what happens? **
......!!!!!!!!...................................
RESPONSE --> They will sell 300, 250 and 200 widgets self critique assessment: 1
.................................................
......!!!!!!!!...................................
22:24:28 query problem 11. quadratic function depth(t) = .01 t^2 - 2t + 100 representing water depth vs. What is the equation of the straight line connecting the t = 20 point of the graph to the t = 60 point?
......!!!!!!!!...................................
RESPONSE --> depth(t) = .01t^2 - 2t + 100 depth(t) = .01(20^2) - 2(20) + 100 depth(t) = 4 - 40 + 100 depth(t) = 136 y = .01t^2 - 2t + 100 y = .01(60^2) - 2(60) + 100 y = 36 - 120 + 100 y = 16 Points equal = (20, 136) (60, 16) Slope = (16-136)/(60-20) = -120 / 40 = -3 confidence assessment: 1
.................................................
......!!!!!!!!...................................
22:26:10 ** The t = 20 point is (20,64) and the t = 60 point is (60, 16), so the slope is (-48 / 20) = -1.2. This can be plugged into the form y = m t + b to get y = -1.2 t + b. Then plugging in the x and y coordinates of either point you get b = 88. y = -1.2 t + 88 **
......!!!!!!!!...................................
RESPONSE --> The t = 20 point is (20,64) and the t = 60 point is (60, 16), so the slope is (-48 / 20) = -1.2. This can be plugged into the form y = m t + b to get y = -1.2 t + b. Then plugging in the x and y coordinates of either point you get b = 88. y = -1.2 t + 88 I calculated incorrectly self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:29:22 query problem 13. quadratic depth function y = depth(t) = .01 t^2 - 2t + 100. What is `dy / `dt based on the two time values t = 30 sec and t = 40 sec.
......!!!!!!!!...................................
RESPONSE --> y = .01(30^2) - 2(30) + 100 y = 9 - 60 + 100 y = 49 y= .01(40^2) - 2(40) + 100 y = 16 - 80 + 100 y = 36 (36-49) / (40-30) = -13/10 = -1.3 confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:29:30 ** For t = 30 we have y = 49 and for t = 40 we have y = 36. The slope between (30, 49) and (40,36) is (36 - 49) / (40 - 30) = -1.3. This tells you that the depth is changing at an average rate of -1.3 cm / sec. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:30:15 ** Based on t = 30 and t = 31 the value for `dy / `dt is -1.39, following the same steps as before **
......!!!!!!!!...................................
RESPONSE --> 'dy / `dt is -1.39 self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:30:51 ** STUDENT RESPONSE: The value for 'dy / `dt based on t = 30 sec and t = 30.1 sec is -1.4 INSTRUCTOR COMMENT: ** Right if you round off the answer. However the answer shouldn't be rounded off. Since you are looking at a progression of numbers (-1.3, -1.39, and this one) and the differences in these numbers get smaller and smaller, you have to use a precision that will always show you the difference. Exact values are feasible here and shoud be used. I believe that this one comes out to -1.399. **
......!!!!!!!!...................................
RESPONSE --> -1.399 self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:31:28 STUDENT RESPONSE: An even more and more accurate slope value. I don't think it would have to continue to decrease. INSTRUCTOR COMMENT **If you look at the sequence -1.3, -1.39, -1.399, ..., what do you think happens? It should be apparent that the limiting value is -1.4 **
......!!!!!!!!...................................
RESPONSE --> If you look at the sequence -1.3, -1.39, -1.399, ..., what do you think happens? It should be apparent that the limiting value is -1.4 self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:31:48 ** The function tells you that at any clock time t the rate of depth change is given by the function .02 t - 2. For t = 30, for example, this gives us .02 * 30 - 2 = -1.4, which is the rate we obtained from the sequence of calculations above. **
......!!!!!!!!...................................
RESPONSE --> The function tells you that at any clock time t the rate of depth change is given by the function .02 t - 2. For t = 30, for example, this gives us .02 * 30 - 2 = -1.4, which is the rate we obtained from the sequence of calculations above. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:36:44 .
......!!!!!!!!...................................
RESPONSE --> I got coordinates (1, 8.46), (2, 8.83), (3, 9.2), (4, 9.57), (5, 9.94). The pattern is that the y increases by .37 each time. confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:37:11 ** The first five terms are 8.46, 8.83, 9.2, 9.57, and 9.94 **
......!!!!!!!!...................................
RESPONSE --> 8.46, 8.83, 9.2, 9.57, and 9.94 self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:37:26 What is the pattern of these numbers?
......!!!!!!!!...................................
RESPONSE --> increase in .37 confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:37:34 ** These numbers increase by .37 at each interval. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:39:00 If you didn't know the equation for the function, how would you go about finding the 100th member of the sequence? How can you tell your method is valid?
......!!!!!!!!...................................
RESPONSE --> You could multiply .37 by 100 and add it to 8.09. confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:39:19 ** You could find the 100th member by noting that you have 99 ‘jumps’ between the first number and the 100 th, each ‘jump’ being .37. Multiplying 99 times .37 and then adding the result to the 'starting value' (8.46). STUDENT RESPONSE: simply put 100 as the x in the formula .37x +8.09 INSTRUCTOR COMMENT: That's what you do if you have the equation. Given just the numbers you could find the 100th member by multiplying 99 times .37 and then adding the result to the first value 8.46. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:42:47 for quadratic function y = g(x) = .01 x^2 - 2x + 100 what are the first five terms of the basic sequence {g(n), n = 1, 2, 3, ...}?
......!!!!!!!!...................................
RESPONSE --> 98.01, 96.04, 94.09, 92.16, 90.25 confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:42:57 ** We have g(1) = .01 * 1^2 - 2 * 1 + 100 = 98.01 g(2) = .01 * 2^2 - 2 * 2 + 100 = 96.04, etc. The first 5 terms are therefore {98.01, 96.04, 94.09, 92.16, 90.25}
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:44:14 What is the pattern of these numbers?
......!!!!!!!!...................................
RESPONSE --> With each number you start to lose .01 * the number used starting from 2.00 confidence assessment: 3
.................................................
......!!!!!!!!...................................
22:44:28 ** The changes in these numbers are -1.97, -1.95, -1.93, -1.91. With each interval of x, the change in y is .02 greater than for the previous interval. **
......!!!!!!!!...................................
RESPONSE --> The changes in these numbers are -1.97, -1.95, -1.93, -1.91. With each interval of x, the change in y is .02 greater than for the previous interval. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:44:48 ** According to the pattern established above, the next three changes are -1.89, -1.87, -1.85. This gives us g(6) = g(5) - 1.89, g(7) = g(6) - 1.87, g(7) = g(6) - 1.85. **
......!!!!!!!!...................................
RESPONSE --> According to the pattern established above, the next three changes are -1.89, -1.87, -1.85. This gives us g(6) = g(5) - 1.89, g(7) = g(6) - 1.87, g(7) = g(6) - 1.85 self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:45:17 ** You can verify the result using the original formula; if you evaluate it at 5, 6 and 7 it should confirm your results. That's the best answer that can be given at this point. You should understand, though that even if you verified it for the first million terms, that wouldn't really prove it (who knows what might happen at the ten millionth term, or whatever). It turns out that to prove it would require calculus or the equivalent. **
......!!!!!!!!...................................
RESPONSE --> You can verify the result using the original formula; if you evaluate it at 5, 6 and 7 it should confirm your results. That's the best answer that can be given at this point. You should understand, though that even if you verified it for the first million terms, that wouldn't really prove it (who knows what might happen at the ten millionth term, or whatever). It turns out that to prove it would require calculus or the equivalent. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:46:28 ** You get a(1+1) = a(1) + .4, or a(2) = a(1) + .4. Knowing a(1) = 5 you get a(2) = 5.4. **
......!!!!!!!!...................................
RESPONSE --> The difference equation a(n+1) = a(n) + .4, a(1) = 5 If you substitute n = 1 into a(n+1) = a(n) + .4, how do you determine a(2)? ** You get a(1+1) = a(1) + .4, or a(2) = a(1) + .4. Knowing a(1) = 5 you get a(2) = 5.4. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:46:51 ** You have to do the substitution. You get a(2+1) = a(2) + .4, or since 2 + 1 = 3, a(3) = a(2) + .4 Then knowing a(2) = 5.4 you get a(3) = 5.4 + .4 = 5.8. **
......!!!!!!!!...................................
RESPONSE --> You get a(2+1) = a(2) + .4, or since 2 + 1 = 3, a(3) = a(2) + .4 Then knowing a(2) = 5.4 you get a(3) = 5.4 + .4 = 5.8. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:47:09 ** We get a(4) = a(3) +.4 = 5.8 + .4 = 6.2 **
......!!!!!!!!...................................
RESPONSE --> We get a(4) = a(3) +.4 = 5.8 + .4 = 6.2 self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:47:28 ** a(100) would be equal to a(1) plus 99 jumps of .4, or 5 + 99*.4 = 44.6. **
......!!!!!!!!...................................
RESPONSE --> a(100) would be equal to a(1) plus 99 jumps of .4, or 5 + 99*.4 = 44.6. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:47:50 ** You get a(2) = a(1) + 2 * 1 = 4 + 2 = 6, then a(3) = a(2) + 2 * 3 = 6 + 6 = 12 then a(4) = a(3) + 2 * 4 = 12 + 8 = 20; etc. The sequence is 6, 12, 20, 30, 42, ... . **
......!!!!!!!!...................................
RESPONSE --> You get a(2) = a(1) + 2 * 1 = 4 + 2 = 6, then a(3) = a(2) + 2 * 3 = 6 + 6 = 12 then a(4) = a(3) + 2 * 4 = 12 + 8 = 20; etc. The sequence is 6, 12, 20, 30, 42 self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:48:28 ** The differences of the sequence are 6, 8, 10, 12, . . .. The difference change by the same amount each time, which is a property of quadratic functions. **
......!!!!!!!!...................................
RESPONSE --> The difference change by the same amount each time, which is a property of quadratic functions. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:48:50 ** The slope = slope equation sets the slope between two given points equal to the slope between one of those points and the variable point (x, y). Since all three points lie on the same straight line, the slope between any two of the three points must be equal to the slope between any other pair. **
......!!!!!!!!...................................
RESPONSE --> The slope = slope equation sets the slope between two given points equal to the slope between one of those points and the variable point (x, y). Since all three points lie on the same straight line, the slope between any two of the three points must be equal to the slope between any other pair. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:49:18 ** The rate at which streamRange changes is change in streamRange / change in t = -10 cm / (5 sec) = -2 cm/s. This will be the slope m of the graph. Since streamRange is 50 cm when t = 20 sec the point (20, 50) lies on the graph. So the graph passes through (20, 50) and has slope -2. The function is therefore of the form y = m t + b with m = -2, and b such that 50 = -2 * 20 + b. Thus b = 90. The function is therefore y = -2 t + 90, or using the meaningful name of the function steamRange(t) = -2t + 90 You need to use function notation. y = f(x) = -2x + 90 would be OK, or just f(x) = -2x + 90. The point is that you need to give the funcion a name. Another idea here is that we can use the 'word' streamRange to stand for the function. If you had 50 different functions and, for example, called them f1, f2, f3, ..., f50 you wouldn't remember which one was which so none of the function names would mean anything. If you call the function streamRange it has a meaning. Of course shorter words are sometimes preferable; just understand that function don't have to be confined to single letters and sometimes it's not a bad idea to make the names easily recognizable. STUDENT RESPONSE: y = -2x + 50 INSTRUCTOR COMMENT: ** At t = 20 sec this would give you y = -2 * 20 + 50 = 10. But y = 50 cm when t = 20 sec. Slope is -10 cm / (5 sec) = -2 cm/s, so you have y = -2 t + b. Plug in y = 50 cm and t = 20 sec and solve for b. You get b = 90 cm. The equation is y = -2 t + 90, or streamRange(t) = -2t + 90. **
......!!!!!!!!...................................
RESPONSE --> The rate at which streamRange changes is change in streamRange / change in t = -10 cm / (5 sec) = -2 cm/s. This will be the slope m of the graph. Since streamRange is 50 cm when t = 20 sec the point (20, 50) lies on the graph. So the graph passes through (20, 50) and has slope -2. The function is therefore of the form y = m t + b with m = -2, and b such that 50 = -2 * 20 + b. Thus b = 90. The function is therefore y = -2 t + 90, or using the meaningful name of the function steamRange(t) = -2t + 90 You need to use function notation. y = f(x) = -2x + 90 would be OK, or just f(x) = -2x + 90. The point is that you need to give the funcion a name. Another idea here is that we can use the 'word' streamRange to stand for the function. If you had 50 different functions and, for example, called them f1, f2, f3, ..., f50 you wouldn't remember which one was which so none of the function names would mean anything. If you call the function streamRange it has a meaning. Of course shorter words are sometimes preferable; just understand that function don't have to be confined to single letters and sometimes it's not a bad idea to make the names easily recognizable. STUDENT RESPONSE: y = -2x + 50 INSTRUCTOR COMMENT: ** At t = 20 sec this would give you y = -2 * 20 + 50 = 10. But y = 50 cm when t = 20 sec. Slope is -10 cm / (5 sec) = -2 cm/s, so you have y = -2 t + b. Plug in y = 50 cm and t = 20 sec and solve for b. You get b = 90 cm. The equation is y = -2 t + 90, or streamRange(t) = -2t + 90. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:49:36 ** Using the correct equation streamRange(t) = -2t + 90, you would set streamRange(t) = 12 and solve 12 = -2t + 90, obtaining t = 39 sec. **
......!!!!!!!!...................................
RESPONSE --> Using the correct equation streamRange(t) = -2t + 90, you would set streamRange(t) = 12 and solve 12 = -2t + 90, obtaining t = 39 sec. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:49:55 ** You have to get the whole equation. y = m t + b is now y = -1.88 t + b. You have to solve for b. Plug in the coordinates of the t = 7 point and find b. You get 90.9 = -1.88 * 7 + b so b = 104, approximately. Find the correct value. The equation will end up something like y = depth(t) = -1.88 t + 104. **
......!!!!!!!!...................................
RESPONSE --> You have to get the whole equation. y = m t + b is now y = -1.88 t + b. You have to solve for b. Plug in the coordinates of the t = 7 point and find b. You get 90.9 = -1.88 * 7 + b so b = 104, approximately. Find the correct value. The equation will end up something like y = depth(t) = -1.88 t + 104. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:50:15 ** The deviations are for t = 3, 4, 5, 6, 7, & 8 as follows: .08, .03. 0. -.01, 0, .03. **
......!!!!!!!!...................................
RESPONSE --> The deviations are for t = 3, 4, 5, 6, 7, & 8 as follows: .08, .03. 0. -.01, 0, .03. self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:50:34 ** Not counting t= 5 and t = 7, which are 0, the next closest t value is t = 6, the deviation for this is -.01. **
......!!!!!!!!...................................
RESPONSE --> Not counting t= 5 and t = 7, which are 0, the next closest t value is t = 6, the deviation for this is -.01 self critique assessment: 3
.................................................
......!!!!!!!!...................................
22:50:49 ** On the t = 4 side the approximation is closer. The quadratic function curves away on the positive x side. **
......!!!!!!!!...................................
RESPONSE --> On the t = 4 side the approximation is closer. The quadratic function curves away on the positive x side. self critique assessment: 3
.................................................