Jeremiah

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course Mth 271

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Question: `q001. Explain the difference between x - 2 / x + 4 and (x - 2) / (x + 4). Then evaluate each expression for x = 2.

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Your solution:

If you do the order of operations 2-2/2+4 the answer comes out to be 5.

But with the parentheses the order changes (2-2)=0 and (2+4)=6 resulting in 0/6 which equals 0.

confidence rating #$&*: 3

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Given Solution:

`aThe order of operations dictates that grouped expressions must be evaluated first, that exponentiation must be done before multiplication or division, which must be done before addition or subtraction.

It makes a big difference whether you subtract the 2 from the x or divide the -2 by 4 first. If there are no parentheses you have to divide before you subtract. Substituting 2 for x we get

2 - 2 / 2 + 4

= 2 - 1 + 4 (do multiplications and divisions before additions and subtractions)

= 5 (add and subtract in indicated order)

If there are parentheses you evaluate the grouped expressions first:

(x - 2) / (x + 4) = (2 - 2) / ( 2 + 4 ) = 0 / 6 = 0.

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Self-critique (if necessary):

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Self-critique Rating:ok

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Question: `q002. Explain the difference between 2 ^ x + 4 and 2 ^ (x + 4). Then evaluate each expression for x = 2.

Note that a ^ b means to raise a to the b power. This process is called exponentiation, and the ^ symbol is used on most calculators, and in most computer algebra systems, to represent exponentiation.

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Your solution:

2^2+4 is 2 to the 2nd power plus 4 which gives us 8.

2^(2+4) is 2 to the 6th power which gives us 64.

confidence rating #$&*: 3

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Given Solution:

`a2 ^ x + 4 indicates that you are to raise 2 to the x power before adding the 4.

2 ^ (x + 4) indicates that you are to first evaluate x + 4, then raise 2 to this power.

If x = 2, then

2 ^ x + 4 = 2 ^ 2 + 4 = 2 * 2 + 4 = 4 + 4 = 8.

and

2 ^ (x + 4) = 2 ^ (2 + 4) = 2 ^ 6 = 2*2*2*2*2*2 = 64.

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Self-critique (if necessary):

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Question: `q003. What is the numerator of the fraction in the expression x - 3 / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x? What is the denominator? What do you get when you evaluate the expression for x = 2?

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Your solution:

The numerator is the top and the denominator is the bottom. And after evaluating the expression for 2 the numerator is (2)-(-3)=-1 and the denominator [(2(2)-5)^2*3(2)+1]-2+7(2) gives us 19.

confidence rating #$&*:

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Given Solution:

`aThe numerator is 3. x isn't part of the fraction. / indicates division, which must always precede subtraction. Only the 3 is divided by [ (2x-5)^2 * 3x + 1 ] and only [ (2x-5)^2 * 3x + 1 ] divides 3.

If we mean (x - 3) / [ (2x-5)^2 * 3x + 1 ] - 2 + 7x we have to write it that way.

The preceding comments show that the denominator is [ (2x-5)^2 * 3x + 1 ]

Evaluating the expression for x = 2:

- 3 / [ (2 * 2 - 5)^2 * 3(2) + 1 ] - 2 + 7*2 =

2 - 3 / [ (4 - 5)^2 * 6 + 1 ] - 2 + 14 = evaluate in parenthese; do multiplications outside parentheses

2 - 3 / [ (-1)^2 * 6 + 1 ] -2 + 14 = add inside parentheses

2 - 3 / [ 1 * 6 + 1 ] - 2 + 14 = exponentiate in bracketed term;

2 - 3 / 7 - 2 + 14 = evaluate in brackets

13 4/7 or 95/7 or about 13.57 add and subtract in order.

The details of the calculation 2 - 3 / 7 - 2 + 14:

Since multiplication precedes addition or subtraction the 3/7 must be done first, making 3/7 a fraction. Changing the order of the terms we have

2 - 2 + 14 - 3 / 7 = 14 - 3/7 = 98/7 - 3/7 = 95/7.

COMMON STUDENT QUESTION: ok, I dont understand why x isnt part of the fraction? And I dont understand why only the brackets are divided by 3..why not the rest of the equation?

INSTRUCTOR RESPONSE: Different situations give us different algebraic expressions; the situation dictates the form of the expression.

If the above expression was was written otherwise it would be a completely different expression and most likely give you a different result when you substitute.

If we intended the numerator to be x - 3 then the expression would be written (x - 3) / [(2x-5)^2 * 3x + 1 ] - 2 + 7x, with the x - 3 grouped.

If we intended the numerator to be the entire expression after the / the expression would be written x - 3 / [(2x-5)^2 * 3x + 1 - 2 + 7x ].

STUDENT COMMENT: I wasn't sure if the numerator would be 3 or -3. or is the subtraction sign just that a sign in this case?

INSTRUCTOR RESPONSE: In this case you would regard the - sign as an operation to be performed between the value of x and the value of the fraction, rather than as part of the numerator. That is, you would regard x - 3 / [ (2x-5)^2 * 3x + 1 ] as a subtraction of the fraction 3 / [ (2x-5)^2 * 3x + 1 ] from the term x.

STUDENT QUESTION: There was another question I had about this problem that wasn’t addressed. At the end when you changed the order of operation from

2 – 2 + 14 – 3/7 = 14 – 3/7

where did the 98/7 – 3/7 come into play before the end solution of 95/7? I must have forgotten how to do this part.

INSTRUCTOR RESPONSE: It's not clear how you can get 95/7 without this step.

To do the subtraction 14 - 3/7 both terms must be expressed in terms of a common denominator. The most convenient common denominator is 7.

So 14 must be expressed with denominator 7. This is accomplished by multiplying 14 by 7 / 7, obtaining 14 * 7 / 7 = 98 / 7. Since 7/7 = 1, we have just multiplied 14 by 1. We chose to use 7 / 7 in order to give us the desired denominator 7.

Thus our subtraction is

14 - 3/7 =

98/7 - 3/7 =

(98 - 3) / 7 =

95 /7.

STUDENT COMMENT

It took me a while to think thru this one especially when I got to working with the fraction. Fractions have always been my

weak spot. Any tips to make working with fractions a little easier is greatly appreciated.

INSTRUCTOR RESPONSE

Fractions are seriously undertaught in our schools, so your comment is not unusual.

I have to focus my attention on the subject matter of my courses, and while I do address it to a point, I don't have time to do justice to the subject of fractions. In any case , to do so would be redundant on my part, since there are a lot of excellent resources on the Internet.

I suggest you search the Web using something like 'review of fractions', and find something appropriate to your needs. You should definitely review the topic, as should 95% of all students entering your course.

STUDENT COMMENT

I think I am confused on why the Numerator is not the top portion and denominator the bottom portion of the problem.

INSTRUCTOR RESPONSE

Everything is on one line so there is no top or bottom in the given expression. A numerator and denominator are determined by a division of two expressions.

As we know, a denominator divides a numerator. In the given expression the division sign occurs between the 3 and the [ (2x-5)^2 * 3x + 1 ], so 3 is the numerator and [ (2x-5)^2 * 3x + 1 ] is the denominator.

x is not divided by the denominator, since the division occurs before the subtraction. For the same reason the -2 + 7x is not involved in the division. So neither the x nor the -2 + 7 x is part of the fractional expression.

STUDENT COMMENT

Didn’t know that 3 / 7 was 3/7 as a

fraction.

INSTRUCTOR RESPONSE

3/7 is treated as a fraction because of the order of operations. 3 must be divided by 7 before any other operation is applied to either number, and 3 divided by 7 is the fraction 3/7.

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Self-critique (if necessary):

I was obviously very lost in that problem because my answers were nothing like they were supposed to be

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Self-critique Rating:2

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Question: `q004. Explain, step by step, how you evaluate the expression (x - 5) ^ 2x-1 + 3 / x-2 for x = 4.

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Your solution:

Plug in the 4 (4-5)^2(4)-1+3/4-2 and do the order of operations. The answer is 7/4

confidence rating #$&*: 3

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Given Solution:

`aWe get

(4-5)^2 * 4 - 1 + 3 / 4 - 2

= (-1)^2 * 4 - 1 + 3 / 4 - 2 evaluating the term in parentheses

= 1 * 4 - 1 + 3 / 4 - 2 exponentiating (2 is the exponent, which is applied to -1 rather than multiplying the 2 by 4

= 4 - 1 + 3/4 - 2 noting that 3/4 is a fraction and adding and subtracting in order we get

= 1 3/4 = 7 /4 (Note that we could group the expression as 4 - 1 - 2 + 3/4 = 1 + 3/4 = 1 3/4 = 7/4).

COMMON ERROR:

(4 - 5) ^ 2*4 - 1 + 3 / 4 - 2 =

-1 ^ 2*4 - 1 + 3 / 4-2 =

-1 ^ 8 -1 + 3 / 4 - 2.

INSTRUCTOR COMMENTS:

There are two errors here. In the second step you can't multiply 2 * 4 because you have (-1)^2, which must be done first. Exponentiation precedes multiplication.

Also it isn't quite correct to write -1^2*4 at the beginning of the second step. If you were supposed to multiply 2 * 4 the expression would be (-1)^(2 * 4).

Note also that the -1 needs to be grouped because the entire expression (-1) is taken to the power. -1^8 would be -1 because you would raise 1 to the power 8 before applying the - sign, which is effectively a multiplication by -1.

STUDENT QUESTION: if it's read (-1)^8 it would be 1 or would you apply the sign afterward even if it is grouped and it be a -1?

INSTRUCTOR RESPONSE: The 8th power won't occur in this problem, of course, but you ask a good question.

-1^8 would require raising 1 to the 8th power, then applying the negative sign, and the result would be -1.

(-1)^8 would be the 8th power of -1, which as you see would be 1.

STUDENT COMMENT: I think it would be easier to visualize what your trying to raise to the exponent if you actually put parenthesis around the 2, that part seems to get tricky on the computer.

INSTRUCTOR RESPONSE: The expression was intentionally written to be misleading and make the point that, to avoid ambiguity, order of operations apply strictly, no matter what the expression looks like.

Normally, for clarity, the parentheses would be included. They aren't necessary, but when helpful it's a good idea to include them. You can, of course, have too many parentheses in an expression, making it harder than necessary to sort out. In practice we try to strike a balance.

The original expression was

(x - 5) ^ 2x-1 + 3 / x-2

White spaces make no difference in how an expression is evaluated, but they can help show the structure; e.g.,

(x - 5)^2 * x - 1 + 3 / x -2

is a visual improvement over the original. The * between the 2 and the x is not strictly necessary, but is also helpful.

((((x - 5) ^ 2)) * x) - 1 + (3 / x) - 2

verges on having too many parentheses at the beginning; it does help clarify the 3 / x.

STUDENT COMMENT

Although I read through your explanation and do see the point you are making, that 2x is actually 2 * x, I still think that

(-1) should be raised to 2x rather than 2. Kaking the answer -11/4, not 7/4.

INSTRUCTOR RESPONSE

When the expression (x - 5) ^ 2x-1 + 3 / x-2 is copied and pasted into a computer algebra system it is translated as

This notation is universal and unambiguous. Any deviation from strict interpretation (which does occur among some authors and among manufacturers of some calculators) tends to result in ambiguity and confusion.

STUDENT COMMENT

While I do understand what you are trying to relate, I will continue to make these mistakes on more than one occasion and will not penalize myself for not rewriting years of mathematics because of a syntax issue in an online class.

INSTRUCTOR RESPONSE

I don't penalize errors in typed notation when the intent is clear (though I will sometimes point out these errors), and when you take your tests you'll be writing them out by hand and this won't be an issue.

However this is not a syntax issue in an online class. This is the order of operations, as it has been since algebra was developed hundreds of years ago, and it's completely consistent with the mathematics you appear to know (quite well).

As stated here, if you use the wrong syntax in any computer algebra system, your expression will not be interpreted correctly. For this reason alone you need to understand the notation.

For this and other valid reasons you need to understand how the order of operations are represented in 'linear' fashion (i.e., 'typewriter notation') and to correctly interpret expressions written in this notatation.

Any mathematics that has been learned correctly is completely consistent with the order of operations and with the notation used in this course. If the mathematics you've learned was inconsistent with the order of operations (and I don't believe this is so in your case, but it is with many students), then you would need to adjust your thinking. Fortunately this is very easy to do. Interpret expressions literally, assume nothing, and everything works out.

You will also find that the notation quickly becomes easy to read and use, and that it expands your comprehension of all mathematical notation.

STUDENT COMMENT

I used -1^(2*4). I didn't realize that was doing multiplication before exponents. All of this typewriter notations seems ambiguous to me but I think that had I seen the expression in standard notation I would probably have made the same mistake in this instance. If I were writing this expression I would probably use a parenthesis or * to show the necessary separation.

INSTRUCTOR RESPONSE

Parentheses, even when they aren't strictly necessary, are often useful to clarify the expression. An parentheses, even when not necessary, are part of the order of operations.

Spacing is not part of the order of operations. An expression has the same meaning even if all spaces are removed.

However as long as an expression is correctly formed, spacing as well as parentheses can certainly be used to make it more readable.

I don't go to any trouble in this exercise to make the expressions readable, since my goal here is to make the point about order of operations, which give an expression its unambiguous meaning.

However in most of the documents you will be working with, I do make an effort to clarify the meanings of expressions through their formatting, often using unnecessary parentheses and spacing to help clarify meaning.

Certainly I encourage you to do the same.

STUDENT QUESTION

I didn’t separate the ¾ as a stand alone fraction, I am confused about why you don’t treat it as an equation that the

denominator isn’t treated as a denominator.

INSTRUCTOR RESPONSE

Your work was good throughout most of this problem. You did forget to copy down a -1 in one of the early steps, but otherwise followed the order of operations correctly until nearly the last step.

However near the end you said that 4+3/4-2=7/2.

You appear to have performed the addition 4 + 3 and the subtraction 4 - 2 before dividing. However the division has to be done first.

The division sign is between the 3 and the 4, so the division is 3/4, and that gives you the fraction 3/4.

Therefore the expression 4+3/4-2 tells you to 'add 3/4 to 4 then subtract 2'.

When actually writing this out we would probably include parentheses. That wasn't done here, as it would have defeated the point being made about order of operations, but for clarity we might have written

4 + (3/4) - 2.

The parentheses are not necessary around the 3/4, since the order of operations is sufficient to unambiguously define the result, but they do make the expression easier to read and reduce the likelihood of error.

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Self-critique (if necessary):ok

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Question: `q005. At the link

http://www.vhcc.edu/dsmith/genInfo/introductory problems/typewriter_notation_examples_with_links.htm

(copy this path into the Address box of your Internet browser; alternatively use the path

http://vhmthphy.vhcc.edu/ > General Information > Startup and Orientation (either scroll to bottom of page or click on Links to Supplemental Sites) > typewriter notation examples

and you will find a page containing a number of additional exercises and/or examples of typewriter notation.Locate this site, click on a few of the links, and describe what you see there.

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Your solution:

Problems for extra exercise

confidence rating #$&*:

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Given Solution:

`aYou should see a brief set of instructions and over 30 numbered examples. If you click on the word Picture you will see the standard-notation format of the expression. The link entitled Examples and Pictures, located in the initial instructions, shows all the examples and pictures without requiring you to click on the links. There is also a file which includes explanations.

The instructions include a note indicating that Liberal Arts Mathematics students don't need a deep understanding of the notation, Mth 173-4 and University Physics students need a very good understanding,

while students in other courses should understand the notation and should understand the more basic simplifications.

There is also a link to a page with pictures only, to provide the opportunity to translated standard notation into typewriter notation.

end program

STUDENT COMMENT (not quite correct)

I see a collection of typewriter problems, after looking at some of them I see that the slash mark is to create a fraction rather than to denote division.

INSTRUCTOR CORRECTION

A fraction is a division of the numerator by the denominator. The slash mark indicates division, which can often be denoted by a fraction.

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Question: `q006 Standard mathematics notation is easier to look at; it's easier to see the meaning of the expressions.

However it's very important to understand order of operations, and students do get used to this way of doing it.

You should of course write everything out in standard notation when you work it on paper.

It is likely that you will at some point use a computer algebra system, and when you do you will probably have to enter expressions using a keyboard, so it is well worth the trouble to get used to this notation.

Indicate your understanding of why it is important to understand this notation.

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Your solution:

We need to understand this notation to communicate our responses to the problems to you.

confidence rating #$&*:3

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Self-critique (if necessary):

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&#Good work. Let me know if you have questions. &#

Jeremiah

#$&*

course Mth 271

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002. Describing Graphs*********************************************

Question: `q001. You will frequently need to describe the graphs you have constructed in this course. This exercise is designed to get you used to some of the terminology we use to describe graphs. Please complete this exercise and email your work to the instructor. Note that you should do these graphs on paper without using a calculator. None of the arithmetic involved here should require a calculator, and you should not require the graphing capabilities of your calculator to answer these questions.

Problem 1. We make a table for y = 2x + 7 as follows: We construct two columns, and label the first column 'x' and the second 'y'. Put the numbers -3, -2, -1, -, 1, 2, 3 in the 'x' column. We substitute -3 into the expression and get y = 2(-3) + 7 = 1. We substitute -2 and get y = 2(-2) + 7 = 3. Substituting the remaining numbers we get y values 5, 7, 9, 11 and 13. These numbers go into the second column, each next to the x value from which it was obtained. We then graph these points on a set of x-y coordinate axes. Noting that these points lie on a straight line, we then construct the line through the points.

Now make a table for and graph the function y = 3x - 4.

Identify the intercepts of the graph, i.e., the points where the graph goes through the x and the y axes.

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Your solution:

X Y

-3 -5

-2 -2

-1 1

0 4

1 7

2 10

3 13

Intercepts are (0,-4)(4/3,0)

confidence rating #$&*: 3

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Given Solution:

`aThe graph goes through the x axis when y = 0 and through the y axis when x = 0.

The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3.

The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4).

Your graph should confirm this.

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Question: `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes.

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Your solution:

It doesn’t change because the slope stays the same, which can be determined by rise over run and that equals 3.

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Given Solution:

`aThe graph forms a straight line with no change in steepness.

STUDENT COMMENT

Ok, I may not understand what exactly it meant by steepness, I was thinking since it was

increasing it would also be getting steeper?????

INSTRUCTOR RESPONSE

A graph can increase while getting steeper and steeper; or it can increase while getting less and less steep. Or it can increase with no change in steepness.

Analogies:

When you walk up a hill, typically as you approach the top the slope starts to level off--it gets less steep.

When you go up a ramp the steepness stays the same until you get to the end of the ramp.

When you start climbing a hill, typically it gets steeper for awhile, the stays at about a constant slope, then gets less steep toward the top.

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Question: `q003. What is the slope of the graph of the preceding two exercises (the function is y = 3x - 4;slope is rise / run between two points of the graph)?

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Your solution:

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Given Solution:

`aBetween any two points of the graph rise / run = 3.

For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3.

Note that 3 is the coefficient of x in y = 3x - 4.

Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change.

EXPANDED EXPLANATION

Any student who has completed Algebra I and Algebra II should be familiar with slope calculations. Most students are. However a number of students appear to be very fuzzy on the concept, and I suspect that not all prerequisite courses cover this concept adequately (though I am confident that it's done well at VHCC). Also a number of students haven't taken a math course in awhile, and might simply be a bit rusty with this idea. In any case the following expanded explanation might be helpful to some students:

Slope = rise / run.

The rise between two graph points is the change in the y coordinate. The run is the change in the x coordinate.

Our function is y = 3 x - 4.

When x = 2, we substitute 2 for x to get y = 3 * 2 - 4, which is equal to 2.

When x = 8, we substitute 8 for x to get y = 3 * 8 - 4, which is equal to 20.

The graph therefore contains the points (2, 2) and (8, 20).

You should have made a graph showing these points. If not you should do so now.

As you go from point to point your y coordinate goes from 2 to 20. So the 'rise' between the points is 20 - 2 = 18.

Your x coordinate goes from 2 to 8. So the 'run' between the points is 8 - 2 = 6.

The slope is rise / run = 18 / 6 = 3.

The numbers 2 and 8, which were used for the x values, were chosen arbitrarily. Any other two x values would have given you

different coordinates, likely with different rise and run. However whatever two x values you use, you will get the same slope. The slope of this graph is constant, and is equal to 3.

STUDENT QUESTION

Am I not allowed to utilize my calculus tools, yet?

Couldn't I have just taken the derivative for the function, y = 3x -4 to obtain 3 as the slope?

However, I do know how to do both ways. Which is the more preferred method?

INSTRUCTOR RESPONSE

This exercise develops a language for describing some aspects of graphs, and does not assume calculus tools.

Of course it's fine to use the calculus tools if you have them, as long as you understand the problem at the more basic level as well.

Unfortunately, not every student who has had a calculus course would know how to apply those tools to this situation (for example, I've had students from other institutions who have made A's in Applied Calculus courses from other (not particularly reputable) institutions, who didn't understand the concept of a slope).

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Question: `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate?

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Your solution:

X Y

0 0

1 1

2 4

3 9

The graph is increasing at an increasing rate.

confidence rating #$&*:

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Given Solution:

`aGraph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right.

The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate

STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate?

INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1?

In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&.

STUDENT QUESTION: I am a little hazy on what the steepness is

INSTRUCTOR RESPONSE: The hill analogy I used above might be helpful.

Formally, steepness could be defined as the magnitude of the slope, i.e., the absolute value of the slope.

Two graphs with respective slopes 4 and -4 would be equally steep; both would have slope of magnitude 4. Both of these graphs would be steeper than, say a graph with slope 3 or -3.

NOTE FOR STUDENT WITH CALCULUS BACKGROUND (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus)

In terms of the calculus, the derivative function is easily seen to be y ' = 2 x, which is positive and increasing, and which therefore implies an increasing slope.

Since in this case the slope is positive, which implies that the function is increasing, the increasing slope therefore implies that the value of the function is increasing at an increasing rate.

Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as increasing and concave upward.

This could also be explained in terms of the second derivative, y '' = 2, which is positive everywhere. The positive second derivative implies that the graph is concave up.

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Question: `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

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Your solution:

X Y

-3 -9

-2 -4

-1 -1

0 0

The graph is decreasing at a decreasing rate.

confidence rating #$&*: 3

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Given Solution:

`aFrom left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing.

Thus the graph is decreasing, but more and more slowly. We therefore say that the graph on this interval is decreasing at a decreasing rate.

NOTE FOR STUDENT WITH CALCULUS BACKGROUND (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus)

In terms of the calculus, the derivative function is easily seen to be y ' = 2 x, which is positive and increasing, and which therefore implies an increasing slope.

Since in this case the slope is negative, which implies that the function is decreasing, the increasing slope therefore implies that the rate of decrease is decreasing. The value of the function is therefore decreasing at a decreasing rate.

Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as decreasing and concave upward.

This could also be explained in terms of the second derivative, y '' = 2, which is positive everywhere. The positive second derivative implies that the graph is concave up.

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Question: `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

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Your solution:

X Y

0 0

1 1

2 1.41

3 1.73

The graph is increasing at a decreasing rate.

confidence rating #$&*: 3

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Given Solution:

`aIf you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing.

The graph would be increasing at a decreasing rate.

If the graph represents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing.

If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as described take another look at your plot and make a note in your response indicating any difficulties.

STUDENT QUESTION: I am still unsure why the steepness is decreasing, I see why going

from right to left, but the graph looks linear?

INSTRUCTOR RESPONSE: The y value increases, but it changes by less and less for every succeeding x value. So the graph is increasing, but by less and less with each step. It's increasing but at a decreasing rate.

The graph does not look linear. If it does, then it's probably because your x and/or y axis is not scaled in equal increments.

NOTE FOR CALCULUS-PREPARED STUDENTS (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus)

In terms of the calculus, the derivative function is easily seen to be y ' = 1 / (2 sqrt(x)), which is positive but decreasing, and which therefore implies a decreasing slope.

Since in this case the slope is positive, which implies that the function is increasing, the decreasing slope therefore implies that the rate of increase is decreasing. The value of the function is therefore increasing at a decreasing rate.

Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as increasing and concave downward.

This could also be explained in terms of the second derivative, y '' = -1 / (4 x^(3/2)), which is negative on this interval. The negative second derivative implies that the graph is concave down.

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Question: `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3.

Would you say that the graph is increasing or decreasing?

Does the steepness of the graph change and if so, how?

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Your solution:

X Y

0 5

1 2.5

2 1.25

3 .625

The graph is decreasing at a decreasing rate

confidence rating #$&*: 3

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Given Solution:

`a** From basic algebra recall that a^(-b) = 1 / (a^b).

So, for example:

2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4.

5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc.

The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time.

The graph is therefore decreasing at a decreasing rate. **

STUDENT QUESTION

I don’t understand how the graph decreases at a decreasing rate because it decreases by half every time. The ½ is constant.

INSTRUCTOR RESPONSE

The values decrease by a factor of 1/2 every time. That means each number would be multiplied by 1/2 to get the next.

As a result the numbers we are halving keep decreasing.

Half of 5 is 2.5; half of 2.5 is 1.25; half of 1.25 is .625. The decreases from one number to the next are respectively 2.5, 1.25 and .625.

If the y values 5, 2.5, 1.25, .625 are placed at equal x intervals, it should be clear that the graph is decreasing at a decreasing rate.

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Question: `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster.

If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing?

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Your solution:

It would be increasing at an increasing rate because the car is speeding up while moving away from you.

confidence rating #$&*:3

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Given Solution:

`a** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. **

STUDENT COMMENT

I don’t fully understand a distance vs. time graph.

INSTRUCTOR RESPONSE

If y represents the distance from you to the car and t represents the time in seconds since the car started out, then the graph of y vs. t is a graph of distance vs. clock time.

The car is speeding up, so in any series of equal time intervals it moves further with each new interval.

The distance it moves on an interval is represented by the difference between the y coordinates, so if it move further during an interval the 'rise' of the graph on that interval will be greater. If the intervals are equally spaced along the t axis, the result is an increasing graph with increasing slope.

This is best understood by sketching the graph according to this description.

STUDENT QUESTION

I understand the clock time but could you give me some examples of numbers to sketch a graph. I am drawing a

blank to how to make myself understand.?????

INSTRUCTOR RESPONSE

If the car's velocity for the first second averages 1 ft / sec, then in subsequent second 3 ft / sec, then 5 ft / sec, then 7

ft / sec, it will move 1 foot during the first second, 3 feet during the next, 5 feet during the next and 7 feet during the

next.

A graph of velocity vs. clock time would be a straight line, since the velocity increases by the same amount every second.

However the positions of the car, as measured from the starting point, would be

position 1 foot after 1 second

position 4 feet after 2 seconds (the position changes by 3 feet, started this second at 1 ft, so the car ends up with

position 4 feet)

position 9 feet after 3 seconds (the position changes by 5 feet, started this second at 4 ft, so the car ends up with

position 9 feet)

position 16 feet after 4 seconds (the position changes by 7 feet, started this second at 9 ft, so the car ends up with

position 16 feet)

So the graph of position vs. clock time has positions 0, 1, 4, 9 and 16 feet after 0, 1, 2, 3 and 4 seconds, respectively.

The position vs. clock time graph is therefore increasing at an increasing rate.

Let me know if this doesn't answer your question.

STUDENT QUESTION

I still don’t totally understand why it would necessarily be increasing at an increasing rate. Couldn’t it be a decreasing

or even a standard rate as I mentioned above?

INSTRUCTOR RESPONSE

If the car speeds up then its distance from its starting position increases at an increasing rate.

Its speed might be increasing at an increasing, constant or decreasing rate, but not its position.

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

Jeremiah

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course Mth 271

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005. Calculus*********************************************

Question: `q001. The graph of a certain function is a smooth curve passing through the points (3, 5), (7, 17) and (10, 29).

Between which two points do you think the graph is steeper, on the average?

Why do we say 'on the average'?

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Your solution:

Do the rise over run and between the points of (7,17) and (10,29) the slope is 4.

confidence rating #$&*: 3

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Given Solution:

`aSlope = rise / run.

Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7) =12 / 3 = 4.

The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 / 4 = 3.

The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to point. The slope obtained over the interval is a specific type of average of the slopes of all points between the endpoints.

2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values are taken by the expression 1 / (x - 2)?

1. As the process continues, with x getting closer and closer to 2, what happens to the values of 1 / (x-2)?

2. Will the value ever exceed a billion? Will it ever exceed one trillion billions?

3. Will it ever exceed the number of particles in the known universe?

4. Is there any number it will never exceed?

5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2?

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Your solution:

1. They take the positions of the 10’s 100’s 1000’s and so on.

2. It can

3. It can

4. No

5. It doesn’t exist. There is an asymptote there.

confidence rating #$&*: 3

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Given Solution:

`aFor x = 2.1, 2.01, 2.001, 2.0001 we see that x -2 = .1, .01, .001, .0001. Thus 1/(x -2) takes respective values 10, 100, 1000, 10,000.

It is important to note that x is changing by smaller and smaller increments as it approaches 2, while the value of the function is changing by greater and greater amounts.

As x gets closer in closer to 2, it will reach the values 2.00001, 2.0000001, etc.. Since we can put as many zeros as we want in .000...001 the reciprocal 100...000 can be as large as we desire. Given any number, we can exceed it.

Note that the function is simply not defined for x = 2. We cannot divide 1 by 0 (try counting to 1 by 0's..You never get anywhere. It can't be done. You can count to 1 by .1's--.1, .2, .3, ..., .9, 1. You get 10. You can do similar thing for .01, .001, etc., but you just can't do it for 0).

As x approaches 2 the graph approaches the vertical line x = 2; the graph itself is never vertical. That is, the graph will have a vertical asymptote at the line x = 2. As x approaches 2, therefore, 1 / (x-2) will exceed all bounds.

Note that if x approaches 2 through the values 1.9, 1.99, ..., the function gives us -10, -100, etc.. So we can see that on one side of x = 2 the graph will approach +infinity, on the other it will be negative and approach -infinity.

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Question: `q003. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.

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Your solution:

It is easily seen by drawing, but it can also be determined mathematically through solving for area. The second trapezoid is much greater in area.

confidence rating #$&*: 3

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Given Solution:

`aYour sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area.

To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area.

This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.

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Question: `q004. If f(x) = x^2 (meaning 'x raised to the power 2') then which is steeper, the line segment connecting the x = 2 and x = 5 points on the graph of f(x), or the line segment connecting the x = -1 and x = 7 points on the same graph? Explain the basisof your reasoning.

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Your solution:

You just plug in the numbers for the first and second. The first number is your x value and the one that you get with the equation is your y value. After doing the math the first ones slope is larger than the second. (2,4) and (5,25)= (25-4)/(5-2)= 7 the second is (-1,-1) and (7,49)= (49+1)/(7+1)=6.25

confidence rating #$&*:3

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Given Solution:

`aThe line segment connecting x = 2 and the x = 5 points is steeper: Since f(x) = x^2, x = 2 gives y = 4 and x = 5 gives y = 25. The slope between the points is rise / run = (25 - 4) / (5 - 2) = 21 / 3 = 7.

The line segment connecting the x = -1 point (-1,1) and the x = 7 point (7,49) has a slope of (49 - 1) / (7 - -1) = 48 / 8 = 6.

The slope of the first segment is greater.

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Question: `q005. Suppose that every week of the current millenium you go to the jewler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time ( this is so), that the the gold remains undisturbed (maybe, maybe not so), that no other source adds gold to your backyard (probably so), and that there was no gold in your yard before..

1. If you construct a graph of y = the number of grams of gold in your backyard vs. t = the number of weeks since Jan. 1, 2000, with the y axis pointing up and the t axis pointing to the right, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly?

2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week.

{}3. Answer the same question assuming that every week you bury half the amount you did the previous week.

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Your solution:

1. It would be a straight line

2. The graph will increase at an increasing rate

3. The graph will increase at a decreasing rate

confidence rating #$&*: 3

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Given Solution:

`a1. If it's the same amount each week it would be a straight line.

2. Buying gold every week, the amount of gold will always increase. Since you buy more each week the rate of increase will keep increasing. So the graph will increase, and at an increasing rate.

3. Buying gold every week, the amount of gold won't ever decrease. Since you buy less each week the rate of increase will just keep falling. So the graph will increase, but at a decreasing rate. This graph will in fact approach a horizontal asymptote, since we have a geometric progression which implies an exponential function.

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Question: `q006. Suppose that every week you go to the jewler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time, that the the gold remains undisturbed, and that no other source adds gold to your backyard.

1. If you graph the rate at which gold is accumulating from week to week vs. tne number of weeks since Jan 1, 2000, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly?

2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week.

3. Answer the same question assuming that every week you bury half the amount you did the previous week.

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Your solution:

1. It is a straight line

2. It is and increasing straight line

3. It is a decreasing straight line that falls more and more slowly

confidence rating #$&*: 3

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Given Solution:

`aThis set of questions is different from the preceding set. This question now asks about a graph of rate vs. time, whereas the last was about the graph of quantity vs. time.

Question 1: This question concerns the graph of the rate at which gold accumulates, which in this case, since you buy the same amount eact week, is constant. The graph would be a horizontal straight line.

Question 2: Each week you buy one more gram than the week before, so the rate goes up each week by 1 gram per week. You thus get a risingstraight line because the increase in the rate is the same from one week to the next.

Question 3. Since half the previous amount will be half of a declining amount, the rate will decrease while remaining positive, so the graph remains positive as it decreases more and more slowly. The rate approaches but never reaches zero.

STUDENT COMMENT: I feel like I am having trouble visualizing these graphs because every time for the first one I picture an increasing straight line

INSTRUCTOR RESPONSE: The first graph depicts the amount of gold you have in your back yard. The second depicts the rate at which the gold is accumulating, which is related to, but certainly not the same as, the amount of gold.

For example, as long as gold is being added to the back yard, the amount will be increasing (though not necessarily on a straight line). However if less and less gold is being added every year, the rate will be decreasing (perhaps along a straight line, perhaps not).

FREQUENT STUDENT RESPONSE

This is the same as the problem before it. No self-critique is required.

INSTRUCTOR RESPONSE

This question is very different that the preceding, and in a very significant and important way. You should have

self-critiqued; you should go back and insert a self-critique on this very important question and indicate your insertion by

preceding it with ####. The extra effort will be more than worth your trouble.

These two problems go to the heart of the Fundamental Theorem of Calculus, which is the heart of this course, and the extra effort will be well worth it in the long run. The same is true of the last question in this document.

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``q007. If the depth of water in a container is given, in centimeters, by 100 - 2 t + .01 t^2, where t is clock time in seconds, then what are the depths at clock times t = 30, t = 40 and t = 60? On the average is depth changing more rapidly during the first time interval or the second?

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Your solution:

It is changing more rapidly in the second because it goes from 49 cm to 36 cm in the first but in the second it goes from 36cm to 16cm

confidence rating #$&*: 3

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Given Solution:

`aAt t = 30 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 30 + .01 * 30^2 = 49.

At t = 40 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 40 + .01 * 40^2 = 36.

At t = 60 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 60 + .01 * 60^2 = 16.

49 cm - 36 cm = 13 cm change in 10 sec or 1.3 cm/s on the average.

36 cm - 16 cm = 20 cm change in 20 sec or 1.0 cm/s on the average.

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Self-critique (if necessary):

I didn’t factor in the seconds causing me to misinterpret this problem

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Question: `q008. If the rate at which water descends in a container is given, in cm/s, by 10 - .1 t, where t is clock time in seconds, then at what rate is water descending when t = 10, and at what rate is it descending when t = 20? How much would you therefore expect the water level to change during this 10-second interval?

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Your solution:

It decreases by one every ten seconds. Therefore, I expect the next ten seconds to also decrease by one

confidence rating #$&*:3

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Given Solution:

`aAt t = 10 sec the rate function gives us 10 - .1 * 10 = 10 - 1 = 9, meaning a rate of 9 cm / sec.

At t = 20 sec the rate function gives us 10 - .1 * 20 = 10 - 2 = 8, meaning a rate of 8 cm / sec.

The rate never goes below 8 cm/s, so in 10 sec the change wouldn't be less than 80 cm.

The rate never goes above 9 cm/s, so in 10 sec the change wouldn't be greater than 90 cm.

Any answer that isn't between 80 cm and 90 cm doesn't fit the given conditions..

The rate change is a linear function of t. Therefore the average rate is the average of the two rates, or 8.5 cm/s.

The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of 85 cm.

STUDENT RESPONSES

The following, or some variation on them, are very common in student comments. They are both very good questions. Because of the importance of the required to answer this question correctly, the instructor will typically request for a revision in response to either student response:

• I don't understand how the answer isn't 1 cm/s. That's the difference between 8 cm/s and 9 cm/s.

• I don't understand how the answer isn't 8.5 cm/s. That's the average of the 8 cm/s and the 9 cm/s.

INSTRUCTOR RESPONSE

A self-critique should include a full statement of what you do and do not understand about the given solution. A phrase-by-phrase analysis of the solution is not unreasonable (and would be a good idea on this very important question), though it wouldn't be necessary in most situations.

An important part of any self-critique is a good question, and you have asked one. However a self-critique should if possible go further. I'm asking that you go back and insert a self-critique on this very important question and indicate your insertion by preceding it with ####, before submitting it. The extra effort will be more than worth your trouble.

This problem, along with questions 5 and 6 of this document, go to the heart of the Fundamental Theorem of Calculus, which is the heart of this course, and the extra effort will be well worth it in the long run.

You should review the instructions for self-critique, provided at the link given at the beginning of this document.

STUDENT COMMENT

The question is worded very confusingly. I took a stab and answered correctly. When answering, """"How much would you

therefore expect the water level to change during this 10-second interval?"""" It is hard to tell whether you are asking for

what is the expected change in rate during this interval and what is the changing """"water level."""" But now, after looking at

it, with your comments, it is clearer that I should be looking for the later. Thanks!

INSTRUCTOR RESPONSE

'Water level' is clearly not a rate. I don't think there's any ambiguity in what's being asked in the stated question.

The intent is to draw the very important distinction between the rate at which a quantity changes, and the change in the quantity.

It seems clear that as a result of this question you understand this and will be more likely to make such distinctions in your subsequent work.

This distinction is at the heart of the calculus and its applications. It is in fact the distinction between a derivative and an integral.

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Jeremiah

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course Mth 271

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001. RatesNote that there are 10 questions in this assignment. The questions are of increasing difficulty--the first questions are fairly easy but later questions are very tricky. The main purposes of these exercises are to refine your thinking about rates, and to see how you process challenging information. Most students in most courses would not be expected to answer all these questions correctly; all that's required is that you do your best and follows the recommended procedures for answering and self-critiquing your work.

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Question: If you make $50 in 5 hr, then at what rate are you earning money?

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Your solution:

50/5=10 10 dollars and hr

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Given Solution:

The rate at which you are earning money is the number of dollars per hour you are earning. You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour. It is very likely that you immediately came up with the $10 / hour because almosteveryone is familiar with the concept of the pay rate, the number of dollars per hour. Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it. Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it.

You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes.

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Question: `q003.If you make $60,000 per year then how much do you make per month?

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Your solution:

60,000/12=5000

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Given Solution:

Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month. Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it.

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Question: `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month?

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Your solution:

Makes an average of 5000 per month

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Given Solution:

Small businesses do not usually make the same amount of money every month. The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others. Therefore it is much more appropriate to say that the business makes and average of $5000 per month.

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Question: `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate?

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Your solution:

50 miles and hr. Because there may be times where you are going faster or slower, or not even moving.

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Given Solution:

The average rate is 50 miles per hour, or 50 miles / hour. This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtaining ave rate = 300 miles / ( 6 hours) = 50 miles / hour. Note that the rate at which distance is covered is called speed. The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities.

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Question: `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled?

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Your solution:

60/1200= .05 gallons per mile

confidence rating #$&*:3

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Given Solution:

The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second. As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it.

By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled. In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile.

Note that this question didn't ask for miles per gallon. Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used. Be sure you see the difference.

Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time. This rate is calculated with respect to the amount of fuel used. We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover t miles. Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that result in more or fewer miles covered with a certain amount of fuel.

It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms.

In this case gallons / mile tells you how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult.

STUDENT COMMENT

Very Tricky! I thought I had a rhythm going. I understand where I messed up. I am comfortable with the calculations.

INSTRUCTOR RESPONSE

There's nothing wrong with your rhythm.

As I'm sure you understand, there is no intent here to trick, though I know most people will (and do) tend to give the answer you did.

My intent is to make clear the important point that the definition of the terms is unambiguous and must be read carefully, in the right order.

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Question: `q007. The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added. Why is it that we are calculating average rates but we aren't adding anything?

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Your solution:

Because everything that is applied doesn’t go at a constant rate, it can usually vary so that’s why we use the term average, which would be the means to the variations that are happening.

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Given Solution:

The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month. In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles. Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate.

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Question: `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup?

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Your solution:

The second group did 40 more pushups and gained 15lbs of strength. 15/40=.375lbs per pushup

confidence rating #$&*: 3

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Given Solution:

The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first. The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup.

STUDENT COMMENT:

I have a question with respect as to how the question is interpreted. I used the interpretation given in the solution

to question 008 to rephrase the question in 009, but I do not see how this is the correct interpretation of the question as

stated.

INSTRUCTOR RESPONSE:

This exercise is designed to both see what you understand about rates, and to challenge your understanding a bit with concepts that aren't always familiar to students, despite their having completed the necessary prerequisite courses.

The meaning of the rate of change of one quantity with respect to another is of central importance in the application of mathematics. This might well be your first encounter with this particular phrasing, so it might well be unfamiliar to you, but it is important, unambiguous and universal.

You've taken the first step, which is to correctly apply the wordking of the preceding example to the present question.

You'll have ample opportunity in your course to get used to this terminology, and plenty of reinforcement.

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Question: `q009. In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight. At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds. At what average rate did lifting strength increase with respect to the added shoulder weight?

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Your solution:

17/20= .85 pounds per pound on their shoulders

confidence rating #$&*: 3

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Given Solution:

The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight. The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound. The strength advantage was .85 lifting pounds per pound of added weight, on the average.

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Question: `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions?

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Your solution:

10 meters per second

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Given Solution:

The runner traveled 100 meters between the two positions, and required 10 seconds to do so. The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second. Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds.

STUDENT QUESTION

Is there a formula for this is it d= r*t or distance equal rate times time??????????????????

INSTRUCTOR RESPONSE

That formula would apply in this specific situation.

The goal is to learn to use the general concept of rate of change. The situation of this problem, and the formula you quote, are just one instance of a general concept that applies far beyond the context of distance and time.

It's fine if the formula helps you understand the general concept of rate. Just be sure you work to understand the broader concept.

Note also that we try to avoid using d for the name of a variable. The letter d will come to have a specific meaning in the context of rates, and to use d as the name of a variable invite confusion.

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Question: `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the intervening 100 meter distance?

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Your solution:

9.5 meters per second on average. The 100 will be run in 10.5 seconds and the 200 will be run in 21 seconds.

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Given Solution:

At 10 meters/sec, the runner would require 10 seconds to travel 100 meters. However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters. We don't know what the runner's average speed is, we only know that it goes from 10 m/s to 9 m/s. The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s ) / 2 = 9.5 m/s. Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec, approx.. Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds. However we were not given this information, and we don't add extraneous assumptions without good cause. So the approximation we used here is pretty close to the best we can do with the given information.

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Question: `q012. We just averaged two quantities, adding them and dividing by 2, to find an average rate. We didn't do that before. Why we do it now?

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Your solution:

Because we didn’t have enough information and by doing this we made an average rate allowing us to answer the question to the full extent.

confidence rating #$&*: 3

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Given Solution:

In previous examples the quantities weren't rates. We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons). Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question. Within this context, averaging the 2 rates was an appropriate tactic.

STUDENT QUESTION:

I thought the change of an accumulating quantity was the rate?

INSTRUCTOR RESPONSE:

Quick response: The rate is not just the change in the accumulating quantity; if we're talking about a 'time rate' it's the change in the accumulating quantity divided by the time interval (or in calculus the limiting value of this ratio as the time interval approaches zero).

More detailed response: If quantity A changes with respect to quantity B, then the average rate of change of A with respect to B (i.e., change in A / change in B) is 'the rate'. If the B quantity is clock time, then 'the rate' tells you 'how fast' the A quantity accumulates. However the rate is not just the change in the quantity A (i.e., the change in the accumulating quantity), but change in A / change in B.

For students having had at least a semester of calculus at some level: Of course the above generalizes into the definition of the derivative. y ' (x) is the instantaneous rate at which the y quantity changes with respect to x. y ' (x) is the rate at which y accumulates with respect to x.

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#(*!

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

Jeremiah

#$&*

course Mth 271

1/12 5

001. Areas*********************************************

Question: `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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Your solution:

4*3= 12 sq meters

confidence rating #$&*: 3

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Given Solution:

`aA 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2.

The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2.

Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.

FREQUENT STUDENT ERRORS

The following are the most common erroneous responses to this question:

4 * 3 = 12

4 * 3 = 12 meters

INSTRUCTOR EXPLANATION OF ERRORS

Both of these solutions do indicate that we multiply 4 by 3, as is appropriate.

However consider the following:

4 * 3 = 12.

4 * 3 does not equal 12 meters.

4 * 3 meters would equal 12 meters, as would 4 meters * 3.

However the correct result is 4 meters * 3 meters, which is not 12 meters but 12 meters^2, as shown in the given solution.

To get the area you multiply the quantities 4 meters and 3 meters, not the numbers 4 and 3. And the result is 12 meters^2, not 12 meters, and not just the number 12.

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

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Your solution:

4m*3m=12m sq. and since it’s a triangle we divide by 2 getting 6 m sq

confidence rating #$&*: 3

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Given Solution:

`aA right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters.

The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2.

The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h.

STUDENT QUESTION

Looking at your solution I think I am a bit rusty on finding the area of triangles. Could you give me a little more details

on how you got your answer?

INSTRUCTOR RESPONSE

As explained, a right triangle is half of a rectangle.

There are two ways to put two right triangles together, joining them along the hypotenuse. One of these ways gives you a rectangle. The common hypotenuse thus forms a diagonal line across the rectangle.

The area of either triangle is half the area of this rectangle.

If this isn't clear, take a blade or a pair of scissors and cut a rectangle out of a piece of paper. Make sure the length of the rectangle is clearly greater than its width. Then cut your rectangle along a diagonal, to form two right triangles.

Now join the triangles together along the hypotenuse. They will either form a rectangle or they won't. Either way, flip one of your triangles over and again join them along the hypotenuse. You will have joined the triangles along a common hypotenuse, in two different ways. If you got a rectangle the first time, you won't have one now. And if you have a rectangle now, you didn't have one the first time.

It should be clear that the two triangles have equal areas (allowing for a little difference because we can't really cut them with complete accuracy).

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Question: `q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?

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Your solution:

5*2= 10m sq

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Given Solution:

`aA parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h.

The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2.

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Question: `q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?

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Your solution:

5cm * 2cm= 10cm sq

10cm sq/2= 5cm sq

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Given Solution:

`aIt is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2.

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Question: `q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?

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Your solution:

4.0km*5.0km= 20km sq

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Given Solution:

`aAny trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2.

STUDENT SOLUTION ILLUSTRATING NEED TO USE UNITS IN ALL STEPS

A=Base time average altitude therefore………A=4 *5= 20 km ^2

INSTRUCTOR COMMENT

A = (4 km) * (5 km) = 20 km^2.

Use the units at every step. km * km = km^2, and this is why the answer comes out in km^2.

Try to show the units and how they work out in every step of the solution.

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Question: `q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?

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Your solution:

4.0cm*5.5cm=22cm sq

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Given Solution:

`aThe area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.

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Question: `q007. What is the area of a circle whose radius is 3.00 cm?

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Your solution:

3.14*(3)^2= 28.2 cm sq

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Given Solution:

`aThe area of a circle is A = pi * r^2, where r is the radius. Thus

A = pi * (3 cm)^2 = 9 pi cm^2.

Note that the units are cm^2, since the cm unit is part r, which is squared.

The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius.

Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere.

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Self-critique (if necessary):

I couldn’t understand where the 9 come from, but it’s the answer you get divided by pi.

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Question: `q008. What is the circumference of a circle whose radius is exactly 3 cm?

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Your solution:

2(3.14)*3=18.85/3.14= 6 pi cm

confidence rating #$&*: 3

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Given Solution:

`aThe circumference of this circle is

C = 2 pi r = 2 pi * 3 cm = 6 pi cm.

This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.8 cm.

Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong.

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Question: `q009. What is the area of a circle whose diameter is exactly 12 meters?

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Your solution:

3.14*6^2=113.1=36 pi m sq

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Given Solution:

`aThe area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is

A = pi ( 6 m )^2 = 36 pi m^2.

This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.

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Question: `q010. What is the area of a circle whose circumference is 14 `pi meters?

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Your solution:

14 pi m/ 2 pi= 7m

Pi*7m^2= 153.93= 49 pi m sq

confidence rating #$&*: 2

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Given Solution:

`aWe know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r.

We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that

r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m.

We use this to find the area

A = pi * (7 m)^2 = 49 pi m^2.

STUDENT QUESTION:

Is the answer not 153.86 because you have multiply 49 and pi????

INSTRUCTOR RESPONSE

49 pi is exact and easier to connect to radius 7 (i.e., 49 is clearly the square of 7) than the number 153.86 (you can't look at that number and see any connection at all to 7).

You can't express the exact result with a decimal. If the radius is considered exact, then only 49 pi is an acceptable solution.

If the radius is considered to be approximate to some degree, then it's perfectly valid to express the result in decimal form, to an appropriate number of significant figures.

153.86 is a fairly accurate approximation.

However it's not as accurate as it might seem, since you used only 3 significant figures in your approximation of pi (you used 3.14). The first three figures in your answer are therefore significant (though you need to round); the .86 in your answer is pretty much meaningless.

If you round the result to 154 then the figures in your answer are significant and meaningful.

Note that a more accurate approximation (though still just an approximation) to 49 pi is 153.93804. An approximation to 5 significant figures is 153.94, not 153.86.

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Question: `q011. What is the radius of circle whose area is 78 square meters?

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Your solution:

A= pi r^2

R= sqrt A/pi

R=5m

confidence rating #$&*: 2

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Given Solution:

`aKnowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ).

Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution.

Now we substitute A = 78 m^2 to obtain

r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{}

Approximating this quantity to 2 significant figures we obtain r = 5.0 m.

STUDENT QUESTION

Why after all the squaring and dividing is the final product just meters and not meters squared????

INSTRUCTOR RESPONSE

It's just the algebra of the units.

sqrt( 78 m^2 / pi) = sqrt(78) * sqrt(m^2) / sqrt(pi). The sqrt(78) / sqrt(pi) comes out about 5.

The sqrt(m^2) comes out m.

This is a good thing, since radius is measured in meters and not square meters.

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Question: `q012. Summary Question 1: How do we visualize the area of a rectangle?

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Your solution:

A=L*W and we can visualize the final units in squares inside the retangle

confidence rating #$&*: 3

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Given Solution:

`aWe visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W.

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Self-critique (if necessary):

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Self-critique Rating:ok

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Question: `q013. Summary Question 2: How do we visualize the area of a right triangle?

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Your solution:

A=L*W*1/2 because a right triangle is half of a square

confidence rating #$&*: 3

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Given Solution:

`aWe visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h.

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Self-critique (if necessary):

I remember it better the way I put it, which both are right, but my letters aren’t the same.

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Self-critique Rating:ok

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Question: `q014. Summary Question 3: How do we calculate the area of a parallelogram?

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Your solution:

A=B*A

confidence rating #$&*:3

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Given Solution:

`aThe area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.

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Self-critique (if necessary):

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Self-critique Rating:ok

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Question: `q015. Summary Question 4: How do we calculate the area of a trapezoid?

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Your solution:

A= average altitude * w

confidence rating #$&*:3

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Given Solution:

`aWe think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.

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Self-critique (if necessary):

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Self-critique Rating:ok

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Question: `q016. Summary Question 5: How do we calculate the area of a circle?

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Your solution:

A=pi r^2

confidence rating #$&*: 3

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Given Solution:

`aWe use the formula A = pi r^2, where r is the radius of the circle.

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Self-critique (if necessary):

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Self-critique Rating:ok

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Question: `q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?

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Your solution:

C=2 pi r

Circumferences are not measured in sq units

confidence rating #$&*: 3

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Given Solution:

`aWe use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units.

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Self-critique (if necessary):

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Self-critique Rating:ok

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Question: `q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

Everything has its own formula, and it was just a refreshment of earlier maths that I had once before.

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Self-critique (if necessary):

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Self-critique Rating:

Jeremiah

&#Good work. Let me know if you have questions. &#

Jeremiah

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

Jeremiah

&#Good work. Let me know if you have questions. &#

Jeremiah

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

Jeremiah

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