Assignment 1

course MTH 271

Ṭxj؆sassignment #001

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

001. `Query 1

Applied Calculus I

09-16-2007

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12:47:06

Section 0.1.26 solve x/2-x/3>5

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RESPONSE -->

x/2 - x/3 > 5

Multiply each fraction by the least common multiple

6 * (x/2 - x/3) >5 * 6

= 3x - 2x > 30

Eliminate the the smallest variable which is -2x

= 5x > 32

Simplify

x > 27

confidence assessment: 1

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12:48:47

09-16-2007 12:48:47

It's easiest to avoid denominators where possible. So the preferred first step is to multiply both sides of the original equation by the common denominator 6:

6(x/2) - 6(x/3) = 6 * 5, which gives you3x - 2x = 6 * 5 which gives you

x > 6 * 5 which simplifies to

x > 30.

The interval associated with this solution is 30 < x < infinity, or (30, infinity).

To graph you would make an arrow starting at x = 30 and pointing to the right, indicating by an open circle that x = 30 isn't included.**

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NOTES -------> I see where I messed up. Since 3 nd 2 are values of X I should have just subtracted the two of them.

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12:54:48

Section 0.1.28 solve 2x^2+1<9x-3

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RESPONSE -->

I couldn't figure this problem out. Here is what I did...

2x^2+1<9x-3

Remove the smallest value

= 2x^2 < 8x - 4

= 2x^2 < 12x

I can not work it any farther.

confidence assessment: 1

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12:57:10

09-16-2007 12:57:10

The given inequality rearranges to give the quadratic 2x^2 - 9 x + 4 < 0.

The left-hand side has zeros at x = .5 and x = 4, as we see by factoring [ we get (2x-1)(x-4) = 0 which is true if 2x-1 = 0 or x - 4 = 0, i.e., x = 1/2 or x = 4. ]

The left-hand side is a continuous function of x (in fact a quadratic function with a parabola for a graph), and can change sign only by passing thru 0. So on each interval x < 1/2, 1/2 < x < 4, 4 < x the function must have the same sign.

Testing an arbitrary point in each interval tells us that only on the middle interval is the function negative, so only on this interval is the inequality true.

Note that we can also reason this out from the fact that large negative or positive x the left-hand side is greater than the right because of the higher power. Both intervals contain large positive and large negative x, so the inequality isn't true on either of these intervals.

In any case the correct interval is 1/2 < x < 4.

ALTERNATE BUT EQUIVALENT EXPLANATION:

The way to solve this is to rearrange the equation to get

2 x^2 - 9 x + 4< 0.

The expression 2 x^2 - 9 x + 4 is equal to 0 when x = 1/2 or x = 4. These zeros can be found either by factoring the expression to get ( 2x - 1) ( x - 4), which is zero when x = 1/2 or 4, or by substituting into the quadratic formula. You should be able to factor basic quadratics or use the quadratic formula when factoring fails.

The function can only be zero at x = 1/2 or x = 4, so the function can only change from positive to negative or negative to positive at these x values. This fact partitions the x axis into the intervals (-infinity, 1/2), (1/2, 4) and (4, infinity). Over each of these intervals the quadratic expression can't change its sign.

If x = 0 the quadratic expression 2 x^2 - 9 x + 4 is equal to 4. Therefore the expression is positive on the interval (-infinity, 1/2).

The expression changes sign at x = 1/2 and is therefore negative on the interval (1/2, 4).

It changes sign again at 4 so is positive on the interval (4, infinity).

The solution to the equation is therefore the interval (1/2, 4), or in inequality form 1/2 < x < 4. **

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NOTES -------> I would have never come to that.

Quadratic inequalities are challenging.

Here's a response I sent to another student over the Summer. You'll get tired of the phrase 'Do you understand', and I'm not trying to beat you to death with the phrase, but this illustrates (in a whole lot of detail) the way a given solution might be deconstructed.

In analyzing the solution you should have asked yourself questions much like the following, and indicated for each one what you do and do not understand.

Of course few students would be able to construct this entire sequence of questions, so what follows is an idealized example. However, this example illustrates what you should strive for in the self-critique process:

The given inequality rearranges to give the quadratic 2x^2 - 9 x + 4 < 0.

Do you understand how this is equivalent to the given inequality?

The left-hand side has zeros at x = .5 and x = 4, as we see by factoring [ we get (2x-1)(x-4) = 0 which is true if 2x-1 = 0 or x - 4 = 0, i.e., x = 1/2 or x = 4. ]

Do you understand the factoring of the left-hand side 2x^2 - 9 x + 4?

Do you understand that in the factored form x = 1/2 or x = 4 makes the expression zero?

The left-hand side is a continuous function of x (in fact a quadratic function with a parabola for a graph), and can change sign only by passing thru 0. So on each interval x < 1/2, 1/2 < x < 4, 4 < x the function must have the same sign.

Do you understand that the left-hand side could be graphed to see where it is zero, where it is greater than zero and where it is less than zero?

Do you understand that the graph of a quadratic function is a parabola?

Do you understand that an expression changes sign when it goes from positive to negative or from negative to positive?

Do you understand that the only way an expression can change sign is therefore to pass through zero?

Do you understand that only points at which the left-hand side can be zero are x = 1/2 and x = 4?

Do you understand that these two points break the number line into three intervals?

Do you understand that x < 1/2 is an interval, as is 1/2 < x < 4 and as is 4 < x? Could you graph these intervals on the number line?

Do you understand that the function cannot therefore change sign except at these x values, so it must be either positive or negative on each of the intervals?

Testing an arbitrary point in each interval tells us that only on the middle interval is the function negative, so only on this interval is the inequality true.

What does it mean to 'test an arbitrary point' in an interval?

Select a point (a value of x) in each of the three intervals and see whether the expression 2x^2 - 9 x + 4 is less than or greater than zero.

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end of document

Your work has not been reviewed.

Please notify your instructor of the error, using the Submit Work form, and be sure to include the date 09-16-2007.

Good.

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Let me know if you have questions. &#

Assignment 1

course MTH 271

yVw_owr|assignment #001

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

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

09-09-2007

Ȃml

assignment #001

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

09-09-2007

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21:42:36

`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 -->

confidence assessment:

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21:42:42

The first period (4 months) was shorter than the second (5 months), and the value changed by more during the shorter first period (increase of $300) than during the longer second perios (increase of $200). A greater increase in a shorter period implies a greater rate of change. So the rate was greater during the first period.

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self critique assessment:

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21:42:46

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

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RESPONSE -->

confidence assessment:

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21:42:51

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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21:46:56

`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 -->

The water level is changing more quickly between clock times t=10 sec and 40 sec. There was a difference of 40 cm between the two times. The speed at which the water leaves the cylinder slows down between t=40 and t=90. There was only a 20 cm change during that time period.

confidence assessment: 3

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21:47:35

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 -->

This was visible to the naked eye without having to use mathematical equations.

self critique assessment: 2

That's so, but one purpose here is to get you used to doing those calculations, which are very important in a calclulus course (rate-of-change calculations are the basis for the entire course).

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21:47:50

`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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RESPONSE -->

confidence assessment: 0

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21:48:50

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 -->

I could not remember what the previous question was because I completed it yesterday. That is no excuse. I understand the functions of finding the rates.

self critique assessment: 2

It's a good idea to maintain running notes; nothing too extensive but there are lots of problem sequences where results are carried from one question to another. This is done to break the process into smaller and more manageable steps the first time you see it.

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You've got the idea. See my notes and let me know if you have questions.