assignment 0

course MTH 272

JΡKƐ϶Ҽassignment #001

001. typewriter notation

qa initial problems

09-07-2008

دÖaZ

assignment #001

001. Areas

qa areas volumes misc

09-07-2008

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15:57:06

`q001. There are 11 questions and 7 summary questions in this assignment.

What is the area of a rectangle whose dimensions are 4 m by 3 meters.

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RESPONSE -->

12

confidence assessment: 3

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15:58:10

`q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?

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Not sure if I needed so much detail for such a simple problem. There wasnt any real instructions I saw for how to complete so I just answered...

confidence assessment: 2

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纐fewDҔ

assignment #001

001. Depth vs. Clock Time and Rate of Depth Change

09-07-2008

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21:11:55

`qNote 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?

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RESPONSE -->

We can clearly see that over the March-July period that the money was growing faster at about a rate of 75 dollars a month as opposed to 40 dollars a month.

confidence assessment: 3

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21:13:23

`q002. What were the precise average rates of change during these two periods?

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RESPONSE -->

Using division (300/4) and (200/5) we can see that the July period grew at 75 dollars a month vs. 40 dollars a month through December

confidence assessment: 3

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21:13:29

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.

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RESPONSE -->

self critique assessment:

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21:13:50

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.

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RESPONSE -->

No critique needed

self critique assessment: 3

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21:15:49

`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?

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RESPONSE -->

Less quickly.As time progresses the water comes out slower. 40cm/30sec to 20cm/50sec

confidence assessment: 3

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21:17:02

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.

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RESPONSE -->

classic mistake.did not write the problem down and didnt take into account the negative sign

self critique assessment: 2

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21:20:53

`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?

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They are both about average rate of change. Which is necessary to be able to solve derivatives, a huge, fundamental part of calculus. They both use simple math.

confidence assessment: 3

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21:21:14

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.

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RESPONSE -->

self critique assessment:3 not quite as detailed,but i have the idea just didnt right it out.right on with the derivatives though

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&#Your work looks good. Let me know if you have any questions. &#