Calculus

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course Mth 174

6/2 4 ???I have noticed on a couple of past assignments that on certain questions there is no place for self-critic or self-critic rating. I did not add the sections myself because previous instructions have told me not to change document formatting. Are these sections supposed to be missing or are they always supposed to be there??? Thnaks!

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:

According to taking the average slope between the first and second point and then the second and third point I found that the average slope between the second and third point, that is (7,17) and (10,29) to be steeper on the average.

(17 - 5)/(7 - 3) = 12/4 = 3

(29 - 7)/(10 - 7) = 12/3 = 4

We must use phrase ‘on the average’ because we are dealing with a curved line which does not have a constant slope. For this reason we must take the average to say which is steeper.

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:

Let f(x) = 1/(x-2)

F(2.1) = 1/(2.1 - 2) = 1/.1 = 10

F(2.01) = 1/(2.01 - 2) = 1/.01 = 100

F(2.001) = 1/(2.001 - 2) = 1/.001 = 1000

F(2.0001) = 1/(2.0001 - 2) = 1/.0001 = 10000

1. The value of the function will grow without bound as x approaches 2.

2. Yes, the value will exceed a billion and one trillion billions because the function will approach infinity as it gets closer to x = 2 from the right.

3. Yes, same as above, it will exceed the particles in the known universe because it will approach infinity.

4. There is no number the function will never exceed.

5. As x approaches x = 2 from the left the function goes to negative infinity. As the function moves towards x = 2 from the right, the function approaches positive infinity. As the values of x in the function approach negative infinity, the function approaches y = 0 from the negative side. As the values of x approach positive infinity, the function approaches y = 0 from the positive side.

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):

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Self-critique Rating:

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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:

When the trapezoids that formed from the two line segments were drawn on a coordinate plan, it was easy to see that the first trapezoid (that formed from the line connecting (3,5) and (7,9) ) was around twice as high as the trapezoid formed form the line segment connecting points (10,2) and (50,4). In fact the average altitude for the first trapezoid was 5 and the average altitude for the second trapezoid in the y direction was 3. So with the first trapezoid being just under twice as tall as the second I moved my next observation to the width of each trapezoid. The first had a width on the x-axis of 4 from x = 3 to x = 7. The second trapezoid had a width on the x-axis of 40 from x = 10 to x = 50. With this observation I could conclude that because the first trapezoid from line segment (3,5) to (7,9) had nearly twice the altitude, although the second trapezoid from line segment (10,2) to (50,4) had 10 times the width as the first it was plane to see that the second trapezoid formed form the line segment between (10,2) and (50,4) had the larger area.

confidence rating #$&*: 3

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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):

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Self-critique Rating:

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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:

To answer this question I used the function to attain the y-values of the points.

For the line segment connecting x = 2 and x = 5 we have:

at x = 2 f(2) = 2^2 = 4 so (2,4) is our first point

at x = 5 f(5) = 5^2 = 25 so (5,25) is our second point

m = (25 - 4)/(5 - 2) = 21/3 = 7

For the line segment connecting x = -1 and x = 7 we have:

at x = -1 f(-1) = (-1)^2 = 1 so (-1,1) is our first point

at x = 7 f(7) = 7^2 = 49 so (7,49) is our second point.

m = (49 - 1)/(7 + 1) = 48/8 = 6

From the above calculations of the slops of the two line segments it can be determined that the line segment connecting x = 2 and x = 5 as the steeper slope and thus the steeper line segment.

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):

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Self-critique Rating:

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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:

To answer the first question. With all of the assumptions specified in the explanation, I am collecting a certain (set amount) which I assume means the same amount every week. If I attain the same amount every week. The graph of y = number of grams of gold in backyard vs. t = number of weeks since Jan 1, 2000 would increase at a steady constant rate. It would be a rising straight line.

In consideration of the second question. If instead of attaining the same amount of gold per week, I attained one more gram than each previous week, the graph would be a rising line that rises faster and faster with each week. An increasing line at an increasing rate.

In consideration of the third question. If instead of attaining the same amount of gold per week, I attained half the amount that I attained in the previous week, the graph would be a line that rises but more and more slowly each week while maintaining a positive slope.

confidence rating #$&*: 3

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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):

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Self-critique Rating:

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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:

In the first question, from the previous question, I had answered that the graph of the y = number of grams vs. t = number of weeks would be a straight increasing line. This means that this line would have a constant slope or (rate of change). So to answer this question, if the slope is a constant and it is increasing the graph of this function would be a horizontal straight line at y = whatever the constant slope is.

If every week you buy one more gram than you bought the previous week, then the graph of the original function would be a line that is increasing faster and faster or increasing at an increasing rate. If we graphed the rate of this function we would graph the slope or basically first derivative of the original function. In this case since the rate is increasing by one each week, the graph of the rate would be a straight line that is increasing.

To answer the third part of this question. To graph the rate of the third function we have to consider what is happening to the rate when you buy one half of the amount you bought in the previous week each week. The graph of the original function would be increasing less and less each week, although it would still be increasing. Because we are adding gold each week it is still gaining just less and less each time. This means that if we graphed the rate of this function, the values of the rate function would be positive; although they would be less and less positive as time went on. Because we always add gold, the rate function would never be zero, just very close to zero. So it would be a falling line that falls less and less as the weeks go on.

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.

Self-critique (if necessary):

At first I was a bit confused; although because I was pretty sure that you wouldn’t put the same question twice I realized that it said to graph the rate function, not the actual number of grams.

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?

Your solution

At t = 30

Depth = 100 - 2(30) + 0.01(30)^2

= 100 - 60 + .01(900)

= 40 + 9 = 49

At t = 40

Depth = 100 - 80 + 0.01(40)^2

= 100 - 80 + .01(1600)

= 20 + 16 = 36

At t = 60

Depth = 100 - 120 + 0.01(60)^2

= 100 - 120 + .01(3600)

= -20 + 36 = 16

So the slope between first time interval

m = (36 - 49)/(40 - 30) = -13/10 = -1.3

and the slope for the second time interval is

m = (16 - 36)/(60 - 40) = -20/20 = -1

From the above calculations, the conclusion can be drawn that the average depth change is more rapid on average for 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):

The only descrepency I have is that my slopes are negative because the depth is decreasing over the interval. ???Does this mean it is negligible to include the negative signs because we are only talking about which one has the higher magnitude in change???

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Self-critique Rating: 3

@&
If the question asked on which interval the rate was greater, then you would take signs into consideration, as any positive rate would be greater than any negative rate.

However when asked when the depth is changing more quickly, we don't care whether it is increasing or decreasing, and we compare the absolute values of the two rates.
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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:

At t = 10, the water is descending at a rate of

10 - 0.1(10) = 10 - 1 = 9cm/s

At t = 20, the water is descending at a rate of

10 - 0.1(20) = 10 - 2 = 8cm/s

To derive the answer for how much the water level will change over the entire 10 second interval I’ve found the rate of descent at each second over the interval and then added them all together.

10 - 0.1(10) = 10 - 1 = 9cm/s

10 - 0.1(11) = 10 - 11/10 = 100/10 - 11/10 = 8.9cm/s

10 - 0.1(12) = 10 - 12/10 = 8.8cm/s

10 - 0.1(13) = 10 - 13/10 = 8.7cm/s

10 - 0.1(14) = 10 - 14/10 = 8.6cm/s and the pattern continues as I have already worked these on paper.

At t = 15 rate = 8.5 cm/s

At t = 16 rate = 8.4cm/s

At t = 17 rate = 8.3cm/s

At t = 18 rate = 8.2cm/s

At t = 19 rate = 8.1cm/s

If we start at 10 seconds we would only count from t = 11 to t = 20 and we’d have

8.1 + 8.2 + 8.3 + 8.4 + 8.5 + 8.6 + 8.7 + 8.8 + 8.9 + 9.0 for the rates of those corresponding seconds.

This added up would give the total amount displaced over the ten second period and is 85.5 cm

One could also solve this problem with less work by realizing that the rate is constantly declining as the time goes on in that it is a linear function. And since the rate at t = 10 is 9cm/s and the value at t = 20 is 8cm/s. I could find the average rate of change over this interval by adding these two rates together and then dividing by two. This would give 8.5cm/s as the average rate of change over the interval. To get how much the water level changed I could then multiply by 10. Giving around 85 cm.

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):

??? When I first looked at this problem I tried finding the average rate of change of the two rates. Then upon thinking about this I realized that I already had the rates of change and to take their average all I had to do was average the two rates. I put a longer version of the solution as well just to explain things to myself. Is my understanding of this correct in my last paragraph of my answer???

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Self-critique Rating: 3

@&
Your thinking is good throughout, except that if you include the rate at the last instant t = 20 you would have to also include the rate at the first instant when t = 10.

If the rate wasn't constant, you would still include the t = 10 and t = 20 rates, but each with only half the weight of the others.
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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:

At x = 6 the value of h(6) is the product of f(6) and g(6) so it is 4 * 1 = 4

No, the graph of h(x) is at its maximum value at x = 6 because if we move to the right g(x) goes below one. Even though f(x) goes beyond 4, the product of g(x) and f(x) is h(x) so something larger than 4 times something smaller than one is always going to be smaller than 1 * 4 itself. As we move from the left of x = 6, the value of h(x) is getting smaller than 4 as well. Because g(x) and h(x) are constant we can conclude that this will be a constant trend and the further right or left you go from x = 6 will only make h(x) a smaller value.

I would describe the graph of h(x) as a function that is almost parabolic, having a maximum point at x = 6 being (6,4) and decreasing on both sides of x = 6 based on my above observations of the sketch.

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:

Upon formulating a sketch of these to trapezoids on a coordinate plan, I was able to determine that both trapezoids have same width along the x-axis of 4 from x = 3 to x = 7. This means that to find which one had a larger area, the width along the x-axis wouldn’t be a factor. If both line segments were the exact same (a straight line), the average altitudes would mean that they would both have the same exact area, although one is not a straight line. To say that the curve connecting the top of the trapezoid from (3,9) to (7,5) is decreasing at an increasing rate, is saying that it is curved down or concave down. From this information I am able to determine that since the widths are identical, and the average altitudes are the same, the trapezoid with a top that is curved downwards would have a top that is in a sense bulging out and thus would have a larger area. So the trapezoid with the larger area is the one with a curved top.

confidence rating #$&*: 2

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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:

In the first scenario the car is gaining speed at a constant rate, which means its rate of change or slope is a constant positive number. The graph of this car would look like a straight increasing line.

In the second scenario, the car is gaining more and more speed every second due to the change in the steepness of the hill. The graph of this function would look like a curved line that is increasing at an increasing rate. This is because its rate of change is positive and is increasing with every move right.

In the third scenario, the car is coasting down the hill gaining in position, although at a slower and slower rate as time goes on. This graph would look like a curved line that is increasing at a decreasing rate.

In the first graph above of the position of the car, the graph of position vs. time would be a straight increasing line. Because the rate is a constant, the graph of the rate would be simply a horizontal straight line.

In the second graph of position vs. time of the car above, the rate is increasing and the position is increasing. Because the rate is increasing, assuming it is increasing at a constant rate, the graph of the rate of the second function would be an increasing straight line.

In the third graph above of position vs. time of a car, the car is increasing in position although at a slower and slower rate as time goes on. This means that the graph of position vs. time would be a line that is increasing at a decreasing rate. The rate is still always positive, although its values are decreasing. This means that the graph of this rate function would be a line that is decreasing, although decreasing less and less as time goes on and it never touches zero. Until the car stops. So it would be decreasing at a decreasing rate.

confidence rating #$&*: 3

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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:

If at t = 100 seconds the water is flowing at a rate of 1.4 L/s and at t = 150 seconds water is flowing at a rate of 1.0 L/s, the average rate of change over this time interval can be obtained.

The average rate of change over the time interval:

1.0L/s + 1.4 L/s = 2.4 L/s *(1/2) = 1.2L/s

Over this interval there are 50 seconds. So 1.2L/s (average rate of change) * 50 (number of seconds in interval) = 60 liters of water.

confidence rating #$&*: 3

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Calculus

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