Jamie

#$&*

course Mth 173

1/18 6

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.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:

The rate at which you are earning money is $10 per hour. $50/5hr.

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

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Your solution:

There is 12 months per year so your rate per month would be $5000 per month. $60,000/12months.

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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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:

It would be more appropriate to say the business averages 5k per month because business fluctuates.

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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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 have an average rate of 50mph because you are not always going the same rate. People stop for gas, not always going the same speed, etc.

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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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:

If you use 60 gallons of gasoline on a 1200 mile trip then your average use is 20mpg. or .5 gal per mile.

confidence rating #$&*: 3

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Given Solution:

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

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

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

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

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

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

STUDENT COMMENT

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

INSTRUCTOR RESPONSE

There's nothing wrong with your rhythm.

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

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

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

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Your solution:

Because it not always necessary to add various amounts of something to find the average. Average can apply to income per year, interest, mpg, mph a various amount of different things.

confidence rating #$&*: 3

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Given Solution:

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

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

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Your solution:

Since the second group did more daily pushups then they have a greater strength at the end of the year shown by their lifting strength. 15/40 = .375 lbs per day average.

confidence rating #$&*: 3

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Given Solution:

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

STUDENT COMMENT:

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

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

stated.

INSTRUCTOR RESPONSE:

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

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

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

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

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

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Your solution: The average rate increased at .85 stronger because they were able to lift 17lbs more and they had 20 more lbs than the other group.

confidence rating #$&*: 3

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Given Solution:

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

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

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Your solution:

He was going 8.3 m/s during the first segment and 10.0 m/s during the second segment for a total average of 9.09 m/s.

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Given Solution:

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

STUDENT QUESTION

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

INSTRUCTOR RESPONSE

That formula would apply in this specific situation.

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

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

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

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

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Your solution:

10.5 sec because he slows down between the segments and if you average the two together it gives you the best estimate.

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Given Solution:

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

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

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Your solution:

Because we are taking two average quantities and we want to find a total average so we must take these two total quantities and divide them by 2.

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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Given Solution:

STUDENT QUESTION:

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

INSTRUCTOR RESPONSE:

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#(*!

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

Jamie

#$&*

course Mth 173

1/18 6

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

qa areas etc

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:

The area is 12 meters squared.

confidence rating #$&*: 3

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Given Solution:

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

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

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

FREQUENT STUDENT ERRORS

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

4 * 3 = 12

4 * 3 = 12 meters

INSTRUCTOR EXPLANATION OF ERRORS

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

However consider the following:

4 * 3 = 12.

4 * 3 does not equal 12 meters.

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

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

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

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

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Your solution:

The area is equal to 6 m^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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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:

The area is equal to 10m^2.

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Given Solution:

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

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

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

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Your solution:

The area of the triangle is 5cm^2.

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Given Solution:

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

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

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Your solution:

The area of the trapezoid is 20 km^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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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:

The area of the trapezoid is 22cm^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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Question: `q007. What is the area of a circle whose radius is 3.00 cm?

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Your solution:

The area of the circle is 28.27cm^2.

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

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Your solution: The circumference around the circle is 18.8 cm^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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Question: `q009. What is the area of a circle whose diameter is exactly 12 meters?

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Your solution:

The area of this circle is 113.10 m^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. The radius of this circle is half the 12 m diameter, or 6 m. So the area is

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

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

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

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Your solution:

The area of the circle is 49 pi because after squaring the 7 which is half of 7.

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Given Solution:

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

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

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

We use this to find the area

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

STUDENT QUESTION:

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

INSTRUCTOR RESPONSE

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

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

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

153.86 is a fairly accurate approximation.

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

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

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

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

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Your solution:

The radius is 5.0 rounding with sig figs because 78 sq routed and divided by 3.14 is what that equals.

confidence rating #$&*: 3

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Given Solution:

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

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

Now we substitute A = 78 m^2 to obtain

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

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

STUDENT QUESTION

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

INSTRUCTOR RESPONSE

It's just the algebra of the units.

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

The sqrt(m^2) comes out m.

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

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

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Your solution:

We look at the rectangle as a 3d object with space on all four sides and it has area inside of it and in most cases this is what we find.

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

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Your solution:

We visualize the area of a right triangle as one half the base multiplied by the height of the triangle.

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

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Your solution:

By multiplying the base times the height.

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

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Your solution:

We calculate the area of a trapezoid by taking the average of its parallel sides times the width of the trapezoid.

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

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Your solution:

We calculate the area of the circle by taking one half of the distance across the circle = radius and square it then multiply that by pi.

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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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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:

By remembering that circumference is the distance around the circle and area is distance inside. The formula for circumference is 2pi*r

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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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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 organized these principles by memorizing them time and time again throughout the years since grade school.

Self-critique Rating 3

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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 organized these principles by memorizing them time and time again throughout the years since grade school.

Self-critique Rating 3

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

Jamie

#$&*

course Mth 173

1/18 7

002. Volumes*********************************************

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:

The volume of the solid is 35m^2.

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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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:

The volume of the sold is 96m^3. Because we already have 48sq m then we multiply that by the alt which is 2 sq m.

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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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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:

The volume is 800m^3.

confidence rating #$&*:3

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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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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: The volume is 750pi cm^3.

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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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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:

A regular can is about 35 in^3

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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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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:

The volume is 1000 cm ^3 because that is one third of (50sq cm x 60 cm)^2.

confidence rating #$&*:3

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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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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:

One third the area times the altitude is 60 m^3.

confidence rating #$&*:

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

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Your solution:

The volume of a sphere is equal to 4/3 pi r^3, so 4 x 64 is 256 and since this does not come out evenly we leave it as a fraction and divide it by 3 and leave pi on the end of it as well since this is more accurate. 256/3 m^3 pi

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

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Your solution:

The volume is 1,372,000,000,000 / 3 * pi km^3.

confidence rating #$&*:3

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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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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:

We assume the cross section is the same throughout the entire cylinder.

confidence rating #$&*:3

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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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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: We divide them into thirds because they are 3d and we take one third the base times the height.

confidence rating #$&*:3

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

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Your solution:

The formula for a sphere is 4/3 pi r^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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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): These principles have always been the same and I�ve had to memorize them.

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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): These principles have always been the same and I�ve had to memorize them.

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#(*!

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

Jamie

#$&*

course Mth 173

1/19 5

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Question: `q001. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?

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Your solution:

The surface area of a rectangular solid is equal to 2ab+2bc+2ac, therefore we can conclude that the area of this rectangle is 108 meters^2.

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Given Solution:

`aA rectangular solid has six faces (top, bottom, front, back, left side, right side if you're facing it). The pairs top and bottom, right and left sides, and front-back have identical areas. This solid therefore has two faces with each of the following dimensions: 3 m by 4 m, 3 m by 6 m and 4 m by 6 m, areas 12 m^2, 18 m^2 and 24 m^2. Total area is 2 * 12 m^2 + 2 * 18 m^2 + 2 * 24 m^2 = 108 m^2.

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Question: `q002. What is the surface area of the curved sides of a cylinder whose radius is five meters and whose altitude is 12 meters? If the cylinder is closed what is its total surface area?

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Your solution:

The surface area equals the circumference around the cylinder and it�s area of the circle on the end so it would be 2pir*a= 2* pi* 5* 12 = 120pi.

When the cylinder is closed we would add one side so it ends up being 2pir^2* 12 = 170pi.

confidence rating #$&*:

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Given Solution:

The circumference of this cylinder is 2 pi r = 2 pi * 5 m = 10 pi m. If the cylinder was cut by a straight line running up its curved face then unrolled it would form a rectangle whose length and width would be the altitude and the circumference. The area of the curved side is therefore

A = circumference * altitude = 10 pi m * 12 m = 120 pi m^2.

If the cylinder is closed then it has a top and a bottom, each a circle of radius 5 m with resulting area A = pi r^2 = pi * (5 m)^2 = 25 pi m^2. The total area would then be

total area = area of sides + 2 * area of base = 120 pi m^2 + 2 * 25 pi m^2 = 170 pi m^2.

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Question: `q003. What is surface area of a sphere of diameter three cm?

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Your solution:

Surface Area of a Sphere = 4 pi r 2

4*pi*1.5^2 = 9cm^2

confidence rating #$&*: 3

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Given Solution:

`aThe surface area of a sphere of radius r is A = 4 pi r^2. This sphere has radius 3 cm / 2, and therefore has surface area

A = 4 pi r^2 = 4 pi * (3/2 cm)^2 = 9 pi cm^2.

NOTE TO STUDENT:

While your work on most problems has been good, you left this problem blank and didn't self-critique.

You should self-critique here.

� For example you should acknowledge having made note of the formula for the surface area of the sphere, which I expect you didn't know before.

I expect from your previous answers that you are very capable of applying the formula once you have it, and based on this history you probably wouldn't need to self-critique that aspect of the process.

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Question: `q004. What is hypotenuse of a right triangle whose legs are 5 meters and 9 meters?

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Your solution:

A^2+B^2= C^2

Sqrt(106m^2) = 10.30m

confidence rating #$&*:

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Given Solution:

`aThe Pythagorean Theorem says that the hypotenuse c of a right triangle with legs a and b satisfies the equation c^2 = a^2 + b^2. So, since all lengths are positive, we know that

c = sqrt(a^2 + b^2) = sqrt( (5 m)^2 + (9 m)^2 ) = sqrt( 25 m^2 + 81 m^2) = sqrt( 106 m^2 ) = 10.3 m, approx..

Note that this is not what we would get if we made the common error of assuming that sqrt(a^2 + b^2) = a + b; this would tell us that the hypotenuse is 14 m, which is emphatically not so. There is no justification whatsoever for applying a distributive law (like x * ( y + z) = x * y + x * z ) to the square root operator.

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Question: `q005. If the hypotenuse of a right triangle has length 6 meters and one of its legs has length 4 meters what is the length of the other leg?

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Your solution:

Sqrt 20m^2

confidence rating #$&*: 3

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Given Solution:

`aIf c is the hypotenuse and a and b the legs, we know by the Pythagorean Theorem that c^2 = a^2 + b^2, so that a^2 = c^2 - b^2. Knowing the hypotenuse c = 6 m and the side b = 4 m we therefore find the unknown leg:

a = sqrt( c^2 - b^2) = sqrt( (6 m)^2 - (4 m)^2 ) = sqrt(36 m^2 - 16 m^2) = sqrt(20 m^2) = sqrt(20) * sqrt(m^2) = 2 sqrt(5) m,

or approximately 4.4 m.

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Question: `q006. If a rectangular solid made of a uniform, homogeneous material has dimensions 4 cm by 7 cm by 12 cm and if its mass is 700 grams then what is its density in grams per cubic cm?

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Your solution:

Density = m/v

4cm*7cm*12cm = 336cm^3

700grams/336cm^3 = 2.083 grams/cm^3

confidence rating #$&*: 3

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Given Solution:

`aThe volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3.

Its density in grams per cm^3 is the number of grams in each cm^3. We find this quantity by dividing the number of grams by the number of cm^3. We find that

� density = 700 grams / (336 cm^3) = 2.06 grams / cm^3.

Note that the solid was said to be uniform and homogeneous, meaning that it's all made of the same material, which is uniformly distributed. So each cm^3 does indeed have a mass of 2.06 grams.

� Had we not known that the material was uniform and homogeneous we could have said that the average density is 2.06 grams / cm^3, but not that the density is 2.06 grams / cm^3 (for example the object could be made of two separate substances, one with density less than 2.06 grams / cm^3 and the other with density greater than 2.06 g / cm^3, in appropriate proportions; neither substance would have density 2.06 g / cm^3, but the average density could be 2.06 g / cm^3).

NOTE TO STUDENT: (in this note the instructor attempts to clarify the idea of 'demonstrating what you do and do not understand about the statement of the problem' and 'giving a phrase-by-phrase analysis of the given solution')

You did not respond to the question and did not self-critique.

You would be expected to address the question, stating what you do and do not understand.

� For example you should understand what a rectangular solid with dimensions 4 cm by 7 cm by 12 cm is, and how to find its volume and surface area. You might not know what to do with this information (for example you might well not understand that it's the volume and not the surface area that's related to density), but from previous work you should understand this much, and should at least mention something along the lines of 'well, I do know that I can find the volume and/or surface area of that solid' in a partial solution.

� The word 'density' is clearly very important. Even if you don't know what density is, you could note from the statement of the problem that its units here are said to be 'grams per cubic centimeter'.

Having noted these things, you will be much better prepared to understand the information in the given solution.

Then you need to address the information in the given solution. A 'phrase-by-phrase' analysis is generally very beneficial:

� I expect you understand the first statement from previous knowledge (you should have this understanding from prerequisite courses, and if not you encountered it in the preceding 'volumes' exercise): 'The volume of this solid is 4 cm * 7 cm * 12 cm = 336 cm^3.' It would of course be appropriate to ask a question here if necessary.

� It is likely that, as is the case with many students, the concept of density is not that familiar to you. However if this wasn't addressed specifically in prerequisite courses, those courses would be expected to prepare you to understand this concept. The statement 'Its density in grams per cm^3 is the number of grams in each cm^3.' serves as a definition of density. In your self-critique you should have addressed what what this phrase means to you, and what you do or do not understand about it

� The next phrase is 'We find this quantity by dividing the number of grams by the number of cm^3.' You would be expected to understand that this phrase is related to the preceding, and as best you can to address the connection. At this point many students would need to ask a question, and it would be perfectly appropriate to do so (or to have done so regarding previous statements).

� The subsequent phrase 'density = 700 grams / (336 cm^3) = 2.06 grams / cm^3' is an illustration of the ideas and definitions in the preceding statements. A reasonable self-critique would demonstrate your attempt to understand this statement and its connection to the preceding. Once again questions would also be appropriate and welcome.

� The above addresses sufficient information to solve the problem. If you get to this point, you're probably doing OK and you wouldn't necessarily be expected to address the rest of the given solution, which expands on the finer details of the problem and provides additional information. The basic prerequisite courses should have prepared you to understand the information, but students entering Liberal Arts Mathematics, College Algebra and even Precalculus or Applied Calculus (or Physics 121-122) courses probably don't need to address anything beyond the basic solution at this point. Though Precalculus and Applied Calculus students could benefit from doing so, and if time permits would certainly be encouraged to do so, time is also a factor and it would be understandable if these students chose to move on.

� Students entering the Mth 173-4 sequence or the Phy 201-202 or 231-232 sequence would be expected to either completely understand all the details of the given solution, or address them in your self-critique.

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Question: `q007. What is the mass of a sphere of radius 4 meters if its average density is 3,000 kg/cubic meter?

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Your solution:

The volume of a sphere of radius 4 meters is 4/3 pi r^3 so 4/3 pi 4^3 * 3000= 256,000pi kg

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Given Solution:

`aA average density of 3000 kg / cubic meter implies that, at least on the average, every cubic meter has a mass of 3000 kg. So to find the mass of the sphere we multiply the number of cubic meters by 3000 kg.

The volume of a sphere of radius 4 meters is 4/3 pi r^3 = 4/3 * pi (4m)^3 = 256/3 * pi m^3. So the mass of this sphere is

mass = density * volume = 256 / 3 * pi m^3 * 3000 kg / m^3 = 256,000 * pi kg.

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

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Question: `q008. If we build a an object out of two pieces of material, one having a volume of 6 cm^3 at a density of 4 grams per cm^3 and another with a volume of 10 cm^3 at a density of 2 grams per cm^3 then what is the average density of this object?

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Your solution:

First object 24 grams, second object 20 grams = 44grams / total volume which is 16 cm^3 = 2.75 grams per cubic centimeter.

confidence rating #$&*: 3

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Given Solution:

`aThe first piece has a mass of 4 grams / cm^3 * 6 cm^3 = 24 grams. The second has a mass of 2 grams / cm^3 * 10 cm^3 = 20 grams. So the total mass is 24 grams + 20 grams = 44 grams.

The average density of this object is

average density = total mass / total volume = (24 grams + 20 grams) / (6 cm^3 + 10 cm^3) = 44 grams / (16 cm^3) = 2.75 grams / cm^3.

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Question: `q009. In a large box of dimension 2 meters by 3 meters by 5 meters we place 27 cubic meters of sand whose density is 2100 kg/cubic meter, surrounding a total of three cubic meters of cannon balls whose density is 8,000 kg per cubic meter. What is the average density of the material in the box?

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Your solution:

Mass of sand = 27 m^3*2100 = 56,700

Mass of cannonballs = 3m^3 * 8000kg = 24,000

80700/30m^3 = 2690 kg/cubic meter

confidence rating #$&*: 3

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Given Solution:

`aWe find the average density from the total mass and the total volume. The mass of the sand is 27 m^3 * 2100 kg / m^3 = 56,700 kg. The mass of the cannonballs is 3 m^3 * 8,000 kg / m^3 = 24,000 kg.

The average density is therefore

average density = total mass / total volume = (56,700 kg + 24,000 kg) / (27 m^3 + 3 m^3) = 80,700 kg / (30 m^3) = 2,700 kg / m^3, approx..

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Question: `q010. How many cubic meters of oil are there in an oil slick which covers 1,700,000 square meters (between 1/2 and 1 square mile) to an average depth of .015 meters? If the density of the oil is 860 kg/cubic meter the what is the mass of the oil slick?

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Your solution:

Volume of the oil slick is 1,700,000m^2 * .015m = 25,500m^3 * 860kg.m^3 = 21,930,000 kg.

confidence rating #$&*: 3

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Given Solution:

`aThe volume of the slick is V = A * h, where A is the area of the slick and h the thickness. This is the same principle used to find the volume of a cylinder or a rectangular solid. We see that the volume is

V = A * h = 1,700,000 m^2 * .015 m = 25,500 m^3.

The mass of the slick is therefore

mass = density * volume = 860 kg / m^3 * 25,500 m^3 = 21 930 000 kg.

This result should be rounded according to the number of significant figures in the given information.

STUDENT QUESTION

I didn�t round to the most significant figure. ???? How important is this?

INSTRUCTOR RESPONSE

It will be important.

This document is preliminary; the issue of significant figures will be addressed more specifically as we move into the course.

Right now I just want you to be aware of the general idea.

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Question: `q011. Part 1 Summary Question 1: How do we find the surface area of a cylinder?

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Your solution:

Surface Area of a Cylinder = 2 pi r 2 + 2 pi r h

confidence rating #$&*:

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Given Solution:

`aThe curved surface of the cylinder can be 'unrolled' to form a rectangle whose dimensions are equal to the circumference and the altitude of the cylinder, so the curved surface has volume

Acurved = circumference * altitude = 2 pi r * h, where r is the radius and h the altitude.

The top and bottom of the cylinder are both circles of radius r, each with resulting area pi r^2.

{]The total surface area is therefore

Acylinder = 2 pi r h + 2 pi r^2.

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Question: `q012. Part 1 Summary Question 2: What is the formula for the surface area of a sphere?

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Your solution:

Surface Area of a Sphere = 4 pi r 2

confidence rating #$&*: 3

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Given Solution:

`aThe surface area of a sphere is

A = 4 pi r^2,

where r is the radius of the sphere.

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Question: `q013. Part 1 Summary Question 3: What is the meaning of the term 'density'.

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Your solution: density can also be described as the thickness and hardness combined into one word. density of a material is defined as its mass per unit volume.

confidence rating #$&*: 3

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Given Solution:

`aThe average density of an object is its mass per unit of volume, calculated by dividing its total mass by its total volume. If the object is uniform and homogeneous then its density is constant and we can speak of its 'density' as opposed to its 'average density'

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Question: `q014. Part 1 Summary Question 4: If we know average density and mass, how can we find volume?

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Your solution:

Mass = average den*volume

Volume = mass/avg density

confidence rating #$&*: 3

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Given Solution:

Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.

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Given Solution:

Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.

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Given Solution:

Since mass = ave density * volume, it follows by simple algebra that volume = mass / ave density.

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