QA 6

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

This is a copy of the note I just emailed you:

Some of my most frequent notes are automated, and the system sent you the wrong response. My fault all the way, since I'm solely responsible for designing and implementing the system.

You appear to be deleting or otherwise omitting the given solutions and subsequent self-critique (which it turns out you haven't really needed, but may at some point in the future). The system relies on the format of the document to show me all necessary information in quick succession, so I can quickly spot errors and trouble spots, which allows me to respond individually to a large number of students submitting a lot of work.

My note should have asked you to retain all of the information on the query and qa documents, inserting your responses as appropriate but not deleting anything. This also ensures that you don't skip over the self-critique process, in those cases where it's necessary.

This will ensure that I spot errors or incomplete understanding, in the event these things do occur in your work. In other words, when you need it, you'll get better feedback by submitting the document in the prescribed manner.

As it turns out you are clearly doing an excellent job. You made 98% on your test and had excellent mastery of all concepts and procedures.

However on future assignments, please don't delete anything from the original documents.

006. goin' the other way

09-26-2009

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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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&#Good work. See my notes and let me know if you have questions. &#