Calculus Initial Questions

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course Mth 272

6/7 1:00

Question: `q001. There are 12 questions in this document.

The graph of a certain function is a smooth curve passing through the points (3, 5), (7, 17) and (10,

29).

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

Why do we say 'on the average'?

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Your solution:

The slope between the first two coordinates is (17-5)/(7-3) which is 12/4=3, and the slope between the

second and third coordinates is (29-17)/(10-7)=12/3=4. On average, the second set of points is steep

because it has a greater slope. We say ""on average"" because the slope changes and we are only

evaluating this portion of the graph.

confidence rating #$&*: 3

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Given Solution:

`aSlope = rise / run.

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

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

The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to

point. The slope obtained over the interval is a specific type of average of the slopes of all points

between the endpoints.

2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values

are taken by the expression 1 / (x - 2)?

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

(x-2)?

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

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

4. Is there any number it will never exceed?

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

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Your solution:

1. The values will continue to approach but never reach that limit. When you have a function with 1 as

the number divided by something else, that is a good clue that it is the type of graph that will have a

limit.

2. The values will infinitely approach 2 and 0, no matter no close to those values that may be.

3. It could since there is no such thing as the ""smallest number"".

4. No, it will just continue to approach 0 and 2, but it will never reach those numbers.

5. x=2 is the vertical asmyptote

confidence rating #$&*:3

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Given Solution:

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

respective values 10, 100, 1000, 10,000.

It is important to note that x is changing by smaller and smaller increments as it approaches 2, while

the value of the function is changing by greater and greater amounts.

As x gets closer in closer to 2, it will reach the values 2.00001, 2.0000001, etc.. Since we can put as

many zeros as we want in .000...001 the reciprocal 100...000 can be as large as we desire. Given any

number, we can exceed it.

Note that the function is simply not defined for x = 2. We cannot divide 1 by 0 (try counting to 1 by

0's..You never get anywhere. It can't be done. You can count to 1 by .1's--.1, .2, .3, ..., .9, 1. You

get 10. You can do similar thing for .01, .001, etc., but you just can't do it for 0).

As x approaches 2 the graph approaches the vertical line x = 2; the graph itself is never vertical.

That is, the graph will have a vertical asymptote at the line x = 2. As x approaches 2, therefore, 1 /

(x-2) will exceed all bounds.

Note that if x approaches 2 through the values 1.9, 1.99, ..., the function gives us -10, -100, etc..

So we can see that on one side of x = 2 the graph will approach +infinity, on the other it will be

negative and approach -infinity.

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Self-critique (if necessary): The difference between my answer and the given solution is that I did

mention that there is also the values the approach 2 from the left, but will never reach 2.

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Self-critique Rating:3

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Question: `q003. One straight line segment connects the points (3,5) and (7,9) while another connects

the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the

x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with

something more precise than, for example, 'from a sketch I can see that this one is much bigger so it

must have the greater area'.

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Your solution:

The distance formula (x1 + x2)/2 and (y1 + y2) /2 tells us that the distance between the first two

points is (5, 7) and the distance between the second two points is (30, 3). The trapazoid with the

greater dimensions will have a greater area, so we know that the second set of points will have a

trapezoid with a greater area.

confidence rating #$&*: OK

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Given Solution:

`aYour sketch should show that while the first trapezoid averages a little more than double the

altitude of the second, the second is clearly much more than twice as wide and hence has the greater

area.

To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4

units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The

altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average

altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over

twice as high, on the average, as the first. However the second is 10 times as wide, so the second

trapezoid must have the greater area.

This is all the reasoning we need to answer the question. We could of course multiply average altitude

by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second.

However if all we need to know is which trapezoid has a greater area, we need not bother with this

step.

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Self-critique (if necessary): OK

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Self-critique Rating:OK

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Question: `q004. If f(x) = x^2 (meaning 'x raised to the power 2') then which is steeper, the line

segment connecting the x = 2 and x = 5 points on the graph of f(x), or the line segment connecting the

x = -1 and x = 7 points on the same graph? Explain the basis of your reasoning.

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Your solution:

For x=-1 and x=7, the points are (-1,1) and (7,49). The slope of the line connecting those points is

48/8=6.

For x=2 and x=5, the points are (2,4) and (5,25). The slope of the line connecting those points is

21/3=7. Therefore, the segment connecting 2 and 5 as a steeper slope because it has a greater rise/run.

confidence rating #$&*: 3

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Given Solution:

`aThe line segment connecting x = 2 and the x = 5 points is steeper: Since f(x) = x^2, x = 2 gives y =

4 and x = 5 gives y = 25. The slope between the points is rise / run = (25 - 4) / (5 - 2) = 21 / 3 = 7.

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

/ (7 - -1) = 48 / 8 = 6.

The slope of the first segment is greater.

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Self-critique (if necessary): Ok

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Self-critique Rating: OK

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Question: `q005. Suppose that every week of the current millennium you go to the jeweler and obtain a

certain number of grams of pure gold, which you then place in an old sock and bury in your backyard.

Assume that buried gold lasts a long, long time ( this is so), that the the gold remains undisturbed

(maybe, maybe not so), that no other source adds gold to your backyard (probably so), and that there

was no gold in your yard before..

1. If you construct a graph of y = the number of grams of gold in your backyard vs. t = the number of

weeks since Jan. 1, 2000, with the y axis pointing up and the t axis pointing to the right, will the

points on your graph lie on a level straight line, a rising straight line, a falling straight line, a

line which rises faster and faster, a line which rises but more and more slowly, a line which falls

faster and faster, or a line which falls but more and more slowly?

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

week.

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

week.

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Your solution:

1. If the gold doesn't decay or anything, then the graph would be a straight line. For example, y=3

grams of gold.

@&

The amount of gold increases. y = 3 is a horizontal line, not an increasing line.

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2. This graph would appear to be a line which rises faster and faster.The line would not be straight

because the slope is increasing each week.

3. The graph would appear as a decreasing line that falls more and more slowly.

@&

You continue adding gold so the amount would not decrease.

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confidence rating #$&*:OK

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Given Solution:

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

2. Buying gold every week, the amount of gold will always increase. Since you buy more each week the

rate of increase will keep increasing. So the graph will increase, and at an increasing rate.

3. Buying gold every week, the amount of gold won't ever decrease. Since you buy less each week the

rate of increase will just keep falling. So the graph will increase, but at a decreasing rate. This

graph will in fact approach a horizontal asymptote, since we have a geometric progression which implies

an exponential function.

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Self-critique (if necessary): OK

@&

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

&#

*@

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Self-critique Rating:OK

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Question: `q006. Suppose that every week you go to the jeweler and obtain a certain number of grams of

pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts

a long, long time, that the the gold remains undisturbed, and that no other source adds gold to your

backyard.

1. If you graph the rate at which gold is accumulating from week to week vs. the number of weeks since

Jan 1, 2000, will the points on your graph lie on a level straight line, a rising straight line, a

falling straight line, a line which rises faster and faster, a line which rises but more and more

slowly, a line which falls faster and faster, or a line which falls but more and more slowly?

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

week.

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

week.

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Your solution:

1. Whether the line is linear or exponential depends on the rate at which the gold is accumulating. But

since we are assuming not other source is adding or removing gold, then the graph would be a straight

horizontal line with no slope.

2. This is an exponential growth function, which means this is a line which rises father and faster.

3. This is exponential decay so the line will decrease slower and slower.

confidence rating #$&*: 3

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Given Solution:

`aThis set of questions is different from the preceding set. This question now asks about a graph of

rate vs. time, whereas the last was about the graph of quantity vs. time.

Question 1: This question concerns the graph of the rate at which gold accumulates, which in this case,

since you buy the same amount eact week, is constant. The graph would be a horizontal straight line.

Question 2: Each week you buy one more gram than the week before, so the rate goes up each week by 1

gram per week. You thus get a risingstraight line because the increase in the rate is the same from one

week to the next.

Question 3. Since half the previous amount will be half of a declining amount, the rate will decrease

while remaining positive, so the graph remains positive as it decreases more and more slowly. The rate

approaches but never reaches zero.

STUDENT COMMENT: I feel like I am having trouble visualizing these graphs because every time for the

first one I picture an increasing straight line

INSTRUCTOR RESPONSE: The first graph depicts the amount of gold you have in your back yard. The second

depicts the rate at which the gold is accumulating, which is related to, but certainly not the same as,

the amount of gold.

For example, as long as gold is being added to the back yard, the amount will be increasing (though not

necessarily on a straight line). However if less and less gold is being added every year, the rate will

be decreasing (perhaps along a straight line, perhaps not).

FREQUENT STUDENT RESPONSE

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

INSTRUCTOR RESPONSE

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

should have

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

indicate your insertion by

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

These two problems go to the heart of the Fundamental Theorem of Calculus, which is the heart of this

course, and the extra effort will be well worth it in the long run. The same is true of the last

question in this document.

STUDENT COMMENT

Aha! Well you had me tricked. I apparently misread the question. Please don’t do this on a test!

INSTRUCTOR RESPONSE

I don't usually try to trick people, and wasn't really trying to do so here, but I was aware when

writing these two problems that most students would be tricked.

My real goal: The distinction between these two problems is key to understanding what calculus is all

about. I want to at least draw your attention to it early in the course.

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Self-critique (if necessary):

I did not notice the difference between the past two questions, but now that I go back I see that one

graph was related to accumulation of gold while the other is related to the amount of gold. So for

question two on the problem, the graph would be a straight line instead of an exponential because this

is a linear relationship.

@&

Good.

Note in relation to your answers that #2 is not exponential, while #3 is.

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Self-critique Rating: 3

``q007. If the depth of water in a container is given, in centimeters, by 100 - 2 t + .01 t^2, where t

is clock time in seconds, then what are the depths at clock times t = 30, t = 40 and t = 60? On the

average is depth changing more rapidly during the first time interval or the second?

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Your solution:

At t= 30, the depth is 49, at t = 40, the average depth is 36, and at t = 60, the depth is 16. The

slope between the second time interval is (16-36)/(60-40)= -20/20= -1, and the slope between the first

time interval is -13/10=-1.3. That means the average depth is changing more rapidly between the first

time interval.

confidence rating #$&*:

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Given Solution:

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

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

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

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

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

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q008. If the rate at which water descends in a container is given, in cm/s, by 10 - .1 t,

where t is clock time in seconds, then at what rate is water descending when t = 10, and at what rate

is it descending when t = 20? How much would you therefore expect the water level to change during this

10-second interval?

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Your solution:

The equation is y=10-.1t, when t=10 the rate at which the water is descending is 9cm/s and when t=20

the date is 8cm/s. I would expect the water level to change between 80 and 90 cm because 10 seconds is

passing and we know the between that interval it would be 8 cm/s and 9 cm/s.

confidence rating #$&*: 3

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Given Solution:

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

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

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

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

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

The rate change is a linear function of t. Therefore the average rate is the average of the two rates,

or 8.5 cm/s.

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

STUDENT RESPONSES

The following, or some variation on them, are very common in student comments. They are both very good

questions. Because of the importance of the required to answer this question correctly, the instructor

will typically request for a revision in response to either student response:

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

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

INSTRUCTOR RESPONSE

A self-critique should include a full statement of what you do and do not understand about the given

solution. A phrase-by-phrase analysis of the solution is not unreasonable (and would be a good idea on

this very important question), though it wouldn't be necessary in most situations.

An important part of any self-critique is a good question, and you have asked one. However a self-

critique should if possible go further. I'm asking that you go back and insert a self-critique on this

very important question and indicate your insertion by preceding it with ####, before submitting it.

The extra effort will be more than worth your trouble.

This problem, along with questions 5 and 6 of this document, go to the heart of the Fundamental Theorem

of Calculus, which is the heart of this course, and the extra effort will be well worth it in the long

run.

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

this document.

STUDENT COMMENT

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

much would you

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

you are asking for

what is the expected change in rate during this interval and what is the changing """"water level."""" But

now, after looking at

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

INSTRUCTOR RESPONSE

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

stated question.

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

the change in the quantity.

It seems clear that as a result of this question you understand this and will be more likely to make

such distinctions in your subsequent work.

This distinction is at the heart of the calculus and its applications. It is in fact the distinction

between a derivative and an integral.

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Self-critique (if necessary): OK

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Self-critique Rating: OK

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Question: `q009. Sketch the line segment connecting the points (2, -4) and (6, 4), and the line segment

connecting the points (2, 4) and (6, 1). The first of these lines if the graph of the function f(x),

the second is the graph of the function g(x). Both functions are defined on the interval 2 <= x <= 6.

Let h(x) be the function whose value at x is the product of the values of these two functions. For

example, when x = 2 the value of the first function is -4 and the value of the second is 4, so when x =

2 the value of h(x) is -4 * 4 = -16.

Answer the following based just on the characteristics of the graphs you have sketched. (e.g., you

could answer the following questions by first finding the formulas for f(x) and g(x), then combining

them to get a formula for h(x); that's a good skill but that is not the intent of the present set of

questions).

What is the value of h(x) when x = 6?

Is the value of h(x) ever greater than its value at x = 6?

What is your best description of the graph of h(x)?

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Your solution:

If h(x) is the product of f(x) and g(x), when x=6, h(x) is 4, because the points are (6,4) and (6,1).

The value of h(x) is never greater than its value at x=6 because g(x) keeps decreasing and when you

multiply f(x) by g(x) you will keep getting more negative numbers for h(x). My best description for the

graph h(x) is a graph that is decreasing from the interval 2<=x<=4 is getting less steep and increasing

at a decreasing rate from 5 to 6.

@&

Your thinking isn't bad but your conclusions aren't correct.

Consider the following, which is admittedly more than I would expect from a student after a single semester of calculus, but which can be understood by someone with that knowledge:

One reason I don't think the maximum of h(x) is at x = 6:

f ' (x) = 2 at all points.

g ' (x) = -3/4 at all points.

At any point (f * g) ' = f ' g + g ' f.

At x = 6, we get (f * g) ' = 2 * 1 + (-3/4) * 4 = -1.

Thus the graph of f g is decreasing at x = 4.

I don't expect many students to give this argument, though a knowledge of first-year calculus should be sufficient to arrive at this conclusion. However most students will not think of applying the product rule in this manner to the situation.

A numerical investigation would also reveal these trends. For example, using the slopes you could easily establish that g is almost twice as great at x = 5 as at x = 6, while f is more than half as great, proving without any extensive calculation or function modeling that h(5) must be greater than h(6). This argument is in fact nearly equivalent to the previous product-rule argument.

It would also be easy enough to find the linear equations for f and g, multiply them to get the quadratic function f g, find its critical points and second derivative, and draw the conclusion that the function maximizes somewhere between x = 2 and x = 6. However the preceding arguments are more elegant and preferable.

There are as well a number of other ways to answer this question.

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confidence rating #$&*: 3

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Question: `q010. A straight line segment connects the points (3,5) and (7,9), while the points (3, 9)

and (7, 5) are connected by a curve which decreases at an increasing rate. From each of the four points

a line segment is drawn directly down to the x axis, so that the first line segment is the top of a

trapezoid and the second a similar to a trapezoid but with a curved 'top'. Which trapezoid has the

greater area?

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Your solution:

I am not sure about this problem, but I think that the straight line would have a greater area because

it would have more space inside the trapezoid, whereas if the line is decreasing at an increasing rate,

the segment would be smaller. I am having a hard time drawing out these points being connected with a

curve that is decreasing at an increasing rate.

@&

The only way a curve could have that description would be to start out with a gradual downward slope, lying above the straight line of the first trapezoid, with the steepness of descent gradually increasing until it meets that straight line at the right-hand point.

That would give the second figure the greater area.

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confidence rating #$&*: 1

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Question: `q011. Describe the graph of the position of a car vs. clock time, given each of the

following conditions:

The car coasts down a straight incline, gaining the same amount of speed every second

The car coasts down a hill which gets steeper and steeper, gaining more speed every second

The car coasts down a straight incline, but due to increasing air resistance gaining less speed with

every passing second

Describe the graph of the rate of change of the position of a car vs. clock time, given each of the

above conditions.

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Your solution:

By position of the car, do you mean the distance from its starting point?

1. The graph would be increasing at a constant rate and the slope would be constant.

2. The graph would be increasing at an increasing rate, given that is it gaining more speed each

second.

3. The graph would be increasing at a decreasing rate because the air resistance is slowing down the

car.

@&

These would be good descriptions of the rate of change of position vs. clock time.

Position would implicitly be measured from the starting point, though any reference point would give the same answers provided the positive direction is the direction of motion.

*@

@&

You should think a little more about that position vs. clock time graph.

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confidence rating #$&*:2

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Question: `q012. If at t = 100 seconds water is flowing out of a container at the rate of 1.4 liters /

second, and at t = 150 second the rate is 1.0 liters / second, then what is your best estimate of how

much water flowed out during the 50-second interval?

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Your solution:

I would say that is could have been no more than 70 liters and no less than 50 liters, so my best guess

would be that during this 50-second interval 60 liters flowed out of the container. The rate at 100

seconds is 1.4 liters/second, and 1.4 * 50 seconds = 70 liters. The rate at 150 seconds slows down to 1

liter/second and 1 liter * 50 seconds is 50 liters. We don't know when it slowed down to this rate, so

we can only guess that the average was between these two rates that we do know for the 50 second

interval.

@&

Very good.

Your best estimate, without any additional information, would split the difference and conclude that the amount is about 60 liters.

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confidence rating #$&*: 3

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Self-critique Rating: 3"

@&

Very good, but do check my notes.

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Self-critique (if necessary):

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Self-critique rating: