course Mth 271 Sorry these assignments are late, I unfortunately fell behind due to illness. I am planning to take the first Major Quiz Wednesday Evening 02/18/09. ҙg֮˔֠悟assignment #006
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22:53:49 `qNote 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?
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RESPONSE --> I expect the depth to be slightly less than -4 cm/s confidence assessment: 1
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22:55:24 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.
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RESPONSE --> I didn't know how to approach this problem, but I can clearly see now that because there is only a one second interval, we multiply the -4 cm by the 1 second to arrive at -4 cm/s. self critique assessment: 2
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22:57:06 `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?
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RESPONSE --> Because this is a 10 second interval, we multiply the -4 cm/s by the 10 sec interval to arrive at -40 cm. This estimate is less accurate than the first because there is a much larger time interval that we must piece together. confidence assessment: 2
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22:58:09 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.
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RESPONSE --> I forgot to subtract the -40 cm we got from 80 to arrive at positive 40 cm. self critique assessment: 2
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22:59:40 `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?
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RESPONSE --> A lesser change in depth. before, we were driving off of a -4 cm rate, but now that there is a -3 second rate, depth will be changing less rapidly. confidence assessment: 1
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22:59:52 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.
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RESPONSE --> self critique assessment: 3
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23:01:15 `q004. What is your specific estimate of the depth at t = 30 seconds?
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RESPONSE --> Because of the 10 second interval, we multiply 10 by the rate of -3 and arrive at -30, then subtract that from 80, arriving at 50 cm. confidence assessment: 2
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23:02:40 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.
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RESPONSE --> I see now that I should have averaged the two rates to arrive at 3.5, instead of just 3. Other than that I understand the calculations that must be made. self critique assessment: 2
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23:05:39 `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.
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RESPONSE --> t = 20 y = .1(20) - 6 y = -4 t = 30 y = .1(30) - 6 y = -3 confidence assessment: 3
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23:06:04 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.
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RESPONSE --> self critique assessment: 3
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23:07:24 `q006. For the rate function y ' = .1 t - 6, at what clock time does the rate of depth change first equal zero?
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RESPONSE --> y = .1t - 6 0 = .1t - 6 .1t = 6 t = 60 confidence assessment: 2
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23:07:46 The rate of depth change is first equal to zero when y ' = .1 t - 6 = 0. This equation is easily solved to see that t = 60 sec.
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RESPONSE --> self critique assessment: 3
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23:08:14 `q007. How much depth change is there between t = 20 sec and the time at which depth stops changing?
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RESPONSE --> 64 cm. The distance between -4 and 60. confidence assessment: 1
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23:10:20 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.
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RESPONSE --> I can see the right way to solve this problem is to average the rate of change between t = 20 and t = 60. self critique assessment: 2
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