course MTH 173 ??Z?????L????assignment #002002.
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11:46:20 `qNote 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.
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RESPONSE --> slope: m=(y2-y1)/(x2-x1) between point one and point two: m = (5300-5000)/(7-3)= 300/4 = 75 between point two and point three: m= (5500-5300)/(12-7) = 200/5 = 40 confidence assessment: 3
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11:49:49 `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?
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RESPONSE --> I wasn't able to see the graph when I clicked on next picture. The slopes and the average rate of change is the same because the slope measures from the first point to the second and the average rate of change measures that same interval for the points. The graph represents the average rate of change as increasing at a decreasing rate. confidence assessment: 2
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11:50:42 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.
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RESPONSE --> I completely understand this concept. self critique assessment: 3
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11:52:01 `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?
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RESPONSE --> We are only given three points when we should be given 9 points. So our graph is only about a third as accurate as it could be if given all of the points on the graph. confidence assessment: 3
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11:53:28 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.
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RESPONSE --> I understand what you mean. We can't control our intake to be the exact same every month or day so with such variation our averages can only be said to be accurate for that particular interval. self critique assessment: 2
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11:54:24 `q004. From the given information, do you think you can accurately infer the detailed behavior of the stock values over the nine-month period?
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RESPONSE --> No, I wouldn't be able to give an accurate detailed behavior. I could give you an average of the behavior over the given interval with no guarantee. confidence assessment: 2
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11:54:47 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.
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RESPONSE --> I completely understand that it is only an average trend. self critique assessment: 2
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??????k?Q???????assignment #003 003. 02-03-2007
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23:21:40 `qNote 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.
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RESPONSE --> between points one and two slope: m = (40-80)/(40-10) = -40/30 = -1.33 between points two and three slope: m = (20-40)/(90-40) = -20/50 = -.4 the graph is decreasing at a decreasing rate confidence assessment: 2
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23:22:21 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.
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RESPONSE --> I was unable to view the graph when clicking on next picture but I did construct it in my notes from these points. self critique assessment: 2
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23:23:25 `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?
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RESPONSE --> The slope from one point to another is equal to the rate of an interval and the interval is the change. The graph shows that the average rate of change is decreasing at a decreasing rate. confidence assessment: 2
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23:23:55 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.
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RESPONSE --> I completely understand. self critique assessment: 3
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23:27:02 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.
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RESPONSE --> The graph with three points can only provide so much for us. It gives us the average rate of change for these three points only and can only be seen as a trend not a fully accurate depiction. My graph has straight lines drawn from point to point. An actual graph of this model with more points would be more curved. The graph of the actual behavior would be curved more and have more information and provide us with more accurate behavior of the water. self critique assessment: 2
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23:28:58 `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?
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RESPONSE --> There is an equal amount of information provided for each graph. In each instance we were given three points. In both of these it is said that we can only provide a trend to be observed. Our graph depicts our observation not how the model will behave every single time with the exact same details as this. confidence assessment: 2
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23:30:09 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.
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RESPONSE --> I understand what you mean when you say the bucket time would be more accurate than the stock market because it is more predictable. self critique assessment: 2
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??~??????? assignment #002 002. `query 2 Calculus I 02-02-2007
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12:03:45 Were you able to complete the DERIVE exercise?
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RESPONSE --> No, I couldn't find the free trial on the internet and haven't been able to go to the learning lab.
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??????????}? assignment #003 ??~??????? Calculus I 02-03-2007
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23:59:52 Query class notes #04 explain how we can prove that the rate-of-depth-change function for depth function y = a t^2 + b t + c is y' = 2 a t + b
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RESPONSE --> Given this quadratic function, we evaluate for y and t. depth y is a function of clock time t, meaning y(t). between the clock times t and t + 'dt the change in depth is equal to 'dy = y(t + 'dt) -y(t) and change in clock time is 'dt therefore, our average rate of change is 'dy/'dt = [ y(t + `dt) - y(t)] / `dt and when we use the quadratic function in this process we get: `dy / `dt = 2 a t + b + a `dt As time approaches 0, we get dy/dt = 2 at + b
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00:00:22 explain how we know that the depth function for rate-of-depth-change function y' = m t + b must be y = 1/2 m t^2 + b t + c, for some constant c, and explain the significance of the constant c.
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RESPONSE --> I understand and have copied all of this into my notes.
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00:01:39 ** Student Solution: If y = a t^2 + b t + c we have y ' (t) = 2 a t + b, which is equivalent to the given function y ' (t)=mt+b . Since 2at+b=mt+b for all possible values of t the parameter b is the same in both equations, which means that the coefficients 2a and m must be equal also. So if 2a=m then a=m/2. The depth function must therefore be y(t)=(1/2)mt^2+bt+c. c is not specified by this analysis, so at this point c is regarded as an arbitrary constant. c depends only on when we start our clock and the position from which the depth is being measured. **
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RESPONSE --> All of this has also been documented into my notes.
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00:03:16 Explain why, given only the rate-of-depth-change function y' and a time interval, we can determine only the change in depth and not the actual depth at any time, whereas if we know the depth function y we can determine the rate-of-depth-change function y' for any time.
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RESPONSE --> If we are only give the change in time we can only calculate what the change was not the actual numbers that brought us to this change. If we are given the actual depths we will be able to calculate the 'dy along with actual depths.
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00:04:07 ** Given the rate function y' we can find an approximate average rate over a given time interval by averaging initial and final rates. Unless the rate function is linear this estimate will not be equal to the average rate (there are rare exceptions for some functions over specific intervals, but even for these functions the statement holds over almost all intervals). Multiplying the average rate by the time interval we obtain the change in depth, but unless we know the original depth we have nothing to which to add the change in depth. So if all we know is the rate function, have no way to find the actual depth at any clock time. ANOTHER EXPLANATION: The average rate of change over a time interval is rAve = `dy / `dt. If we know rAve and `dt, then, we can easily find `dy, which is the change in depth. None of this tells us anything about the actual depth, only about the change in depth. If we don't know rAve but know the function r(t) we can't use the process above to get the exact change in depth over a given interval, though we can often make a pretty good guess at what the average rate is (for a quadratic depth function, as the quiz showed, you can actually be exact the average rate is just the rate at the midpoint of the interval; it's also the average of the initial and final rates; and all this is because for a quadratic the rate function is linear--if you think about those statements you see that they characterize a linear function, whose average on an interval occurs at a midpoint etc.). For anything but a linear rate function we can't so easily tell what the average is. However we do know that the rate function is the derivative of the depth function. So if we can find an antiderivative of the rate function, all we have to do to find the change in depth is find the difference in its values from the beginning to the end of the interval. This difference will be the same whichever antiderivative we find, because the only difference that can exist between two antiderivatives of a given rate function is a constant (whose derivative is zero). We have to develop some machinery to prove this rigorously but this is the essence of the Fundamental Theorem of Calculus. You might not understand it completely at this point, but keep coming back to this explanation every week or so and you will soon enough.**
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RESPONSE --> I understand this and all has also been copied to notes.
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00:07:29 In terms of the depth model explain the processes of differentiation and integration.
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RESPONSE --> When we obtain a rate function from a quantity function we have performed differentiation. When we obtain the change of quantity function we have performed integration. With integration we end up with a function whose derivative is the original rate function.
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00:07:48 ** Rate of depth change can be found from depth data. This is equivalent to differentiation. Given rate-of-change information it is possible to find depth changes but not actual depth. This is equivalent to integration. To find actual depths from rate of depth change would require knowledge of at least one actual depth at a known clock time. **
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RESPONSE --> understood
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