mrh2276

#$&*

course Mth 158

5/27 8 pm

001. typewriter notation

Note that there are six questions in this exercise. Be sure to continue scrolling down until you get to the end of the exercise.

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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: In the equation with the parentheses, you must do the things inside of them first, but in this equation it does not make any difference. The answer to both of these equations is 0/6=0

confidence rating #$&*: 2

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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): I did the equation without the parentheses in my head and did not follow the order of operations.

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Self-critique Rating: 2

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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: In the equation with the parentheses you must complete that addition before you do the exponent. In the first equation you raise 2 to the 2nd then add 4. The first equation answer is 2^2+4=8. 2^(2+4)=64.

confidence rating #$&*: 2

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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):Ok

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Self-critique Rating: Ok

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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 denominator is the part of the equation that comes after the / symbol. -1/19.

confidence rating #$&*: 2

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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 missed up the problem from the beginning. I had the numerator and the dominator completely wrong and thus just performed the equation that I thought to be in the numerator then the equation that I thought to be in dominator. After reading your student questions and answers, I feel more confident in my responses.

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Self-critique Rating: 0

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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: (4-5)^2*4-1+3/4-2=(4-5)^8-1+3/4-2

(-1) ^8-1+3/4-2

1-1+3/4-2

-1/2+3/4

3/4 -2/4

1/4

confidence rating #$&*: 2

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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): I committed the first error. I was confused by the typewriter notations. I thought that I understood the order of operations, but I obviously need more practice with this. Also from the previous problem I was a little confused on what was the dominator and numerator and tried to carry that over into this problem, but I am afraid I did that wrong as well.

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Self-critique Rating: 0

@& Most of your steps were correct, but be sure you check out the expressions in the links provided and remain aware of the notation conventions.*@

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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:

You are given an equation in the typewriter form, then you are provided with a picture of the equation in standard form.

confidence rating #$&*: 3

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Given Solution: Okay

`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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Self-critique (if necessary):

I did not say the student question in my solution, but I did assume that while looking at the equations. I have a better understanding now that I have read the instructor correction.

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Self-critique Rating: 2

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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:

If this is the way the problems will be expressed on a test, it is very important that I copy the problem correctly. Also if I need to communicate with my instructor I will have to use the computer and if I mess the problem up then it will delay his answer and I will not get the information I need. I may also use a computer algebra program in my future.

confidence rating #$&*:

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002. Describing Graphs

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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 1

-2 3

-1 5

0 7

1 9

2 11

3 13

Crosses the x at 1.5, 0 and y at 0,4

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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Self-critique (if necessary): My x was a little off but I just did a rough sketch.

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Self-critique Rating: 2

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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:

No, it does not change because as the x increase by the same number so do the y values.

confidence rating #$&*: 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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Self-critique (if necessary): okay

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Self-critique Rating: 3

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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:

The slope is 3/1

confidence rating #$&*: 3

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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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Self-critique (if necessary): I believe I have a better understanding of slope now.

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Self-critique Rating: 2

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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, as is the steepness.

confidence rating #$&*: 3

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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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Self-critique (if necessary):

I better under that steepness and rate, but I need to review this concept a little more on my own.

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Self-critique Rating: 2

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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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Self-critique (if necessary): I understood this problem better after the working the previous problem.

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Self-critique Rating: 3

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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 1.4

3 1.7

The graph is increasing at an increasing 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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Self-critique (if necessary): I understand this concept better now and can see why it is decreasing instead of increasing.

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Self-critique Rating: 2

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

X Y

0 5

1 2.5

2 1.25

3 .625

The graph is decreasing at a constant 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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Self-critique (if necessary):

I was not for sure which of these 2 answers where correct, however, after looking at the solution, it does make sense why it would be decreasing versus constant.

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Self-critique Rating: 2

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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:

The graph would be increasing at an increasing rate. As y increases, t becomes faster and faster (increasing more and more).

confidence rating #$&*: 2

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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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Self-critique (if necessary): Ok

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Self-critique Rating: Ok

003. PC1 questions

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Question: `q001 A straight line connects the points (3, 5) and (7, 17), while another straight line continues on from (7, 17) to the point (10, 29). Which line is steeper and on what basis to you claim your result?

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Your solution:

7-3/17-5=3

10-7/29-17=4

The second is steeper because it has a bigger slope.

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`aThe point (3,5) has x coordinate 3 and y coordinate 5. The point (7, 17) has x coordinate 7 and y coordinate 17. To move from (3,5) to (7, 17) we must therefore move 4 units in the x direction and 12 units in the y direction.

Thus between (3,5) and (7,17) the rise is 12 and the run is 4, so the rise/run ratio is 12/4 = 3.

Between (7,10) and (10,29) the rise is also 12 but the run is only 3--same rise for less run, therefore more slope. The rise/run ratio here is 12/3 = 4.

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Self-critique (if necessary): Ok

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Self-critique Rating: Ok

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Question: `q002. The expression (x-2) * (2x+5) is zero when x = 2 and when x = -2.5. Without using a calculator verify this, and explain why these two values of x, and only these two values of x, can make the expression zero.

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Your solution:

When x=2 the (x-2) will equal 0 and anything multiplied by 0 will equal 0.

When x=-2.5 the (2x+5)=0 and when anything is multiplied by 0 the answer is 0.

confidence rating #$&*: 3

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Given Solution:

`aIf x = 2 then x-2 = 2 - 2 = 0, which makes the product (x -2) * (2x + 5) zero.

If x = -2.5 then 2x + 5 = 2 (-2.5) + 5 = -5 + 5 = 0.which makes the product (x -2) * (2x + 5) zero.

The only way to product (x-2)(2x+5) can be zero is if either (x -2) or (2x + 5) is zero.

Note that (x-2)(2x+5) can be expanded using the Distributive Law to get

x(2x+5) - 2(2x+5). Then again using the distributive law we get

2x^2 + 5x - 4x - 10 which simplifies to

2x^2 + x - 10.

However this doesn't help us find the x values which make the expression zero. We are better off to look at the factored form.

STUDENT QUESTION

I think I have the basic understanding of how x=2 and x=-2.5 makes this equation 0

I was looking at the distributive law and I understand the basic distributive property as stated in algebra

a (b + c) = ab + ac and a (b-c) = ab - ac

but I don’t understand the way it is used here

(x-2)(2x+5)

x(2x+5) - 2(2x+5)

2x^2 + 5x - 4x - 10

2x^2 + x - 10.

Would you mind explaining the steps to me?

INSTRUCTOR RESPONSE

The distributive law of multiplication over addition states that

a (b + c) = ab + ac

and also that

(a + b) * c = a c + b c.

So the distributive law has two forms.

In terms of the second form it should be clear that, for example

(x - 2) * c = x * c - 2 * c.

Now if c = 2 x + 5 this reads

(x-2)(2x+5) = x * ( 2 x + 5) - 2 * (2 x + 5).

The rest should be obvious.

We could also have used the first form.

a ( b + c) = ab + ac so, letting a stand for (x - 2), we have

(x-2)(2x+5) = ( x - 2 ) * 2x + (x - 2) * 5.

This will ultimately give the same result as the previous. Either way we end up with 2 x^2 + x - 10.

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Self-critique (if necessary): Ok

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Self-critique Rating: Ok

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Question: `q003. For what x values will the expression (3x - 6) * (x + 4) * (x^2 - 4) be zero?

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Your solution:

(3x - 6) * (x + 4) * (x^2 - 4)

X+2(3-3)(2)(1^2-2)=0

X+2(0)(2)(1-2)=0

X=0

confidence rating #$&*: 1

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`aIn order for the expression to be zero we must have 3x-6 = 0 or x+4=0 or x^2-4=0.

3x-6 = 0 is rearranged to 3x = 6 then to x = 6 / 3 = 2. So when x=2, 3x-6 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

x+4 = 0 gives us x = -4. So when x=-4, x+4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

x^2-4 = 0 is rearranged to x^2 = 4 which has solutions x = + - `sqrt(4) or + - 2. So when x=2 or when x = -2, x^2 - 4 = 0 and the entire product (3x - 6) * (x + 4) * (x^2 - 4) must be zero.

We therefore see that (3x - 6) * (x + 4) * (x^2 - 4) = 0 when x = 2, or -4, or -2. These are the only values of x which can yield zero.**

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Self-critique (if necessary): I understand how to do this problem now. I was very unsure how to do it and so I just factored out the x and common numbers.

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Self-critique Rating: 2

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Question: `q004. 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:

7-3/9-5=1

50-10/4-2=20 this has the bigger area

confidence rating #$&*:2

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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: `q005. Sketch graphs of y = x^2, y = 1/x and y = `sqrt(x) [note: `sqrt(x) means 'the square root of x'] for x > 0. We say that a graph increases if it gets higher as we move toward the right, and if a graph is increasing it has a positive slope. Explain which of the following descriptions is correct for each graph:

As we move from left to right the graph increases as its slope increases.

As we move from left to right the graph decreases as its slope increases.

As we move from left to right the graph increases as its slope decreases.

As we move from left to right the graph decreases as its slope decreases.

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Your solution:

As the line moves from left to right the graph decreases and the slope increases and a decreasing rate for y=1/x

confidence rating #$&*:2

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Given Solution:

`aFor x = 1, 2, 3, 4:

The function y = x^2 takes values 1, 4, 9 and 16, increasing more and more for each unit increase in x. This graph therefore increases, as you say, but at an increasing rate.

The function y = 1/x takes values 1, 1/2, 1/3 and 1/4, with decimal equivalents 1, .5, .33..., and .25. These values are decreasing, but less and less each time. The decreasing values ensure that the slopes are negative. However, the more gradual the decrease the closer the slope is to zero. The slopes are therefore negative numbers which approach zero.

Negative numbers which approach zero are increasing. So the slopes are increasing, and we say that the graph decreases as the slope increases.

We could also say that the graph decreases but by less and less each time. So the graph is decreasing at a decreasing rate.

For y = `sqrt(x) we get approximate values 1, 1.414, 1.732 and 2. This graph increases but at a decreasing rate.

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Self-critique (if necessary): I didn’t read the problem carefully enough and did not finish the problem. But I did go back and complete it and gained the correct answers.

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Self-critique Rating:2

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Question: `q006. If the population of the frogs in your frog pond increased by 10% each month, starting with an initial population of 20 frogs, then how many frogs would you have at the end of each of the first three months (you can count fractional frogs, even if it doesn't appear to you to make sense)? Can you think of a strategy that would allow you to calculate the number of frogs after 300 months (according to this model, which probably wouldn't be valid for that long) without having to do at least 300 calculations?

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Your solution:

(20*.10)=x , x*.10=y, y*.1=z x+y+z= the number of frogs in 3 months which is 26.62. (20*.1)^300= after 300 months.

confidence rating #$&*:3

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Given Solution:

`aAt the end of the first month, the number of frogs in the pond would be (20 * .1) + 20 = 22 frogs. At the end of the second month there would be (22 * .1) + 22 = 24.2 frogs while at the end of the third month there would be (24.2 * .1) + 24.2 = 26.62 frogs.

The key to extending the strategy is to notice that multiplying a number by .1 and adding it to the number is really the same as simply multiplying the number by 1.1. We therefore get

20 * 1.1 = 22 frogs after the first month

22 * 1.1 = 24.2 after the second month

etc., multiplying by for 1.1 each month.

So after 300 months we will have multiplied by 1.1 a total of 300 times. This would give us 20 * 1.1^300, whatever that equals (a calculator, which is appropriate in this situation, will easily do the arithmetic).

A common error is to say that 300 months at 10% per month gives 3,000 percent, so there would be 30 * 20 = 600 frogs after 30 months. That doesn't work because the 10% increase is applied to a greater number of frogs each time. 3000% would just be applied to the initial number, so it doesn't give a big enough answer.

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Self-critique (if necessary):OK

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Self-critique Rating: OK

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Question: `q007. Calculate 1/x for x = 1, .1, .01 and .001. Describe the pattern you obtain. Why do we say that the values of x are approaching zero? What numbers might we use for x to continue approaching zero? What happens to the values of 1/x as we continue to approach zero? What do you think the graph of y = 1/x vs. x looks for x values between 0 and 1?

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Your solution:

As you add a 0 to the denominator, you also add a 0 to your answer. We say that the values are approaching 0 because as the number is decreasing. The graph of y=1/x would be stepper because of the values that we obtained in 1/x whereas the numbers between 0 and 1 are greater than those that are approaching 0.

confidence rating #$&*:2

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Given Solution:

`aIf x = .1, for example, 1 / x = 1 / .1 = 10 (note that .1 goes into 1 ten times, since we can count to 1 by .1, getting.1, .2, .3, .4, ... .9, 10. This makes it clear that it takes ten .1's to make 1.

So if x = .01, 1/x = 100 Ithink again of counting to 1, this time by .01). If x = .001 then 1/x = 1000, etc..

Note also that we cannot find a number which is equal to 1 / 0. Deceive why this is true, try counting to 1 by 0's. You can count as long as you want and you'll ever get anywhere.

The values of 1/x don't just increase, they increase without bound. If we think of x approaching 0 through the values .1, .01, .001, .0001, ..., there is no limit to how big the reciprocals 10, 100, 1000, 10000 etc. can become.

The graph becomes steeper and steeper as it approaches the y axis, continuing to do so without bound but never touching the y axis.

This is what it means to say that the y axis is a vertical asymptote for the graph .

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Self-critique (if necessary):ok

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Self-critique Rating:ok

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Question: `q008. At clock time t the velocity of a certain automobile is v = 3 t + 9. At velocity v its energy of motion is E = 800 v^2. What is the energy of the automobile at clock time t = 5?

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Your solution:

V=3*5+9=24

800(24^2)=460800

confidence rating #$&*:3

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Given Solution:

`aFor t=5, v = 3 t + 9 = (3*5) + 9 = 24. Therefore E = 800 * 24^2 = 460800.

•

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Question: `q009. Continuing the preceding problem, can you give an expression for E in terms of t?

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Your solution:

T=(v-9)/3 E = 800 v^2

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`aSince v = 3 t + 9 the expression would be E = 800 v^2 = 800 ( 3t + 9) ^2. This is the only answer really required here.

For further reference, though, note that this expression could also be expanded by applying the Distributive Law:.

Since (3t + 9 ) ^ 2 = (3 t + 9 ) * ( 3 t + 9 ) = 3t ( 3t + 9 ) + 9 * (3 t + 9) = 9 t^2 + 27 t + 27 t + 81 = 9 t^2 + 54 t + 81, we get

E = 800 ( 9 t^2 + 54 t + 81) = 7200 t^2 + 43320 t + 64800 (check my multiplication because I did that in my head, which isn't always reliable).

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Self-critique (if necessary):ok. I understand the concept but I didn’t understand the question completely.

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Self-critique Rating: 2

001. Rates

Note 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 per hour

confidence rating #$&*: 3

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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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Self-critique (if necessary): okay

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Self-critique Rating: 3

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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 per month

confidence rating #$&*: 3

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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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Self-critique (if necessary): ok

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Self-critique Rating: ok

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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:

The answer is a diffident not an average unless the $60000 is not exactly what the business makes per year.

confidence rating #$&*: 3

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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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Self-critique (if necessary): The answer that was given would make more sense than the answer I had given.

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Self-critique Rating: 2

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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: You do not always cover the same amount of miles per hour. 300/6=50 miles per hour

confidence rating #$&*: 3

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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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Self-critique (if necessary): Ok

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Self-critique Rating: Ok

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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:

1200/60=20 miles per gallon of gas

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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Self-critique (if necessary):

I understand where I messed up and need to read the problem more carefully.

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Self-critique Rating: 3

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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:

We are just averaging rates of one set of data, not the average of several numbers of data.

confidence rating #$&*: 2

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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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Self-critique (if necessary): ok

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Self-critique Rating: ok

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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:

10/147=.068

50/162=.31

.31+.068=.378

.378/2=.189

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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Self-critique (if necessary): I understand what I did wrong.

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Self-critique Rating:2

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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:

( 188-171)=17

17/20=.85

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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Self-critique (if necessary): I did not explain my calculations.

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Self-critique Rating:2

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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:

22-12=10

10/100=.1 second per meter

confidence rating #$&*:3

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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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Self-critique (if necessary):ok

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Self-critique Rating: ok

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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:

10-9=1

½=.5

10-.5=9.5

confidence rating #$&*: 2

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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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Self-critique (if necessary): Ok

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Self-critique Rating: Ok

@& Not bad, but it isn't clear what you mean by this. Could be 9.5 m/s, or 9.5 sec.*@

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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:

We do that now because we have 2 values, not just one.

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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Self-critique (if necessary):

My answer wasn’t as detailed as the solution, but I understood the question.

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Self-critique Rating: ok

001. Areas

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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 m^2

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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Self-critique (if necessary): Okay

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Self-critique Rating: Okay

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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*3*=12 m*

12/2= 6m

confidence rating #$&*: 2

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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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Self-critique (if necessary): need to put m^2

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Self-critique Rating: 3

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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^2

confidence rating #$&*: 3

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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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Self-critique (if necessary):

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Self-critique Rating: Ok

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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:

5*2=10cm^2

10/2=5cm^2

confidence rating #$&*: 3

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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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Self-critique (if necessary):

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Self-critique Rating: Ok

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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*5=20km^2

confidence rating #$&*:2

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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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Self-critique (if necessary): ok

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Self-critique Rating: ok

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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*3*8=96 cm^2

confidence rating #$&*: 2

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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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Self-critique (if necessary): I understand what I did wrong.

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Self-critique Rating: 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:

3^2=9cm^2

confidence rating #$&*: 3

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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 left out the pi out.

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Self-critique Rating: 3

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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:

3*2pi= 9pi cm^2

confidence rating #$&*:2

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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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Self-critique (if necessary): ok

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Self-critique Rating: ok

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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:

12/2=6

6^2 pi= 36 pi meters^2

confidence rating #$&*:2

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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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Self-critique (if necessary): ok

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Self-critique Rating: ok

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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:

14/2=7

7 pi meters^2

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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Self-critique (if necessary): I did not complete the problem by squaring the 7

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Self-critique Rating: 2

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

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Your solution:

(Sqrt78)*2

confidence rating #$&*: 1

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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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Self-critique (if necessary):

I understand what I did wrong and the concepts behind it better now.

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Self-critique Rating: 2

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

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Your solution:

Two triangles together equal one square. The area of a square is width times length. So the area of a triangle will be half the area of a square.

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): I can understand that too.

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Self-critique Rating: 3

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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 right triangle has one angle that is 90 degrees so we know for a fact that it is half of a square. So the area is 1/2b*h.

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): Ok

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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:

B*h

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): The altitude is the same as the height.

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Self-critique Rating: 3

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

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Your solution:

Width*altitude.

confidence rating #$&*:2

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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):

I didn’t say the average.

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Self-critique Rating:2

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

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Your solution:

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): ok

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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

A= pi r^2

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):

No square roots in the C.

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Self-critique Rating: 2

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

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Self-critique (if necessary):

I have been able to connect different shapes and tell the difference between areas and C.

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Self-critique Rating: 2

002. Volumes

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Question: `q001. There are 9 questions and 4 summary questions in this assignment.

What is the volume of a rectangular solid whose dimensions are exactly 3 meters by 5 meters by 7 meters?

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Your solution:

3*5*7=105 meter^3

confidence rating #$&*: 3

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Given Solution:

`aIf we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2.

Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3.

The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3.

This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore

V = A * h,

where A is the area of the base and h the altitude.

This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.

STUDENT QUESTION

I guess I am confused at what the length and the width are???? I drew a rectangle I made the top length 5

and the bottom lenghth 7 then the side 3. So the 7 and the 5 are both width and the 3 is the height??????

INSTRUCTOR RESPONSE

You can orient this object in any way you choose. The given solution orients it so that the base is 5 cm by 7 cm. The area of the base is then 35 cm^2. In this case the third dimension, 3 cm, is the height and we multiply the area of the base by the height to get 105 cm^3.

Had we oriented the object so that it rests on the 3 cm by 5 cm rectangle, the area of the base would be 15 cm^2. The height would be the remaining dimension, 7 cm. Multiplying the base by the height we would be 15 cm^2 * 7 cm = 105 cm^3.

We could also orient the object so its base is 3 cm by 7 cm, with area 21 cm^2. Multiplying by the 5 cm height we would again conclude that the volume is 105 cm^3.

All these results can be visualized in terms of 1-cm squares and 1-cm cubes, as explained in the given solution.

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Self-critique (if necessary): okay

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Self-critique Rating: Okay

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Question: `q002. What is the volume of a rectangular solid whose base area is 48 square meters and whose altitude is 2 meters?

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Your solution:

48*2=96 meters ^3

confidence rating #$&*: 3

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Given Solution:

`aUsing the idea that V = A * h we find that the volume of this solid is

V = A * h = 48 m^2 * 2 m = 96 m^3.

Note that m * m^2 means m * (m * m) = m * m * m = m^2.

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Self-critique (if necessary): ok

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Self-critique Rating: ok

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Question: `q003. What is the volume of a uniform cylinder whose base area is 20 square meters and whose altitude is 40 meters?

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Your solution:

20*40= 800 meters^2

confidence rating #$&*: 2

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Given Solution:

`aV = A * h applies to uniform cylinders as well as to rectangular solids. We are given the altitude h and the base area A so we conclude that

V = A * h = 20 m^2 * 40 m = 800 m^3.

The relationship V = A * h applies to any solid object whose cross-sectional area A is constant. This is the case for uniform cylinders and uniform prisms.

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Self-critique (if necessary): it is ^3 not ^2 I understand why it is third and not squared.

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Self-critique Rating: 2

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Question: `q004. What is the volume of a uniform cylinder whose base has radius 5 cm and whose altitude is 30 cm?

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Your solution:

pi 5^2=25

25 pi*30=750 pi cm^3

confidence rating #$&*:2

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Given Solution:

`aThe cylinder is uniform, which means that its cross-sectional area is constant. So the relationship V = A * h applies.

The cross-sectional area A is the area of a circle of radius 5 cm, so we see that A = pi r^2 = pi ( 5 cm)^2 = 25 pi cm^2.

Since the altitude is 30 cm the volume is therefore

V = A * h = 25 pi cm^2 * 30 cm = 750 pi cm^3.

Note that the common formula for the volume of a uniform cylinder is V = pi r^2 h. However this is just an instance of the formula V = A * h, since the cross-sectional area A of the uniform cylinder is pi r^2. Rather than having to carry around the formula V = pi r^2 h, it's more efficient to remember V = A * h and to apply the well-known formula A = pi r^2 for the area of a circle.

STUDENT QUESTION

why do we not calculate the pi times the radius and then the height or calculate the pi after the height

why do we just leave the pi in the answer?

INSTRUCTOR RESPONSE

pi cannot be written exactly in decimal form; it's an irrational number and any decimal representation is going to have round-off error.

750 pi cm^3 is the exact volume of a cylinder with radius 5 cm and altitude 30 cm.

750 pi is approximately 2356. However 2356 has two drawbacks:

• 2356 is a 4-significant-figure approximation of 750 pi. It's not exact. This might or might not be a disadvantage, but we're better off expressing the result as a multiple of pi, which we can then calculate to any desired degree of precision, than in using 2356, which already contains a roundoff error.

• It's hard to look at 2356 and see how it's related to 5 and 30. You probably can't calculate that in your head. However it's not difficult to see that 30 * 5^2 is 30 * 25 or 750.

When in doubt, we use the exact expression rather than the approximation. It's fine to give an answer like the following:

The volume is 750 pi cm^3, which is approximately 2356 cm^3.

STUDENT QUESTION

I should have stated that my answer was an approximate. ???? When using pi, should I calculate this out or just leave pi in the solution?

INSTRUCTOR RESPONSE

I would say to do both when in doubt.

If the given dimensions are known to be approximate, and when the numbers aren't simple in the first place, it's appropriate to just multiply everything out and use an appropriate number of significant figures.

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Self-critique (if necessary): ok

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Self-critique Rating: ok

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Question: `q005. Estimate the dimensions of a metal can containing food. What is its volume, as indicated by your estimates?

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Your solution:

V=A*h

confidence rating #$&*:1

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Given Solution:

`aPeople will commonly estimate the dimensions of a can of food in centimeters or in inches, though other units of measure are possible (e.g., millimeters, feet, meters, miles, km). Different cans have different dimensions, and your estimate will depend a lot on what can you are using.

A typical can might have a circular cross-section with diameter 3 inches and altitude 5 inches. This can would have volume V = A * h, where A is the area of the cross-section. The diameter of the cross-section is 3 inches so its radius will be 3/2 in.. The cross-sectional area is therefore A = pi r^2 = pi * (3/2 in)^2 = 9 pi / 4 in^2 and its volume is

V = A * h = (9 pi / 4) in^2 * 5 in = 45 pi / 4 in^3.

Approximating, this comes out to around 35 in^3.

Another can around the same size might have diameter 8 cm and height 14 cm, giving it cross-sectional area A = pi ( 4 cm)^2 = 16 pi cm^2 and volume V = A * h = 16 pi cm^2 * 14 cm = 224 pi cm^3.

STUDENT QUESTION

Should my in^3 come after the total solution even though it is associated with the 9? As in your example the in^3 is

associated with 224 but you have it at the end of the solution.

INSTRUCTOR RESPONSE

I wouldn't be picky at this point of the course, but the generally used order has the numbers first and the units last.

This is what most readers will expect. It's a lot like using good grammar, which makes everything easier to understand.

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Self-critique (if necessary): ok

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Self-critique Rating: 2

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Question: `q006. What is the volume of a pyramid whose base area is 50 square cm and whose altitude is 60 cm?

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Your solution:

50*60=300 cm^3

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`aWe can't use the V = A * h idea for a pyramid because the thing doesn't have a constant cross-sectional area--from base to apex the cross-sections get smaller and smaller. It turns out that there is a way to cut up and reassemble a pyramid to show that its volume is exactly 1/3 that of a rectangular solid with base area A and altitude h. Think of putting the pyramid in a box having the same altitude as the pyramid, with the base of the pyramid just covering the bottom of the box. The apex (the point) of the pyramid will just touch the top of the box. The pyramid occupies exactly 1/3 the volume of that box.

So the volume of the pyramid is V = 1/3 * A * h. The base area A is 30 cm^2 and the altitude is 60 cm so we have

V = 1/3 * 50 cm^2 * 60 cm = 1000 cm^3.

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Self-critique (if necessary):I understand why you need to multiply by 1/3

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Self-critique Rating: 3

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Question: `q007. What is the volume of a cone whose base area is 20 square meters and whose altitude is 9 meters?

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Your solution:

I am not sure how to find the volume of a cone but the units would be cubed.

confidence rating #$&*: 0

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Given Solution:

`aJust as the volume of a pyramid is 1/3 the volume of the 'box' that contains it, the volume of a cone is 1/3 the volume of the cylinder that contains it. Specifically, the cylinder that contains the cone has the base of the cone as its base and matches the altitude of the cone. So the volume of the cone is 1/3 A * h, where A is the area of the base and h is the altitude of the cone.

In this case the base area and altitude are given, so the volume of the cone is

V = 1/3 A * h = 1/3 * 20 m^2 * 9 m = 60 m^3.

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Self-critique (if necessary): After reading the given solution, I understand how to do the area.

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Self-critique Rating: 3

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Question: `q008. What is a volume of a sphere whose radius is 4 meters?

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Your solution:

Pi 4^2*4= 64 meters ^3

confidence rating #$&*: 2

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Given Solution:

`aThe volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere. In this case r = 4 m so

V = 4/3 pi * (4 m)^3 = 4/3 pi * 4^3 m^3 = 256/3 pi m^3.

STUDENT QUESTION:

How does a formula come up with multiplying by pi? I understand how to work a formula, but don’t know how to

calculate the formula. Does that make sense?

INSTRUCTOR RESPONSE: It makes perfect sense to ask that question.

However the answer is beyond the scope of your course.

(one answer, which will not make sense to anyone until at least the midway point of their third semester of a challenging calculus sequence, is that the volume of a sphere of radius R is the integral of rho^2 sin (phi) cos(theta) from rho = 0 to R, phi from 0 to pi and theta from 0 to 2 pi; also the surface area of a sphere of radius R is double the double integral of r / secant(theta), integrated in polar coordinates from r = 0 to R and theta from 0 to 2 pi) .

(there is another way of figuring this out using solid geometry, a topic with which few students are familiar).

In other words, at this point your best recourse is to just learn the formulas.

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Self-critique (if necessary):

That was my exact question. I understand sooo much better now.

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Self-critique Rating: 3

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Question: `q009. What is the volume of a planet whose diameter is 14,000 km?

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Your solution:

14,000/2=7,000

¾ pi 7,000^2

confidence rating #$&*:2

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Given Solution:

`aThe planet is presumably a sphere, so to the extent that this is so the volume of this planet is V = 4/3 pi r^3, where r is the radius of the planet. The diameter of the planet is 14,000 km so the radius is half this, or 7,000 km. It follows that the volume of the planet is

V = 4/3 pi r^3 = 4/3 pi * (7,000 km)^3 = 4/3 pi * 343,000,000,000 km^3 = 1,372,000,000,000 / 3 * pi km^3.

This result can be approximated to an appropriate number of significant figures.

STUDENT QUESTION

How did we go from 343,000,000,000 to 1,372,000,000,000?

INSTRUCTOR RESPONSE

We go from 4/3 pi * 343,000,000,000 to 1,372,000,000,000 / 3 * pi by multiplying 343 000 000 000 by 4. Like a lot of thing, this is fairly obvious once you see it, hard to see until you do.

Let me know if after thinking about it for a few minutes, then if necessary giving it a rest for awhile (say, a day) and coming back to it, you don't see it.

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Self-critique (if necessary): I love these students questions because they are exactly what I do not understand.

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Self-critique Rating: 3

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Question: `q010. Summary Question 1: What basic principle do we apply to find the volume of a uniform cylinder of known dimensions?

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Your solution:

V=A*h

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`aThe principle is that when the cross-section of an object is constant, its volume is V = A * h, where A is the cross-sectional area and h the altitude. Altitude is measure perpendicular to the cross-section.

STUDENT QUESTION

What does it mean “when the cross-section of an object is constant”? When would it not be

constant?

INSTRUCTOR RESPONSE

For example the cross-sectional area of a cone, which tapers, is not constant; nor is the cross-sectional area of a sphere.

STUDENT QUESTION

And why is altitude measured perpendicular to the cross-section?

INSTRUCTOR RESPONSE

This is for essentially the same reason the altitude of a parallelogram is measured perpendicular to its base.

If you imagine nailing four sticks together to make a rectangle, then imagine partially 'collapsing' the rectangle into a parallelogram, you will see that the altitude of the resulting parallelogram is less than that of the original rectangle, and its area is correspondingly less.

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Self-critique (if necessary): I understand

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Self-critique Rating: 3

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Question: `q011. Summary Question 2: What basic principle do we apply to find the volume of a pyramid or a cone?

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Your solution:

1/3 A*h

confidence rating #$&*: 2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`aThe volumes of these solids are each 1/3 the volume of the enclosing figure. Each volume can be expressed as V = 1/3 A * h, where A is the area of the base and h the altitude as measured perpendicular to the base.

STUDENT QUESTION

I thought I had the right idea but I got lost. I’m not sure how to handle the square roots,

even after reading the solution, I am confused about this one.

INSTRUCTOR RESPONSE

Think of a simple example, the equation x^2 = 25.

It should be clear that x = 5 is a solution to this equation, as is x = -5.

Now 5 is the square root of 25, since 25 is the square of 5. In notation, the same sentence would read

5 = sqrt(25) since 25 = 5^2.

So the solutions to this equation are x = sqrt(25) and x = -sqrt(25). We often write that as x = +- sqrt(25), where the '+-' means 'plus or minus'.

More generally, if c is any positive number, the equation x^2 = c has solutions x = +- sqrt(c).

Now sometimes only one of the two solutions makes sense.

In the present problem A radius is a distance, and a distance can't be negative. So after finding the two solutions, we discard the negative solution. However we always find both solutions before discarding everything, in order to make sure we don't throw out something important

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Self-critique (if necessary): ok

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Self-critique Rating: ok

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Question: `q012. Summary Question 3: What is the formula for the volume of a sphere?

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Your solution:

4/3 pi r^3

confidence rating #$&*:3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`aThe volume of a sphere is V = 4/3 pi r^3, where r is the radius of the sphere.

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Self-critique (if necessary):

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Self-critique Rating: ok

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Question: `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

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Self-critique (if necessary):

I now understand not just the formulas but I understand why the formulas apply to each shape.

Self-critique (if necessary):

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Self-critique rating:

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Question: `q013. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

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Self-critique (if necessary):

I now understand not just the formulas but I understand why the formulas apply to each shape.

Self-critique (if necessary):

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Self-critique rating:

#*&!

@& On the whole your work looks good. I've inserted a few comments.

It's usually best to submit a different document for each assignment. No problem here, but try to do so in future submissions. By submitting separate documents you will tend to get better feedback and more specific feedback on your work (again no problem in this case).*@