......!!!!!!!!...................................
14:22:59
`questionNumber 10000 Note that there are four questions in this assignment. `q001. If your stocks are worth $5000 in mid-March, $5300 in mid-July and $5500 in mid-December, during which period, March-July or July-December was your money growing faster?......!!!!!!!!...................................
RESPONSE --> 5300-5000= $300(rise) over 4 month span(run)
5500-5300= $200(rise) over 5 month span(run) so: 300/4= $75 per month(March-July) 200/5= $40 per month(July-December) $75>$40 clearly March-July had a faster rate of growth per month..................................................
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14:24:35
`questionNumber 10000 The first period (4 months) was shorter than the second (5 months), and the value changed by more during the shorter first period (increase of $300) than during the longer second perios (increase of $200). A greater increase in a shorter period implies a greater rate of change. So the rate was greater during the first period.......!!!!!!!!...................................
RESPONSE --> ok
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14:25:20
`questionNumber 10000 `q002. What were the precise average rates of change during these two periods?......!!!!!!!!...................................
RESPONSE --> see #1 ans.
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14:25:40
`questionNumber 10000 From mid-March thru mid-July is 4 months. A change of $300 in four months gives an average rate of change of $300 / (4 months) = $75/month. From mid-July through mid-December is 5 months, during which the value changes by $200, giving an average rate of $200 / (5 months) = $40 / month. Thus the rate was greater during the first period than during the second.......!!!!!!!!...................................
RESPONSE --> yup.
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15:34:28
`questionNumber 10000 `q003. If the water in a uniform cylinder is leaking from a hole in the bottom, and if the water depths at clock times t = 10, 40 and 90 seconds are respectively 80 cm, 40 cm and 20 cm, is the depth of the water changing more quickly or less quickly, on the average, between t = 10 sec and t = 40 sec, or between t = 40 sec and t = 90 sec?......!!!!!!!!...................................
RESPONSE --> 1 A) 80-40=40cm diff in level 40-10=30sec elapsed time
B) 40-20=20cm diff in level 90-40=50sec elapsed time 2 40/30=1.33_ (cm/sec) fall rate A 20/50=.4 (cm/sec) fall rate B 3 1.33_ (cm/sec) > .4 (cm/sec) 4 between t = 10 sec and t = 40 sec is more quickly between t = 40 sec and t = 90 sec is less quickly.................................................
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15:34:46
`questionNumber 10000 Between clock times t = 10 sec and t = 40 sec the depth changes by -40 cm, from 80 cm to 40 cm, in 30 seconds. The average rate is therefore -40 cm / (30 sec) = -1.33 cm/s. Between t = 40 sec and t = 90 sec the change is -20 cm and the time interval is 50 sec so the average rate of change is -20 cm/ 50 sec = -.4 cm/s, approx. The depth is changing more quickly during the first time interval.......!!!!!!!!...................................
RESPONSE --> ok
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15:44:42
`questionNumber 10000 `q004. How are the two preceding questions actually different versions of the same question? How it is the mathematical reasoning the same in both cases?......!!!!!!!!...................................
RESPONSE --> A )because they both ask to determine a rate of change by an average.
B) because they both employ a measurable rate of change on a standard scale that can be averaged out..................................................
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15:45:03
`questionNumber 10000 In each case we were given a quantity that changed with time. We were given the quantities at three different clock times and we were asked to compare the rates of change over the corresponding intervals. We did this by determining the changes in the quantities and the changes in the clock times, and for each interval dividing change in quantity by change in clock time. We could symbolize this process by representing the change in clock time by `dt and the change in the quantity by `dQ. The rate is thus rate = `dQ /`dt.......!!!!!!!!...................................
RESPONSE --> ok.
.................................................
}ϠԛhX Student Name: assignment #002 002.
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16:23:15
`questionNumber 20000 Note that there are four questions in this assignment. `q001. Recall the stock value problem, where March, July and December values were $5000, $5300 and $5500. Construct a graph of stock value vs. number of month (e.g., 1 for Jan, 2 for Feb, etc.). You will have three points on your graph, one corresponding to the March value, one to the July value, and one to the December value. Stock value will be on the y axis and month number on the x axis. Your first point, for example, will be (3, 5000), corresponding to $5000 in March. Connect your three points with straight lines--i.e., connect the first point to the second and the second to the third. What is the slope of your line between the first and second point, and what is the slope of your line between the second in the third point? Recall that slope is rise / run.......!!!!!!!!...................................
RESPONSE --> 1) 300/4 or 75
1) 200/5 or 40.................................................
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16:23:47
`questionNumber 20000 The three points on the graph are (3, 5000), (7, 5300) and (12, 5500). The rise between the first point and the second is from 5000 to 5300, or 300, and the run is from 3 to 7, or 4, so the slope is 300 / 4 = 75. Note that the 300 represents $300 and the 4 represents 4 months, so the slope represents $300 / (4 months) = $75 / month, which is the average rate of change during the first time interval. The rise between the second point and the third is from 5300 to 5500, or 200, and the run from 7 to 12 is 5, so the slope is 200 / 5 = 40. This slope represents the $40/month average rate of change during the second time interval. Click on 'Next Picture' to see the graph.......!!!!!!!!...................................
RESPONSE --> ok
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16:28:49
`questionNumber 20000 `q002. Look at your results for the slopes, and look the results for the average rates of change. What do you notice? In what way then does the graph represent the average rate of change?......!!!!!!!!...................................
RESPONSE --> 1) the top triangle is shorter and wider the bottom one. 2) as the top triangle having a slower rate of change.
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16:29:11
`questionNumber 20000 We see from this example that the slope of a graph of value vs. clock time represents the rate at which value is changing with respect to clock time.......!!!!!!!!...................................
RESPONSE --> ok
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16:30:53
`questionNumber 20000 `q003. To what extent do you think your graph, consisting of 3 points with straight line segments between them, accurately depicts the detailed behavior of the stocks over the 9-month period?......!!!!!!!!...................................
RESPONSE --> to the extent of a ave. 4.5 month scale- so it is not accurate to the month.
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16:31:12
`questionNumber 20000 Stocks can do just about anything from day to day-they can go up or down more in a single day than their net change in a month or even a year. So based on the values several months apart we can't say anything about what happens from day to day or even from month to month. We can only say that on the average, from one time to another, the stocks changed at a certain rate.......!!!!!!!!...................................
RESPONSE --> yup
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16:31:42
`questionNumber 20000 `q004. From the given information, do you think you can accurately infer the detailed behavior of the stock values over the nine-month period?......!!!!!!!!...................................
RESPONSE --> no
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16:32:01
`questionNumber 20000 Not on a day-to-day basis, and not even on a month-to-month basis. All we can see from the given information is what might be an average trend.......!!!!!!!!...................................
RESPONSE --> yup
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}aΫBͤm~F~ Student Name: assignment #003 003.
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16:43:13
`questionNumber 30000 Note that there are four questions in this assignment. `q001. Sketch a graph similar to that you constructed for the stock values, this time for the depth of the water vs. clock time (depths 80, 40, 20 at clock times 10, 40, 90). Your first point, for example, will be (10, 80). Connect these points with straight lines and determine the slopes of the lines.......!!!!!!!!...................................
RESPONSE --> uhhh...on paper right?? slopes r same as assy# 1 problem.
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16:46:38
`questionNumber 30000 The three points are (10, 80), (40, 40) and (90, 20). From the first point to the second the rise is from 80 to 40, or -40, and the run is from 10 to 40, or 30. So the slope is -40 / 30 = -1.33. From the second point to the third the rise is from 40 to 20, or -20, and the run is from 40 to 90, or 50, so the slope is -20 / 50 = -.4. Click on 'Next Picture' to see graph.......!!!!!!!!...................................
RESPONSE --> ok. i note the (-) signs on your slopes (were not on mine)
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17:04:39
`questionNumber 30000 `q002. Look at your results for the slopes, and look the results for the average rates of change. What do you notice? In what way then does the graph represent the average rate of change?......!!!!!!!!...................................
RESPONSE --> 1) that the triangles are flipped over from the last graph 2) the "rise" is actually dropping. so it shows it is at a neg. rate.
OK.... but i thought it would be in a diff quadrent in relation to the orgin because of the (-) ??????????.................................................
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17:04:52
`questionNumber 30000 The slopes and the rates of change are numerically equal. For example between the second and third points the rise of -20 represents the -20 cm change in depth and the run of 50 represents the 50 seconds required to make this change, so the slope represents the -20 cm / (50 sec) average rate of change over the second time interval. We therefore see that slope represents average rate of change.......!!!!!!!!...................................
RESPONSE --> ok
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17:10:04
`questionNumber 30000 `q003. To what extent do you think your graph with three points and straight line segments between them accurately depicts the detailed behavior of the water over the 80-second period of observation? How do you think the actual behavior of the system differs from that of the graph? How do you think the graph of the actual behavior of the system would differ from that of the graph you made?......!!!!!!!!...................................
RESPONSE --> 1) to the extent of every 20 sec.
2) i think it wound have a continously changing slope. 3) i think it would form a continous curve..................................................
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17:10:17
`questionNumber 30000 The straight line segments would indicate a constant rate of change of depth. It is fairly clear that as depth decreases, the rate of change of depth will decrease, so that the rate of change of depth will not be constant. The graph will therefore never be straight, but will be a curve which is decreasing at a decreasing rate.......!!!!!!!!...................................
RESPONSE --> ok
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17:14:04
`questionNumber 30000 `q004. From the given information, do you think you can accurately infer the detailed behavior of the water depth over the 80-second period? Do you think you can infer the detailed behavior better than you could the values of the stocks? Why or why not?......!!!!!!!!...................................
RESPONSE --> 1) no
2) no 3) because again, there are not enough points from which to infer any real curve..................................................
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17:24:28
`questionNumber 30000 It will turn out that three data points will be sufficient to infer the detailed behavior, provided the data are accurate. However you might or might not be aware of that at this point, so you could draw either conclusion. However it should be clear that the behavior of the water depth is much more predictable than the behavior of the stock market. We don't know on a given day whether the market will go up or down, but we do know that if we shoot a hole in the bottom of a full bucket the water level will decrease, and we expect that identical holes in identical buckets should result in the same depth vs. clock time behavior.......!!!!!!!!...................................
RESPONSE --> yes, but with less water, there will be less press. forcing the water out of the hole, so an exact flow rate lessens on a continiously changing scale.
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ޢ鏺פy˹M Student Name: assignment #004 004. 㪙ЬNϦ͟溱l߀ Student Name: assignment #005 005.
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17:19:58
`questionNumber 50000 Note that there are 9 questions in this assignment. `q001. We see that the water depth vs. clock time system likely behaves in a much more predictable detailed manner than the stock market. So we will focus for a while on this system. An accurate graph of the water depth vs. clock time will be a smooth curve. Does this curve suggest a constantly changing rate of depth change or a constant rate of depth change? What is in about the curve at a point that tells you the rate of depth change?......!!!!!!!!...................................
RESPONSE --> 1) a changing rate of depth change
2) uuhh...sorry not sure what u r asking 4-?.................................................
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17:20:54
`questionNumber 50000 The steepness of the curve is continually changing. Since it is the slope of the curve then indicates the rate of depth change, the depth vs. clock time curve represents a constantly changing rate of depth change.......!!!!!!!!...................................
RESPONSE --> oooooooooohhh. ok
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17:45:05
`questionNumber 50000 `q002. As you will see, or perhaps have already seen, it is possible to represent the behavior of the system by a quadratic function of the form y = a t^2 + b t + c, where y is used to represent depth and t represents clock time. If we know the precise depths at three different clock times there is a unique quadratic function that fits those three points, in the sense that the graph of this function passes precisely through the three points. Furthermore if the cylinder and the hole in the bottom are both uniform the quadratic model will predict the depth at all in-between clock times with great accuracy. Suppose that another system of the same type has quadratic model y = y(t) = .01 t^2 - 2 t + 90, where y is the depth in cm and t the clock time in seconds. What are the depths for this system at t = 10, t = 40 and t = 90?......!!!!!!!!...................................
RESPONSE --> y(10)=.01(10)^2-2(10)+90 y=[.01x100-20+90]/10 y=[1+70]/10 y=71/10 y=7.1 cm y(40)=.01(40)^2-2(40)+90 y=[.01x1600-80+90]/10 y=[16+10]/10 y=26/10 y=2.6 cm y(90)=.01(90)^2-2(90)+90 y=[.01x8100-180+90]/10 y=[81-90]/10 y= -9/10 y=-.9 cm
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17:48:40
`questionNumber 50000 At t=10 the depth is y(10) = .01(10^2) + 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=20 the depth is y(20) = .01(20^2) - 2(20) + 90 = 4 - 40 + 90 = 54, representing a depth of 54 cm. At t=90 the depth is y(90) = .01(90^2) - 2(90) + 90 = 81 - 180 + 90 = -9, representing a depth of -9 cm.......!!!!!!!!...................................
RESPONSE --> damn. my mistake so y(t) is y of t like f(x) ok
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18:12:32
`questionNumber 50000 `q003. For the preceding situation, what are the average rates which the depth changes over each of the two-time intervals?......!!!!!!!!...................................
RESPONSE --> oh yeah also i tries copy&paste on that last one and guess i forgot something.
anyway... 17 cm diff /10 sec diff = 1.7 cm/sec change 63 cm diff /70 sec diff = .9 cm/sec change.................................................
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18:17:31
`questionNumber 50000 From 71 cm to 54 cm is a change of 54 cm - 71 cm = -17 cm; this change takes place between t = 10 sec and t = 20 sec, so the change in clock time is 20 sec - 10 sec = 10 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -17 cm / 10 sec = -1.7 cm/s. From 54 cm to -9 cm is a change of -9 cm - 54 cm = -63 cm; this change takes place between t = 40 sec and t = 90 sec, so the change in clock time is a9 0 sec - 40 sec = 50 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -63 cm / 50 sec = -1.26 cm/s.......!!!!!!!!...................................
RESPONSE --> dammit. yeah i forgot - sign again it is a drop (duh) also, but it said 20 sec, not 40-?
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߸S輪JzڜhȘW Student Name: assignment #005 005.
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16:20:03
`questionNumber 50000 Note that there are 9 questions in this assignment. `q001. We see that the water depth vs. clock time system likely behaves in a much more predictable detailed manner than the stock market. So we will focus for a while on this system. An accurate graph of the water depth vs. clock time will be a smooth curve. Does this curve suggest a constantly changing rate of depth change or a constant rate of depth change? What is in about the curve at a point that tells you the rate of depth change?......!!!!!!!!...................................
RESPONSE --> sss
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16:20:09
`questionNumber 50000 The steepness of the curve is continually changing. Since it is the slope of the curve then indicates the rate of depth change, the depth vs. clock time curve represents a constantly changing rate of depth change.......!!!!!!!!...................................
RESPONSE --> sss
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16:20:16
`questionNumber 50000 `q002. As you will see, or perhaps have already seen, it is possible to represent the behavior of the system by a quadratic function of the form y = a t^2 + b t + c, where y is used to represent depth and t represents clock time. If we know the precise depths at three different clock times there is a unique quadratic function that fits those three points, in the sense that the graph of this function passes precisely through the three points. Furthermore if the cylinder and the hole in the bottom are both uniform the quadratic model will predict the depth at all in-between clock times with great accuracy. Suppose that another system of the same type has quadratic model y = y(t) = .01 t^2 - 2 t + 90, where y is the depth in cm and t the clock time in seconds. What are the depths for this system at t = 10, t = 40 and t = 90?......!!!!!!!!...................................
RESPONSE --> sss
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16:20:23
`questionNumber 50000 At t=10 the depth is y(10) = .01(10^2) + 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=20 the depth is y(20) = .01(20^2) - 2(20) + 90 = 4 - 40 + 90 = 54, representing a depth of 54 cm. At t=90 the depth is y(90) = .01(90^2) - 2(90) + 90 = 81 - 180 + 90 = -9, representing a depth of -9 cm.......!!!!!!!!...................................
RESPONSE --> sss
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16:20:28
`questionNumber 50000 `q003. For the preceding situation, what are the average rates which the depth changes over each of the two-time intervals?......!!!!!!!!...................................
RESPONSE --> sss
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16:20:52
`questionNumber 50000 From 71 cm to 54 cm is a change of 54 cm - 71 cm = -17 cm; this change takes place between t = 10 sec and t = 20 sec, so the change in clock time is 20 sec - 10 sec = 10 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -17 cm / 10 sec = -1.7 cm/s. From 54 cm to -9 cm is a change of -9 cm - 54 cm = -63 cm; this change takes place between t = 40 sec and t = 90 sec, so the change in clock time is a9 0 sec - 40 sec = 50 sec. The average rate of change between these to clock times is therefore ave rate = change in depth / change in clock time = -63 cm / 50 sec = -1.26 cm/s.......!!!!!!!!...................................
RESPONSE --> sss
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17:17:24
`questionNumber 50000 `q004. What is the average rate at which the depth changes between t = 10 and t = 11, and what is the average rate at which the depth changes between t = 10 and t = 10.1?......!!!!!!!!...................................
RESPONSE --> y = y(t) = .01 t^2 - 2 t + 90
y = y(t) = .01 (10)^2 - 2 (10) + 90 y=1-20+90 y=71cm/10s y = y(t) = .01 (10.1)^2 - 2 (10.1) + 90 y=1.02-20.2+90 y=70.82cm/10.1s y = y(t) = .01 (11)^2 - 2 (11) + 90 y=1.21-22+90 y=69.21cm/11s rate from 10 to 11: 11-10=1s 69.21-71=-1.79cm or -1.79cm/sec rate from 10 to 10.1: 10.1-10=.1s 70.82-71=-.18cm .1x10=1s so -.18x10=-1.8cm or -1.8cm/sec.................................................
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17:19:37
`questionNumber 50000 At t=10 the depth is y(10) = .01(10^2) - 2(10) + 90 = 1 - 20 + 90 = 71, representing a depth of 71 cm. At t=11 the depth is y(11) = .01(11^2) - 2(11) + 90 = 1.21 - 22 + 90 = 69.21, representing a depth of 69.21 cm. The average rate of depth change between t=10 and t = 11 is therefore change in depth / change in clock time = ( 69.21 - 71) cm / [ (11 - 10) sec ] = -1.79 cm/s. At t=10.1 the depth is y(10.1) = .01(10.1^2) - 2(10.1) + 90 = 1.0201 - 20.2 + 90 = 70.8201, representing a depth of 70.8201 cm. The average rate of depth change between t=10 and t = 10.1 is therefore change in depth / change in clock time = ( 70.8201 - 71) cm / [ (10.1 - 10) sec ] = -1.799 cm/s. We see that for the interval from t = 10 sec to t = 20 sec, then t = 10 s to t = 11 s, then from t = 10 s to t = 10.1 s the progression of average rates is -1.7 cm/s, -1.79 cm/s, -1.799 cm/s. It is important to note that rounding off could have hidden this progression. For example if the 70.8201 cm had been rounded off to 70.82 cm, the last result would have been -1.8 cm and the interpretation of the progression would change. When dealing with small differences it is important not around off too soon.......!!!!!!!!...................................
RESPONSE --> ooohhh aaahhhh ok. will do./not do i mean
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17:25:25
`questionNumber 50000 `q005. What do you think is the precise rate at which depth is changing at the instant t = 10?......!!!!!!!!...................................
RESPONSE --> y = y(t) = .01 t^2 - 2 t + 90
y=.01(10)^2-2(10)+90 y=.01(100)-20+90 y=1+70 y=71cm @10s 71/10=7.1cm/sec?.................................................
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17:29:37
`questionNumber 50000 The progression -1.7 cm/s, -1.79 cm/s, -1.799 cm/s corresponds to time intervals of `dt = 10, 1, and .1 sec, with all intervals starting at the instant t = 10 sec. That is, we have shorter and shorter intervals starting at t = 10 sec. We therefore expect that the progression might well continue with -1.7999 cm/s, -1.79999 cm/s, etc.. We see that these numbers approach more and more closely to -1.8, and that there is no limit to how closely they approach. It therefore makes sense that at the instant t = 10, the rate is exactly -1.8.......!!!!!!!!...................................
RESPONSE --> oooooooohhhh thats what u meant.
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17:40:11
`questionNumber 50000 `q006. In symbols, what are the depths at clock time t = t1 and at clock time t = t1 + `dt, where `dt is the time interval between the two clock times?......!!!!!!!!...................................
RESPONSE --> uuuhhh.......i guessing what u want
d=d1 and d=d1+dd-???.................................................
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17:45:54
`questionNumber 50000 At clock time t = t1 the depth is y(t1) = .01 t1^2 - 2 t1 + 90 and at clock time t = t1 + `dt the depth is y(t1 + `dt) = .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90.......!!!!!!!!...................................
RESPONSE --> aaaaahhhhhhh......ok. if u had put the formula on there with it i would have figured out what u wanted me to do.
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17:52:10
`questionNumber 50000 `q007. What is the change in depth between these clock times?......!!!!!!!!...................................
RESPONSE --> y(dt) = .01 (dt)^2 - 2 (dt) + 90.--???
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18:15:02
`questionNumber 50000 The change in depth is .01 (t1 + `dt)^2 - 2 (t1 + `dt) + 90 - (.01 t1^2 - 2 t1 + 90) = .01 (t1^2 + 2 t1 `dt + `dt^2) - 2 t1 - 2 `dt + 90 - (.01 t1^2 - 2 t1 + 90) = .01 t1^2 + .02 t1 `dt + .01`dt^2 - 2 t1 - 2 `dt + 90 - .01 t1^2 + 2 t1 - 90) = .02 t1 `dt + - 2 `dt + .01 `dt^2.......!!!!!!!!...................................
RESPONSE --> ok.im lost but trying to figure it out. this is hard on a screen.
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18:22:32
`questionNumber 50000 `q008. What is the average rate at which depth changes between these clock time?......!!!!!!!!...................................
RESPONSE --> .02 t1 dt + - 2 dt + .01 dt^2 /dt ?
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18:24:41
`questionNumber 50000 The average rate is ave rate = change in depth / change in clock time = ( .02 t1 `dt + - 2 `dt + .01 `dt^2 ) / `dt = .02 t1 - 2 + .01 `dt. Note that as `dt shrinks to 0 this expression approaches .02 t1 - 2.......!!!!!!!!...................................
RESPONSE --> ok at least i was close. i didn't reduce
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18:39:05
`questionNumber 50000 `q009. What is the value of .02 t1 - 2 at t1 = 10 and how is this consistent with preceding results?......!!!!!!!!...................................
RESPONSE --> .02(10)-2= -1.8 it is basically the same as 1.99?
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18:39:53
`questionNumber 50000 At t1 = 10 we get .02 * 10 - 2 = .2 - 2 = -1.8. This is the rate we conjectured for t = 10.......!!!!!!!!...................................
RESPONSE --> yeah.
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ޝΗߏR룦 Student Name: assignment #006 006. goin' the other way ovVUe~I Student Name: assignment #006 006. goin' the other way
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19:04:35
`questionNumber 60000 Note that there are 7 questions in this assignment. `q001. If the water and a certain cylinder is changing depth at a rate of -4 cm/sec at the t = 20 second instant, at which instant the depth is 80 cm, then approximately what do you expect the depth will be at the t = 21 second instant?......!!!!!!!!...................................
RESPONSE --> y=mx+b 20=-4(80)+? 20=-320+? ?=340
so y=-4x+340 21=-4x+340 -319=-4x 79.95=x 79.95 cm this has to be close, but i forgot the correct equation (shoulda written it down).................................................
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19:07:38
`questionNumber 60000 At a rate of -4 cm/s, for the 1-second interval between t = 20 s and t = 21 s the change in depth would be -4 cm/s * 1 sec = -4 cm. If the depth was 80 cm at t = 20 sec, the depth at t = 21 sec would be 80 cm - 4 cm/ = 76 cm.......!!!!!!!!...................................
RESPONSE --> damn. i feel stupid. overlooked that
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19:20:47
`questionNumber 60000 `q002. Using the same information, what the you expect the depth will be depth at the t = 30 sec instant? Do you think this estimate is more or less accurate than the estimate you made for the t = 21 second instant?......!!!!!!!!...................................
RESPONSE --> 20-30=-10 -10x(-4)=40 depth is approx. 40 cm
less accurate.because it is a constant rate calculation, and the further it goes, the curve progressively gets farther away from it..................................................
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19:20:57
`questionNumber 60000 At - 4 cm/s, during the 10-second interval between t = 20 sec and t = 30 sec we would expect a depth change of -4 cm/sec * 10 sec = -40 cm, which would result in a t = 30 sec depth of 80 cm - 40 cm = 40 cm.......!!!!!!!!...................................
RESPONSE --> yup.
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19:33:58
`questionNumber 60000 `q003. If you know that the depth in the preceding example is changing at the rate of -3 cm/s at the t = 30 sec instant, how will this change your estimate for the depth at t = 30 seconds--i.e., will your estimate be the same as before, will you estimate a greater change in depth or a lesser change in depth?......!!!!!!!!...................................
RESPONSE --> because -3 is the rate multiple for the depth, and its abs value is less than that of -4, the product will be less. therefore, the difference in water level will be less, indicating a slower drop in it. and this is of course would be closer to the true graph, the curve,at the lower level, but less accurate at a higher level.
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19:34:23
`questionNumber 60000 Since the rate of depth change has changed from -4 cm / s at t = 20 s to -3 cm / s at t = 30 s, we conclude that the depth probably wouldn't change as much has before.......!!!!!!!!...................................
RESPONSE --> ok
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19:44:45
`questionNumber 60000 `q004. What is your specific estimate of the depth at t = 30 seconds?......!!!!!!!!...................................
RESPONSE --> 20-30=-10 -10x(-3)= 30 so about 50cm
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19:45:54
`questionNumber 60000 Knowing that at t = 20 sec the rate is -4 cm/s, and at t = 30 sec the rate is -3 cm/s, we could reasonably conjecture that the approximate average rate of change between these to clock times must be about -3.5 cm/s. Over the 10-second interval between t = 20 s and t = 30 s, this would result in a depth change of -3.5 cm/s * 10 sec = -35 cm, and a t = 30 s depth of 80 cm - 35 cm = 45 cm.......!!!!!!!!...................................
RESPONSE --> oooohhhhh so i was supposed to ave. the 2 rates.ok.
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19:55:32
`questionNumber 60000 `q005. If we have a uniform cylinder with a uniformly sized hole from which water is leaking, so that the quadratic model is very nearly a precise model of what actually happens, then the prediction that the depth will change and average rate of -3.5 cm/sec is accurate. This is because the rate at which the water depth changes will in this case be a linear function of clock time, and the average value of a linear function between two clock times must be equal to the average of its values at those to clock times. If y is the function that tells us the depth of the water as a function of clock time, then we let y ' stand for the function that tells us the rate at which depth changes as a function of clock time. If the rate at which depth changes is accurately modeled by the linear function y ' = .1 t - 6, with t in sec and y in cm/s, verify that the rates at t = 20 sec and t = 30 sec are indeed -4 cm/s and -3 cm/s.......!!!!!!!!...................................
RESPONSE --> y ' = .1 t - 6
y ' = .1 (20) -6 y ' = 2 - 6 y ' = -4 verified. y ' = .1 (30) - 6 y ' = 3 -6 y ' = -3 verified..................................................
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19:55:40
`questionNumber 60000 At t = 20 sec, we evaluate y ' to obtain y ' = .1 ( 20 sec) - 6 = 2 - 6 = -4, representing -4 cm/s. At t = 30 sec, we evaluate y' to obtain y' = .1 ( 30 sec) - 6 = 3 - 6 = -3, representing -3 cm/s.......!!!!!!!!...................................
RESPONSE --> ok
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20:06:41
`questionNumber 60000 `q006. For the rate function y ' = .1 t - 6, at what clock time does the rate of depth change first equal zero?......!!!!!!!!...................................
RESPONSE --> y ' = .1 t - 6
0 ' = .1 t - 6 6 = .1 t 60=t 60 sec.................................................
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20:07:18
`questionNumber 60000 The rate of depth change at t = 20 sec is - 4 cm/s; at t = 60 sec the rate is 0 cm/s. The average rate at which depth changes during this 40-second interval is therefore the average of -4 cm/s and 0 cm/s, or -2 cm/s. At an average rate of -2 cm/s for 40 s, the depth change will be -80 cm. Starting at 80 cm when t = 20 sec, we see that the depth at t = 60 is therefore 80 cm - 80 cm = 0.......!!!!!!!!...................................
RESPONSE --> ok..
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Rɨ¾븴Μoܖ| Student Name: assignment #007 007. Depth functions and rate functions. bړõf\ Student Name: assignment #007 007. Depth functions and rate functions. LUCwމ Student Name: assignment #007 007. Depth functions and rate functions.
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13:53:24
`questionNumber 70000 Note that there are 9 questions in this assignment. `q001. The function y = .05 t^2 - 6 t + 100 is related to the rate function y ' = .1 t - 6 in that if y = .05 t^2 - 6 t + 100 represents the depth, then the depth change between any two clock times t is the same as that predicted by the rate function y ' = .1 t - 6. We saw before that for y ' = .1 t - 6, the depth change between t = 20 and t = 30 had to be 35 cm. Show that for the depth function y = .05 t^2 - 6t + 100, the change in depth between t = 20 and t = 30 is indeed 35 cm.......!!!!!!!!...................................
RESPONSE --> y = .05 t^2 - 6t + 100 y = .05 (20)^2 - 6(20) + 100 y = .05(400) - 120 + 100 y = 20 - 220 y=-200 cm
y = .05 t^2 - 6t + 100 y = .05 (30)^2 - 6(30) + 100 y = .05 (900) - 180 + 100 y = 45 - 280 y= -235 cm so -200 cm-( -235 cm)= 35 cm..................................................
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13:58:19
`questionNumber 70000 The depth at t = 20 will be .05(20^2) - 6(20) + 100 = 20 - 120 + 100 = 0. The depth at t = 30 will be .05(30^2) - 6(30) + 100 = 45 - 180 + 100 = -35. Thus the depth changes from 0 cm to -35 cm during the 10-second time interval between t = 20 s and t = 30 s. This gives us and average rate of ave rate = change in depth / change in clock time = -35 cm / (10 sec) = -3.5 cm/s.......!!!!!!!!...................................
RESPONSE --> damn. sorry, stupid mistake. tried to rush through that one.
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14:51:19
`questionNumber 70000 `q002. What depth change is predicted by the rate function y ' = .1 t - 6 between t = 30 and t = 40? What is the change in the depth function y = .05 t^2 - 6 t + 100 between t = 30 and t = 40? How does this confirm the relationship between the rate function y ' and the depth function y?......!!!!!!!!...................................
RESPONSE --> y ' = .1 (30) - 6 y ' = 3 - 6 y ' = -3
y ' = .1 (40) - 6 y ' = 4 - 6 y ' = -2 so -2-(-3)= 1 cm y = .05 (30)^2 - 6 (30) + 100 y = .05 (900) - 180 + 100 y = 45 - 80 y= -35 cm y = .05 (40)^2 - 6 (40) + 100 y = .05 (1600) -240 + 100 y = 80 - 140 y= -60 cm so -60-(-35)=-25 cm it proves that there is a diffrence in the rates of change in the 2..................................................
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14:54:20
`questionNumber 70000 At t = 30 and t = 40 we have y ' = .1 * 30 - 6 = -3 and y ' = .1 * 40 - 6 = -2. The average of the two corresponding rates is therefore -2.5 cm/s. During the 10-second interval between t = 30 and t = 40 we therefore predict the depth change of predicted depth change based on rate function = -2.5 cm/s * 10 s = -25 cm. At t = 30 the depth function was previously seen to have value -35, representing -35 cm. At t = 40 sec we evaluate the depth function and find that the depth is -60 cm. The change in depth is therefore depth change has predicted by depth function = -60 cm - (-35 cm) = -25 cm. The relationship between the rate function and the depth function is that both should predict a same change in depth between the same two clock times. This is the case in this example.......!!!!!!!!...................................
RESPONSE --> i was wondering why my 1st ans was a pos. #. so yes i see they are aver.
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15:19:57
`questionNumber 70000 `q003. Show that the change in the depth function y = .05 t^2 - 6 t + 30 between t = 20 and t = 30 is the same as that predicted by the rate function y ' = .1 t - 6. Show the same for the time interval between t = 30 and t = 40. Note that the predictions for the y ' function have already been made.......!!!!!!!!...................................
RESPONSE --> y ' = .1 (20) - 6 y ' = 2 - 6 y ' = -4
y ' = .1 (30) - 6 y ' = 3 - 6 y ' = -3 so ave of -3 and -4 = -3.5 cm/sec y = .05 (20)^2 - 6 (20) + 100 y = .05 (400) -120 + 100 y = 20 - 20 y= 0 cm y = .05 (30)^2 - 6 (30) + 100 y = .05 (900) - 180 + 100 y = 45 - 80 y= -35 cm so -35-0 = -35 cm or 3.5 cm/sec y ' = .1 (30) - 6 y ' = 3 - 6 y ' = -3 y ' = .1 (40) - 6 y ' = 4 - 6 y ' = -2 so ave. of -2 and -3 = -2.5 cm/sec y = .05 (30)^2 - 6 (30) + 100 y = .05 (900) - 180 + 100 y = 45 - 80 y= -35 cm y = .05 (40)^2 - 6 (40) + 100 y = .05 (1600) -240 + 100 y = 80 - 140 y= -60 cm so -60-(-35)=-25 cm or -2.5 cm/sec.................................................
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15:20:29
`questionNumber 70000 The prediction from the rate function is a depth change of -35 cm, and has already been made in a previous problem. Evaluating the new depth function at t = 20 we get y = .05(20^2) - 6(20) + 30 = -70, representing -70 cm. Evaluating the same function at t = 30 we get y = -105 cm. This implies the depth change of -105 cm - (-70 cm) = -35 cm. Evaluating the new depth function at t = 40 sec we get y = depth = -130 cm. Thus the change from t = 30 to t = 40 is -130 cm - (-105 cm) = -25 cm. This is identical to the change predicted in the preceding problem for the given depth function.......!!!!!!!!...................................
RESPONSE --> ok.
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15:27:34
`questionNumber 70000 `q004. Why is it that the depth functions y = .05 t^2 - 6 t + 30 and y = .05 t^2 - 6 t + 100 give the same change in depth between two given clock times?......!!!!!!!!...................................
RESPONSE --> because 155-140=15 and 85-70=15. so in other words they cancel out.
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15:27:46
`questionNumber 70000 The only difference between the two functions is a constant number at the end. One function and with +30 and the other with +100. The first depth function will therefore always be 70 units greater than the other. If one changes by a certain amount between two clock times, the other, always being exactly 70 units greater, must also change by the same amount between those two clock times.......!!!!!!!!...................................
RESPONSE --> yup
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15:51:00
`questionNumber 70000 `q005. We saw earlier that if y = a t^2 + b t + c, then the average rate of depth change between t = t1 and t = t1 + `dt is 2 a t1 + b + a `dt. If `dt is a very short time, then the rate becomes very clost to 2 a t1 + b. This can happen for any t1, so we might as well just say t instead of t1, so the rate at any instant is y ' = 2 a t + b. So the functions y = a t^2 + b t + c and y ' = 2 a t + b are related by the fact that if the function y represents the depth, then the function y ' represents the rate at which depth changes. If y = .05 t^2 - 6 t + 100, then what are the values of a, b and c in the form y = a t^2 + b t + c? What therefore is the function y ' = 2 a t + b?......!!!!!!!!...................................
RESPONSE --> a=.05 b=6 and c=100.
y ' = 2 a t + b is the solid line function?.................................................
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15:53:39
`questionNumber 70000 If y = .05 t^2 - 6 t + 100 is of form y = a t^2 + b t + c, then a = .05, b = -6 and c = 100. The function y ' is 2 a t + b. With the given values of a and b we see that y ' = 2 ( .05) t + (-6), which simplifies to y ' = .1 t - 6......!!!!!!!!...................................
RESPONSE --> so thas what u wanted.
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15:56:50
`questionNumber 70000 `q006. For the function y = .05 t^2 - 6 t + 30, what is the function y ' ?......!!!!!!!!...................................
RESPONSE --> what is ' supposed to mean??
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16:01:58
`questionNumber 70000 The values of a, b and c are respectively .05, -6 and 30. Thus y ' = 2 a t + b = 2(.05) t + (-6) = .1 t - 6. This is identical to the y ' function in the preceding example. The only difference between the present y function and the last is the constant term c at the end, 30 in this example and 100 in the preceding. This constant difference has no effect on the derivative, which is related to the fact that it has no effect on the slope of the graph at a point.......!!!!!!!!...................................
RESPONSE --> so does ' mean simplify??????
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16:21:44
`questionNumber 70000 `q007. For some functions y we can find the rate function y ' using rules which we will develop later in the course. We have already found the rule for a quadratic function of the form y = a t^2 + b t + c. The y ' function is called the derivative of the y function, and the y function is called an antiderivative of the y ' function. What is the derivative of the function y = .05 t^2 - 6 t + 130? Give at least two new antiderivative functions for the rate function y ' = .1 t - 6.......!!!!!!!!...................................
RESPONSE --> um, y' = .05 t^2 - 6 t + 99?
y = .1 t - 6? y = .1 t(-6) ?.................................................
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16:36:52
`questionNumber 70000 The derivative of y = .05 t^2 - 6 t + 130 is .1 t - 6; as in the preceding problem this function has a = .05 and b = -6, and differs from the preceding two y functions only by the value of c. Since c has no effect on the derivative, the derivative is the same as before. If y ' = .1 t - 6, then a = 1 / 2 ( .1) = .05 and b = -6 and we see that the function y is y = .05 t^2 - 6 t + c, where c can be any constant. We could choose any two different values of c and obtain a function which is an antiderivative of y ' = .1 t - 6. Let's use c = 17 and c = -54 to get the functions y = .05 t^2 - 6 t + 17 and y = .05 t^2 - 6 t - 54.......!!!!!!!!...................................
RESPONSE --> so i take that ur gonna show me?
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16:50:44
`questionNumber 70000 `q008. For a given function y, there is only one derivative function y '. For a given rate function y ', there is more than one antiderivative function. Explain how these statements are illustrated by the preceding example.......!!!!!!!!...................................
RESPONSE --> the derivative is the rate function??
the antider. is another function of y with the same slope, but different intercept???.................................................
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16:51:41
`questionNumber 70000 The derivative function give the rate at which the original function changes for every value of t, and there can be only one rate for a given t. Thus the values of the derivative function are completely determined by the original function. In the previous examples we saw several different functions with the same derivative function. This occurred when the derivative functions differed only by the constant number at the end. However, for a given derivative function, if we get one antiderivative, we can add any constant number to get another antiderivative. y = .05 t^2 - 6 t +17, y = .05 t^2 - 6 t + 30, and y = .05 t^2 - 6 t + 100, etc. are all antiderivatives of y ' = .1 t - 6.......!!!!!!!!...................................
RESPONSE --> ooooooohhhh
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16:54:40
`questionNumber 70000 `q009. What do all the antiderivative functions of the rate function y ' = .1 t - 6 have in common? How do they differ? How many antiderivative functions do you think there could be for the given rate function?......!!!!!!!!...................................
RESPONSE --> they all have the same everything except for the c-value. this is what differs.
an infinate #..................................................
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16:55:14
`questionNumber 70000 All antiderivatives must contain .05 t^2 - 6 t. They may also contain a nonzero constant term, such as -4, which would give us y = .05 t^2 - 6 t - 4. We could have used any number for this last constant.......!!!!!!!!...................................
RESPONSE --> yup.
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송ȲG兼WޟZ~ Student Name: assignment #008 008. Approximate depth graph from the rate function
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17:06:55
`questionNumber 80000 `q001. Note that there are 5 questions in thie assignment. Sketch a graph of the function y ' = .1 t - 6 for t = 0 to t = 100. Describe your graph.......!!!!!!!!...................................
RESPONSE --> the line follows a very slight slope up and to the right from 6 below the orgin (y-int.)
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17:07:27
`questionNumber 80000 The graph of this function has an intecept on the y' axis at (0,-6) and an intercept on the x axis at (60,0). The graph is a straight line with slope .1. At t = 100 the graph point is (100,4).......!!!!!!!!...................................
RESPONSE --> yup.
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17:24:05
`questionNumber 80000 `q002. Now sketch the graph of a function y which starts at the point (0, 100), and whose slope at every t is equal to the value of y ' at t. Thus, for example, at t = 0 we have y ' = .1 * 0 - 6 = -6, so our graph of y will start off a the t = 0 point (0,100) with a slope of -6, and the graph begins by decreasing rather rapidly. But the slope won't remain at -6. By the time we get to t = 10 the slope will be y ' = .1 * 10 - 6 = -5, and the graph is decreasing less rapidly than before. Then by the time we get to t = 20 the slope will be y ' = . 1 * 20 - 6 = -4, etc.. If you sketch a graph of y vs. t with these calculated slopes at the corresponding t values, what does the graph look like? Will it be increasing or decreasing, and will it be doing so at an increasing, decreasing or constant rate? Is the answer to this question different for different parts of the graph? If so over what intervals of the graph do the different answers apply?......!!!!!!!!...................................
RESPONSE --> im confused is this a step graph??
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17:27:49
`questionNumber 80000 The graph will have slopes -6 at t = 0, -5 at t = 10, -4 at t = 9, etc.. At least for awhile, the slope will remain negative so the graph will be decreasing. The negative slopes will however become less and less steep. So the graph be decreasing at a decreasing rate. It's obvious that the slopes will eventually reach 0, and since y' is the slope of the y vs. t graph it's clear that this will happen when y' = .1 t - 6 becomes 0. Setting .1 t - 6 = 0 we get t = 60. Note, and think about about the fact, that this coincides with the x-intercept of the graph of y' vs. t. At this point the slope will be 0 and the graph will have leveled off at least for an instant. Past t = 60 the values of y' will be positive, so the slope of the y vs. t graph will be positive and graph of y vs. t will therefore be increasing. The values of y' will also be increasing, so that the slopes of the y vs. t graph will be increasing, and we can say that the graph will be increasing at an increasing rate.......!!!!!!!!...................................
RESPONSE --> uuuhhh.....ok
<1picture (or graph) =1000 words>.................................................
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18:22:45
`questionNumber 80000 `q003. The graph of y vs. t corresponding to the given rate function y ' = .1 t - 6 has slope -6 at the point (0,100). This slope immediately begins changing, and becomes -5 by the time t = 10. However, let us assume that the slope doesn't change until we get to the t = 10 point. This assumption isn't completely accurate, but we're going to see how it works out. If the slope remained -6 for t = 0 to t = 10, then starting at (0, 100) what point would we reach when t = 10?......!!!!!!!!...................................
RESPONSE --> y=mx+b y=-6x+100 y=-6(10)+100 y=-60+100 y=40
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18:23:14
`questionNumber 80000 The slope of the graph is the ratio slope = rise / run. If the slope remains at -6 from t = 0 to t = 10, then the difference between 10 is the run. Thus the run is 10 and the slope is -6, so the rise is rise = slope * run = -6 * 10 = -60. The y coordinate of the graph therefore changes by -60, from y = 100 to y = 100 + (-60) = 40. The corresponding point is (10, 40).......!!!!!!!!...................................
RESPONSE --> yup.
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18:41:24
`questionNumber 80000 `q004. We see that we reach the point (10, 40) by assuming a slope of -6 from t = 0 to t = 10. We have seen that at t = 10 the slope will be y ' = .1 * 10 - 6 = -5. If we maintain this slope for the interval t = 10 to t = 20, what will be the coordinates of the t = 20 point?......!!!!!!!!...................................
RESPONSE -->
-5=rise/10 so 10 x -5= -50 (20,-10).................................................
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18:41:59
`questionNumber 80000 The run from t = 10 to t = 20 is 10. With a slope of -5 this implies a rise of rise = slope * run = -5 * 10 = -50. Starting from point (10,40), a rise of -50 and a run of 10 takes the graph to the point (20, -10).......!!!!!!!!...................................
RESPONSE --> ok.
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19:39:15
`questionNumber 80000 `q005. Continue this process up to the t = 70 point, using a 10-unit t interval for each approximation. Describe the graph that results.......!!!!!!!!...................................
RESPONSE --> (20,-10) start, slope -4 -4=rise/10 so 10 x -4= -40
(30,-50) start, slope -3 -3=rise/10 or -30 (40,-80) start, slope -2 -2=rise/10 or -20 (50,-100) start, slope -1 -1=rise/10 or -10 (60, -110)start, slope = 0 0=rise/10 or 0 (70, -110), slope = 0 the graphs slope flattens out at x= 70..................................................
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19:39:42
`questionNumber 80000 The slope at t = 20 is y ' = .1 * 20 - 6 = -4. From t = 20 to t = 30 the run is 10, so the rise is rise = slope * run = -4 * 10 = -40. Starting from (20,-10) a rise of -40 and a run of 10 takes us to (30, -50). The slope at t = 30 is y ' = .1 * 30 - 6 = -3. From t = 30 to t = 40 the run is 10, so the rise is rise = slope * run = -3 * 10 = -30. Starting from (30,-50) a rise of -30 and a run of 10 takes us to (40, -80). The slope at t = 40 is y ' = .1 * 40 - 6 = -2. From t = 40 to t = 50 the run is 10, so the rise is rise = slope * run = -2 * 10 = -20. Starting from (40,-80) a rise of -20 and a run of 10 takes us to (50, -100). The slope at t = 50 is y ' = .1 * 50 - 6 = -1. From t = 50 to t = 60 the run is 10, so the rise is rise = slope * run = -1 * 10 = -10. Starting from (50,-100) a rise of -10 and a run of 10 takes us to (60, -110). The slope at t = 60 is y ' = .1 * 70 - 6 = -0. From t = 60 to t = 70 the run is 10, so the rise is rise = slope * run = -0 * 10 = 0. Starting from (60,-110) a rise of and a run of 10 takes us to (70, -110).......!!!!!!!!...................................
RESPONSE --> yes.
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ІOߚ|rT Student Name: assignment #009 009. Finding the average value of the rate using a predicted point