Rates

#$&*

course Mth 163

Submitted 1/25 9:45am

Copy and paste this document into a text editor, insert your responses and submit using the Submit_Work_Form.

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

.

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.

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

50 per hour, at 5 hours. segment, or divide, 50 by 5 to determine the hourly rate.

50/5 = 10

$10 per hour.

confidence rating #$&*:

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

3

.............................................

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

*********************************************

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

60,000 per year. There are 12 months per year. We need to divide 60,000 into 12 equal units to determine our monthly pay.

60,000/12 = 5000

$5000 / month

confidence rating #$&*:

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

3

.............................................

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.

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

It would be more appropriate to say that the business makes an average of 5000 per month. Some months business might be great, others.. not so much. (income flucutuates from month to month)

confidence rating #$&*:

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

3

.............................................

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.

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

300 miles in 6 hours = 300/6 = 50

50 miles per hour. We call this average because speed varies based on road conditions, speed limits, unforeseen circumstances, etc.

confidence rating #$&*:

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

3

.............................................

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.

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

1200 miles divided up into 60 equal units will tell us about how many miles 1 gallon of gasoline allowed us to travel.

1200/60 = 20 miles per gallon.

we averaged 20 miles per gallon of gasoline. Road conditions, incline, gears, and other factors will affect vehicle fuel efficiency

CORRECTED AFTER REVIEW:

My above answer is correct for MILES / GALLONS. However this question asks for GALLONS / MILES.

60 gallons were used over 1200 miles.

60/1200 = .05 gallons / mile

This means that we used .05 gallons of gasoline for every mile we traveled.

confidence rating #$&*:

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

3

.............................................

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.

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.

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I placed the order in the most common order, not the requested. This asked for gallons per mile, not miles per gallon.

------------------------------------------------

Self-critique Rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Part of the reason is because our solutions are affected by things that can’t be calculated in our equations at this point. Things like speed limits, or fuel efficiency circumstances, for example.

As far as adding things, our problems thus far have already had the necessary values added up: total miles, etc. We simply have been asked to divide that total into varying amounts.

confidence rating #$&*:

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

3

.............................................

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.

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

SOLUTION REVIEWED BELOW IN CRITIQUE SECTION. I DID NOT UNDERSTAND THE PROCESS INITIALLY.

confidence rating #$&*:

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

0

.............................................

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.

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.

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

My initial approach to solving wanted to know how much lifting strength each group started with in the beginning. However, on further review ""equally matched"" implies that they would most likely start at exactly the same strenght.

With that in mind, you can then see that group two increased their lifting strength by 15 more pounds than group 1. Therefore, over the course of the year, the 40 extra pushups pre day increased their strength beyond group 1.

(strength increase is equal up to 10 pushups for the two groups). What we are really analyzing is the effect of those 40 pushups. 40 pushups have given the group 15lbs of extra lifting strength. The question asks for Strength

affected by pushups: 40 / 15 = .375.

NOTE: We are looking at change in strength / change in pushups ( ???? similar to slope?), Which ultimately tells us how each of the 40 pushups increased our strength. We CANNOT take into account the initial 10 pushups, because we don't know the lifting

strength that each group started with. (however we can figure it out now that we have our answer).

@&

Very good insights.

*@

????? I definitely feel as if I am on the right track to understanding, however: I'm having difficulty wrapping my head around exactly why we only divide 15 by 40. The 15 pounds was gained over an entire years worth of pushups, not

just 40 of them. It ALMOST makes sense because we have (change in strength over the year) / (Change in pushups every day over the course of the year), but something is missing for me to make complete sense of it.

@&

The question was phrased in terms of 'daily pushups'.

Over the course of the year, each extra pushup you do per day increases your final strength a bit.

This result is a measure of how much you would increase your strength, by the end of the year, by adding a pushup per day.

*@

------------------------------------------------

Self-critique Rating:

2.5

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(Please pardon any excess commentary, as typing my thoughts helps me to make sense of the concept).

Again as before, each group here we can assume starts with equal strength. Therefore we are once again measuring the change strength between the two groups. The question asks for strength affected by weight:

(Change in strength) / (change in weight).

(188 - 171) / (30 - 10)

17 / 20 : 17 extra pounds of strength was attained by group 2, which resulted from having 20 extra pounds of weight on their shoulders during the pushups.

0.85 pounds of strength increase per added pound of weight.

confidence rating #$&*:

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

2.5

.............................................

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.

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I drew a quick line secment reflecting the two points, as well as a full 2-axis graph with distance on the y axis and time on the x. This helps to see that this is a rate of change problem. As with the two previous problems,

I am now seeing that these rates of change have all been constant rates thus far.

We need to determine the change of distance over time:

The runner covered 100 meters in 10 seconds

100/10

10 meters per second.

Unless we want to look up information and calculate mph, then this is as far as we can answer this question with the given information.

confidence rating #$&*:

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

3

.............................................

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.

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.

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

We know once again that we are measuring distance / time. In this case the runner has traveled 100 meters. Over the course of that distance, they have dropped in speed 1m/sec to 9m/sec.

Their average speed (given that we only have two speed references, this is easy to calculate) can best be estimated as 9.5m/sec. While only our best guess, this will give us a pretty good ""middle of the road"" estimate as to

the total time it took the runner to cover this distance.

Distance / Time

100 meters / 9.5 meters per second

10.52 seconds

confidence rating #$&*:

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

3

.............................................

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.

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

*********************************************

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

We add and divide by 2 in this case, because we have a situation in which we cannot assume the extraneous factors that affect the rate. For example in the example regarding 5000 / month average for a business over the year,

it is pretty easy to see/understand that business will have good months and bad months. In this example however, while the runners speed could reasonably have dropped off early, late, or at a constant rate, it is not nearly as easy

to determine an average in this circumstance, without simply adding up and dividing.

confidence rating #$&*:

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

3

.............................................

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.

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.

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.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

The running question gave us the rates with which we needed to work. (we used the rates to determine time over distance). In prior examples however, we were given exact numbers for points in time (benchmarks), distance, etc. We

DETERMINED the rate in those problems. In this problem, we used the rate to determine one of those ""benchmarks"". In other words, we worked backwards.

------------------------------------------------

Self-critique Rating:

*********************************************

Question: `q013. The volume of water in a container increases from 1400 cm^3 to 1600 cm^3 as the depth of the water in the container changes from 10 cm to 14 cm. At what average rate was the volume changing with respect to depth?

Optional question: What does this rate tell us about the container?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

This is asking for the rate of change for volume as affected by depth

(1600cm^3 - 1400cm^3) / (14cm - 10cm)

200cm^3 / 4cm This part of the equation tells us that for every 4cm of water in the container, the volume increases by 200cm^3

50cm^3 per 1cm increase in water

@&

Good.

*@

This solution tells us that for every 1cm increase in the container, that our volume increases by 50cm^2.

@&

The rate is 50 cm^3 / cm, or 50 cm^2.

If you think about it, this tells you that the cross-sectional area of the container is 50 cm^2.

*@

At first, I reversed this equation, and solved as follows:

(14cm - 10cm) - (1600cm^3 - 1400cm^3)

4cm / 200cm^3 This part tells us that for every 200cm^3 increase in the volume of the container, that the depth increases 4cm

.02cm per 1cm^3 increase in volume This part tells us that the water depth increases .02cm for every 1cm^3 increase in the volume of the container.

This however does not answer the question as proposed. The question asks ""at what average rate was the VOLUME changing with respect to depth"", meaning our answer should be in terms of volume per depth, NOT depth per volume.

confidence rating #$&*:

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

3

*********************************************

Question: `q014. An athlete's rate of doing work increases more or less steadily from 340 Joules / second to 420 Joules / second during a 6-minute event. How many Joules of work did she do during this time?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

This problem is very similar to our last runners problem. We are given 2 rates, and are asked to find the total amount of something, using those two rates. This time however, we are told that the increase in the rate happens

steadily over the given time period.

Our change in joules = 420 - 340 = 80 joules. Our athlete increased work by 80 joules/sec over 6 minutes

Now we need to determine the increase in joules per minute: 80joules / 6 minutes = 13.33 joules per minute.

At this point we now know that over the 6 minute time frame:

-The athlete increased work output by 80 joules.

-Joules increased at a rate of 13.33 joules/minute

Now we can use that to determine joules for each time block of 1 minute:

Minute 0-1: (340 Joules * 60 seconds) = 20400 Joules. (NOTE: we do not take into account the 13.33 increase yet, as it would have been rising to this point throughout the first minute)

Minute 1-2: (340+13.33) * 60 seconds = 21199.8 Joules

Minute 2-3: (353.33+13.33) * 60 sec = 21999.6 Joules

Minute 3-4: (366.66+13.33) *60 sec = 22799.4 Joules

Minute 4-5: (379.99+13.33) * 60 sec = 23599.2 Joules

Minute 5-6: (393.32+13.33) * 60 sec = 24399 Joules

(Self-Note: The equation above is essentially solving for Y: y= (340 + 13.33x) * 60 Where y=joules and x=minute block )

Now we must add the number of joules worked per second (added from the solutions to each minute block above)

Total number of joules worked over 6 minutes: 134397 Joules

@&

This is well done, and your thinking is excellent.

That doesn't mean it's completely right.

If you average the initial and final rates you get the average rate, which is 380 Joules / second. So in the 360-second interval the work done is 380 J/s * 360 s = 136 800 Joules.

Your numbers will add up to something close to this, but not exactly the same. One reason is roundoff error, and its accumulation from one step to the next. The other is that during each minute the rate goes up, so the average rate for the 1-minute interval is greater than the rate at the beginning of the interval, which is what you used.

You in fact note this below, and you are right on target with your note.

*@

NOTE: This is, once again, an average. In this solution I did not determine the average increase of joules by more than minute to minute increases. The further we break this down, the more exact a number we would receive. For example we COULD break this down

and determine the number of joules increased per second, and then get a more precise answer, because we are taking into account the increase at more reference points in time. In my answer, I have given 6 reference points, and

calculated the work for that minute ALL at that reference points workload.

confidence rating #$&*:

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

2

------------------------------------------------

Self-critique Rating:

I feel good about my solution, however I am not positive exactly how far to break down the increase in joules (per minute, per 30 sec intervals, per second, etc...). You specifically asked for ""how many joules worked"", you did not

ask for an average, therefore I am not totally confident that my answer will be appropriate. I do however feel that my process and logic is on track. Please comment as necessary regarding my concern.

????If you would have preferred this to be broken down into the increase in joules per second, then I am not sure how to set up an equation to take that type of growth into account. I would have to set it up as I did above, and calculate

each second individually, then add them all up."

@&

Calculus offers you a way to effectively break this down instant by instant. The insights you have demonstrated here will serve you very well when you begin a calculus course.

Your thinking throughout is excellent. Keep up the good work.

*@