course mth271 9/29,1400*I have been getting repeated feedback that I need to include the questions with my work. Yet, on all the work I have submitted, the questions are there, at least when I see the work. ???What am I doing wrong? And, what can I do to correct this??? assignment #006
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11:44:41 `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 --> (t=21seconds) - (t=20 seconds) = 1 second (-4cm/second) * 1 second =-4 cm 80 cm + -4cm= 76cm confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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11:47:06 `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 --> (t=30 sec) - (t=20 sec) = 10 sec (-4cm/sec) * 10 sec = -40 cm 80 cm + -40cm= 40cm The estimate for t=21 seconds is more accurate because there is likely to be less fluctuation of the rate during that 1 second interval than in the 10 second interval between t=20 and t=30. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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11:49:47 `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 --> The estimate for the depth at t=30 will be greater at a rate of -3cm/sec than at a rate of -4cm/sec. This is because there is less depth change occuring per second. If less depth is lost, the value for the depth will be greater. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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11:52:40 `q004. What is your specific estimate of the depth at t = 30 seconds?
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RESPONSE --> Average of rates between t=20 and t=30: (-4cm/sec + -3cm/sec)/2 = -3.5cm/sec (t=30seconds) - (t=20seconds) = 10 seconds -3.5 cm/sec * 10 sec = -35 cm 80 cm + -35cm = 45cm confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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11:55:27 `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 6= .1t 60=t confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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11:59:40 `q007. How much depth change is there between t = 20 sec and the time at which depth stops changing?
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RESPONSE --> y'=.1t - 6 y'=.1(20) - 6 y'=-4 cm/s Depth stops changing at y'=0, solved in the prior problem to be at t=60. (t=60 sec) - (t=20 sec) =40 sec Take an average of the rates between t=20 sec and t=60 sec: (-4cm/sec + 0cm/sec)/2 = -2cm/sec (-2cm/sec) * 40 sec = -80 cm confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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