query_ac1_070113exe

course Mth 271

Some of the problems you asked me to submit from section 0.5 were not the problems in my book.

Dyك㺻assignment #005

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005. `query 5

Applied Calculus I

02-03-2007

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21:24:54

explain why the slope of a depth vs. time trapezoid represents the average rate of change of the depth with respect to the time during the time interval represented

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RESPONSE -->

The slope is rise/run which calculates the slope of the trapezoid and gives you the average rate of change because you are dividing the two sides of the trapezoid.

confidence assessment: 2

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21:25:21

the specific idea is that ave rate of depth change is [change in depth / change in time] ; rise represents change in depth and run represents change in time so slope = rise/run represents ave rate of depth change. **

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RESPONSE -->

Yes I agree.

self critique assessment: 2

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21:26:35

explain why the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval

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RESPONSE -->

It is calculated based on the specific time therefore it represents the change for that time interval based on the calculated rate.

confidence assessment: 2

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21:27:07

The average altitude represents the avg. velocity. The area of a trapezoid involves the altitude, which represents the avg. velocity, and the width, which represents the change in clock time.

When you multiply ave altitude by width you are representing ave vel * change in clock time, which gives change in position.

This reasoning isn't confined to velocities. For any rate vs. clock time graph, average altitude represents approximate average rate, which multiplied by the change in time (not by the time itself) gives you the change in quantity **

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RESPONSE -->

Yes I agree.

self critique assessment: 2

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21:35:47

text problem 0.5 #10 add x/(2-x) + 2/(x-2)

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RESPONSE -->

This is not the text problem that I had in my book, however the answer is:

x^2-4x+4/2x+4x-2x

confidence assessment: 1

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21:36:55

common denominator could be [ (2-x)(x-2) ]. In this case we have

x / (2-x) + 2 / (x-2)

= [ (x-2) / (x-2) ] * [ x / (2-x) ] + [ (2-x) / (2-x) ] * [ 2 / (x-2) ]

= x(x-2) / [ (2-x)(x-2) ] + 2 (2-x) / [ (2-x)(x-2) ]

= [x(x-2) + 2(2-x) ] / [ (2-x)(x-2) ]

= [ x^2 - 2x + 4 - 2x ] / [ (2-x)(x-2) ]

= (x^2-4x+4) / [ -x^2+4x-4 ]

= (x-2)^2 / [-(x-2)^2]

= -1.

NOTE however that there is a SIMPLER SOLUTION:

We can note that x-2 = -(2-x) so that the original problem is -x/(x-2) + 2 /(x-2) = (-x + 2) / (x-2) = -(x-2)/(x-2) = -1. **

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RESPONSE -->

I have copied this solution and will review how the calculation was done.

self critique assessment: 2

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21:39:58

text problem 0.5 #50 cost = 6 x + 900,000 / x, write as single fraction and determine cost to store 240 units

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RESPONSE -->

This was not a problem that was in my book, however I get the answer to be:

$3,756

confidence assessment: 1

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21:41:13

express with common denominator x:

[x / x] * 6x + 900,000 / x

= 6x^2 / x + 900,000 / x

= (6x^2 + 900,000) / x so

cost = (6x^2+900,000)/x

Evaluating at x = 240 we get cost = (6 * 240^2 + 900000) / 240 = 5190. **

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RESPONSE -->

I was off on my common denominator, however I have made a note of this and will make a correction in my notes.

self critique assessment: 2

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"

I'll check on the problem correspondences with the text. Looks like you're doing fine here, despite the confusion.