Calculus

#$&*

course Phy 231

005. Calculus

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

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'?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

This portion of the graph is steepest, on the average, between (7,17) and (10,29). The average slope is the greatest at that point. We can tell this visually and also by calculating rise over run.

Slope of first portion: (17-5)/(7-3)=12/4=3

Slope of second portion: (29-17)/(10-7)=12/3=4

We say ""on the average"" because the slope is not truly equal to 3 for the entire first portion, or 4 for the entire second portion. On a smooth curve like this, the slope is constantly change and will be different at every point. (The slope at x=7 is very slightly different than the slope at x=7.0001.) So, finding a 'slope' of a curve like this tells us something about the nature of the graph but is not a precise, consistent measure across the entire graph or even across an entire small portion of it.

confidence rating #$&*:

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

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

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x=2.1 y=1/(2.1-2)=1/(.1) =10

x=2.01 y=1/(2.01-2)=1/(.01) =100

x=2.001 y=1/(2.001-2)=1/(.001) =1000

x=2.0001 y=1/(2.0001-2)=1/(.0001) =10000

1. As the value x gets closer to 2 following this pattern , y will continue to increase by a power of 10.

2. There is no limit to how close to 2 the value of x can approach, (how many zeroes can come between the decimal point and the 1), so there is no limit to the value of y. It can exceed one billion, one trillion billions, and continue to increase towards infinity.

3. Yes, it can exceed the number of particles in the known universe.

4. No, it can exceed any number.

5. Around x=2, the graph is an extremely steep curve. The curve resembles, but never actually touches, the line that would be formed by x=2, as it decreases sharply-- nearly but not quite vertically-- before forming a concave upward curve as it approaches x=3. The line x=2 is a vertical asymptote; the function can never equal 2, because that would require us to divide by zero, which can not be done.

confidence rating #$&*: 3

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

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

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.

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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'.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

While the first trapezoid is over twice as tall as the second (with a height of 9, compared to 4), The second trapezoid has a base that is 10 times greater than the base of the first (50-10, or 40, versus 7-3, or 4). From this we can tell that the second trapezoid will have a greater area.

confidence rating #$&*: 3

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

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

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.

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

Self-critique (if necessary):

From the precalculus exercises I just did, I remembered this line of reasoning. My inclination, though is not to trust the assumption that the 10-times-greater base seals the deal here, without actually calculating the precise area. Maybe I have been fooled by too many optical illusion puzzles. I want to see the numbers bear out what seems, logically and visually, to be true.

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

Self-critique Rating:

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

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

x y

2 4

5 25

-1 1

7 49

The line segment between (2,4) and (5,25) has a slope of (25-4)/(5-2) or 21/3 or 7

The line segment between (-1,1) and (7,49) has a slope of (49-1)/[7-(-1)] or 48/8 or 6

So, the segment between (2,4) and (5,25) is steeper.

confidence rating #$&*: 3

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

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

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.

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

1. If I bury the same number of grams each week, the points on the graph will lie along a rising straight line. For every equivalent interval of time (weeks, in this case, serving as the ""run"" of the graph), I add exactly the same amount of gold (in grams, serving as the ""rise""). So, rise/run will be the same at all points along the graph, forming a rising straight line.

2. If I bury one additional gram each week, the graph will increase more with each successive time interval. It will increase by, say, 1 gram the first week, then 2 the second week, 3 the third and so on. The slope of each week-to-week segment will increase by 1 every week. This will form a curve which rises faster and faster.

3. If I bury half as much each week, the graph will still rise-- I'm still adding to the total buried in my yard. But it will increase less and less, so my graph will form a curve that rises but more and more slowly.

confidence rating #$&*: 3

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

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

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.

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

Self-critique (if necessary):

???How would we figure out what the value of the horizontal asymptote would be? Say we started with 100 grams, then added 50, then 25, etc. It looks to me like it will never reach 200, but I'm struggling to express that with an equation.???

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

Self-critique Rating:

@&

Your masses would form the geometric series

100 grams * (1 + 1/2 + 1/4 + 1/8 + ... )

with constant ratio r = 1/2. The convergent value, which the masses will approach as an asymptote, is therefore

100 grams * (1 / (1 - 1/2) ) = 200 grams.

This can be reasoned out more directly. Your new mass will always be twice as close to 200 grams as the previous mass, from which you can reason out that you will never reach 200 grams, but you will approach it asymptotically.

*@

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

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

1. If I bury the same amount each week, say 1 gram, the rate will be the same every single week: 10 g/week. So, if I have time on the x-axis and rate of burial on the y-axis, I will have a flat, horizontal line at y=10.

2. If I bury ten additional grams per week, the rate will increase consistently by ten grams each week (10 g/week at week 1, 20 g/week at week 2, etc.). This graph will form a rising straight line.

3. If I bury half as much each week, the rate will decrease, but will decrease by less and less each week (10 g/week, then 5 g/week, then 2.5 g/week, etc.). This graph will form a curve that falls more and more slowly as we move from left to right.

confidence rating #$&*: 3

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

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

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.

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

Self-critique (if necessary):

I didn't specify that the third graph (where we bury half as much each day) would approach but never reach zero.

(I admit I wondered if you had accidentally inserted the same question twice until I noticed that your description of the conditions--the likelihood of someone else adding gold, for example-- was shorter the second time around. I wouldn't have read more closely, and noticed the key word ""rate"" if I hadn't noticed the other difference first!)

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

Self-critique Rating: 2

@&

Good eye. Most students miss this distinction.

*@

``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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

t=30 d=100-2(30)+(.01)(30^2) =100-60+(.01)(900) =40+9 =49

t=40 d=100-2(40)+(.01)(40^2) =100-80+(.01)(1600) =20+16 =36

t=60 d=100-2(60)+(.01)(60^2) =100-120+(.01)(3600) =-20+36 =16

On average, depth is changing by a rate of 13 cm per 10 seconds during the first interval (1.3 cm/s). On average, depth is changing by a rate of 20 cm per 20 seconds during the second interval (1 cm/s). So the rate is faster during the first time interval.

confidence rating #$&*: 2

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

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

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.

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

t=10 r=10-.1(10) =10-1 =9

t=20 r=10-.1(20) =10-2 =8

At the beginning of this 10-second interval, water is descending at a rate of 9 cm/s. At the end of this 10-second interval, water is descending at a rate of 8 cm/s. I will guess that to find the actual water level change, I will find the average rate during that time, which would be 8.5cm/s, and multiply it by 10 seconds for a level change of -85cm.

confidence rating #$&*:

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

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

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.

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

Self-critique (if necessary):

I got this right, but I felt very shaky on whether 8.5cm/s was in fact the mathematically certain rate during that 10-second interval. Do we know for sure this is true because the relationship is linear? If the function formed a curve on the graph, rather than a line, would 8.5 be accurate, or just somewhere in the ballpark?

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

Self-critique Rating:

@&

The fact that the rate is linear assures us that the average rate for any interval occurs at the midpoint of the interval and is equal to the average of the initial and final rates on that interval.

So it is precisely the linearity of the rate function that makes this result accurate.

If the rate function is not linear, but the interval is short enough that it doesn't deviation much from linear, it is still valid to approximate the change in depth using the midpoint rate, or the average of the initial and final rates (as long as you remain aware that the result is in this case an approximation).

*@

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

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)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

When x=6, h(x)=4*1 =4

I am not certain we can tell for sure using only the graph whether h(x) for this interval will ever exceed its value at x=6. Just glancing at the graph, I see that at x=5, for example, f(x) equals some value just over 2, and g(x) equals some value just under 2. With only that information, it appears that the product of those values MAY exceed 4, but it also may not. In any case, it is very near 4. I'll also estimate that at x=3, the value of h(x) is -2*3.25, or -6.5. These two pieces of information tell me that h(x) rises very rapidly at first and then more slowly (the difference between h(x) at x=2 and x=3 is nearly 10, while the difference between x=5 and x=6 is probably less than 1). This indicates a rising, but concave downward curve.

confidence rating #$&*: 2

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

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

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The trapezoid with a curved top has the greater area.

Since the curve decreases at an increasing rate, that means the slope starts out as less steep than a line segment between the two points, then gradually increases until it is greater than that (imagined, in this case) line segment. This indicates a concave downward curve connecting the two points. The shape beneath that curve is essentially the trapezoid that would be formed if the line were straight, plus an 'extra' curved area.

Since the height of the two trapezoids are the same (9 on the taller side, 5 on the shorter side) and the bases are the same (4 in both cases), the trapezoid with the straight line connecting the two sides is the same as the curved-top shape but without the 'extra' space formed by the curve itself.

confidence rating #$&*: 2

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

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

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

When the car gains the same amount of speed each second, the speed is changing. Therefore, each successive second, the car will cover more and more distance. I graphed altitude of the car (I assumed that's what you meant by position, although I suppose you might mean horizontal distance.) On my graph, altitude of the car is indicated by a concave downward curve, which is decreasing at an increasing rate as the car speeds up. If we're talking about horizontal distance, it would be a concave upward curve which is increasing at an increasing rate as the car speeds up.

On a hill which is steeper and steeper, the car gains speed at an increasing rate each second. So each successive second, the car covers more and more distance, as above, but the curve is even steeper. (On the altitude graph, position would decrease more rapidly than the first example; on the horizontal distance graph, position would increase more rapidly than the first example.)

When the car encounters resistance, its rate of speed still increases, but less than the other two examples. We still have a curve of the same shape, but a less dramatic one. If we're graphing altitude, we still have a concave downward curve that decreases with every second (the car is still moving downward, after all), and continues to decrease more rapidly each second than the second before. Similarly, if we're graphing horizontal distance, the graph increases every second, and increases MORE each second than the previous second. But the curve is slightly less steep at any given second than either of the first two examples.

confidence rating #$&*: 2

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

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

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

If we can assume this relationship is linear (and I'm not certain we can, but I don't think we have enough information to assume it isn't), then we can find the average rate of flow during that interval as (1.4+1.0)/2 or 1.2 liters/second. Since the rate is slowing during that time, that makes sense. It is a 50-second interval, so that means we need to multiply 1.2 by 50 seconds, for a total of 60 liters.

confidence rating #$&*: 3

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

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

Self-critique Rating:"

Self-critique (if necessary):

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

Self-critique rating:

Self-critique (if necessary):

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

Self-critique rating:

#*&!

&#Good work. See my notes and let me know if you have questions. &#