Query 1 revised

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course mth 174

4/17 10:55 pm

Query 1#$&*

course mth 174

4/16 9:50 pm

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

The derivative of y = f(x) with respect to x is the instantaneous rate of change of y with respect to x.

The definite integral of a function y = f(x) over an interval between x = a and x = b is equal to the change in an antiderivative function between x = a and x = b over that interval.

The derivative function is the ‘rate-of-change function’.

An antiderivative function is the ‘change-in-quantity function’, in the sense that if y = f(x) represents the rate of change of some quantity Q with respect to x, then the change in the quantity Q over an interval is equal to the change in the antiderivative function over that interval.

More complete statements:

The derivative function f ‘ (x) of a function y = f(x), with respect to variable x, is the instantaneous rate of change of y with respect to x. The rate of change of y with respect to x at a specific point x = x0 is f ‘ (x0).

The change in the function y = f(x) between specific points x = a and x = b is equal to the change in an antiderivative function F(x) between x = a and x = b, provided an antiderivative function exists on this entire interval. When this is the case, the change in the antiderivative function

• An antiderivative function F(x) is any function F(x) whose derivative is equal to f(x).

• Antiderivative functions are known up to an arbitrary integration constant so if an antiderivative function exists on an interval, there are infinitely many antiderivative functions on that interval.

• To find the value of a definite integral, one specific antiderivative function must be chosen.

• The change in the chosen antiderivative function is (final value) - (initial value) = F(b) - F(a).

For a given function f(x) on a given interval:

• the derivative may be defined everywhere on that interval

• the derivative may be defined nowhere on that interval

• the derivative may be defined at all but a finite number of points

• the derivative may be defined at all but an infinite number of points (either countably infinite or uncountably infinite).

Nearly all of the functions studied in first-year calculus, and most of the functions required to apply calculus to the real world, have derivatives defined on easily-recognized intervals.

• For many of the functions studied in first-year calculus, the derivative is defined for all real numbers.

• For some functions there are specific points or intervals where the derivative is not defined, but between these points and intervals are the easily determined intervals on which the derivative is defined.

• The set of intervals (and/or points) over which a derivative function is defined is the domain of the derivative function.

• If the derivative is defined at all points of an interval, then the function is said to be differentiable on that interval, and the derivative function f ‘ (x) will then give the rate of change of y with respect to x at all points of the interval.

If a function f(x) has an antiderivative F(x) at every point of an interval, then it is said to be integrable over that interval.

• If this is the case, then between any two points of the interval, the change in the antiderivative gives us the definite integral of f(x), with respect to x, between those points.

• Note that for f(x) to have an antiderivative F(x) at every point of an interval, F(x) must be differentiable at every point of the interval.

• So if there is a point (or an interval) between x = a and x = b where the derivative does not exist, the change in an antiderivative function cannot be expected to give us the value of the definite integral; in fact the definite integral would not be defined at all for this interval.

Any function f(x) that can be written down as a series of constant multiples, sums, products, quotients and composites of the basic power, exponential, logarithmic and polynomial functions studied in first-year calculus has a derivative f ' (x) that can be found by correctly using the constant-multiple, sum, product, quotient and chain rules, and the expressions for the derivatives of these basic functions. While the calculations can be messy, they aren't tricky. If the method is followed correctly, it works.

However even though an antiderivative F(x) of a such a function f(x) might exist, it is not always possible to express the antiderivative in terms of the basic functions. When it is possible, it is often tricky, and there's really no limit to how tricky it can get. In second-semester calculus we learn some of the basic tricks, and we learn how to approximate definite integrals when the tricks don't work.

An important basic picture:

If y(t) is the depth of water in a container, then the derivative y ' (t) is the function which gives you the rate of change at any instant.

If r(t) is the rate at which depth is changing, then r(t) = y ' (t). If all you know is r(t) then it follows that y(t) is an antiderivative of r(t).

The change in y between any two t values t = t1 and t = t2 is therefore equal to the change in the antiderivative between those t values: change in y = antiderivative of r(t) evaluated at t2 - antiderivative of r(t) evaluated at t1. That is, the change in y is the definite integral of the rate function from t = t1 to t = t2.

In this example we see a clear illustration of the Fundamental Theorem of Calculus.

Calculus II

Asst # 1

07-13-2001

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Note: Problem numbering is according to the problems as presented in Problem Assignments . The numbering will differ from that in your text.

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Question: Section 6.1, Problem 5 5th edition Problem 14 4th edition Problem 5 [[6.1.5 (previously 6.1 #12)]]

f '(x) =1 for x on the interval (0,2), -1 on (2,3), 2 on (3,4), -2 on (4,6), 1 on (6,7)

f(3) = 0

What was your value for the integral of f '?

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Your solution:

When looking at the graph of f’ in the text I see the following

Between 0 and 2 the area is 2*1=2 so the slope is 2/2=1???

Between x = 0 and x = 2 the value of the function f ' (x) is 1. The value of the function does not change on that interval, so the rise of the graph on that interval is zero and its slope is therefore zero.

The slope will in fact be zero on every interval, as the value of the function on each interval is unchanging.

It could be (in fact I think it's likely) that you are talking about the slope of the graph of the antiderivative function f(x). However the slope of the graph is the value of its derivative. On this first interval the derivative has constant value 1.

So the slope of the f(x) graph is 1.

The interval has length 2, which corresponds to the change in x, i.e., the 'run' of the graph of f(x).

If the slope is 1 and the 'run' is 2, it follows that the 'rise' of the f(x) graph is

rise = slope * 'run' = 1 * 2 = 2.

This calculation corresponds to the area beneath the graph of the f ' (x) graph in a way you need to understand.

On every interval the value of the f ' (x) graph represents the slope of the f(x) graph. This is because f ' (x) is the slope of the f(x) graph.

When you calculate the integral of f ' (x) on an interval, you multiply the value of f ' (x) by the width of the interval. Since the width of the interval is the 'run' of the f(x) graph, and f ' (x) is the slope of the f(x) graph, this calculation is the same as multiplying the slope of the f(x) graph by the 'run', which gives you the 'rise' of the f(x) graph.

The 'rise' of the f(x) graph on an interval is the change in the value of f(x) for that interval.

&&&& I was indeed looking at the graph I drew of the antiderivative graph when describing the slopes. I understand that for every value of f’(x) the slope is zero. I also now understand that the value of the area of the graph under the f’(x) intervals is the value of the corresponding slope on the antiderivative graph. I understand as well what you are saying about slope values. So for this question, “What was your value for the integral of f '?”

Would the correct answer be

Between 0 and 2 the value is 2*1=2

Between 2 and 3 the value is 1*1=-1

Between 3 and 4 the value is 2*1=2

Between 4 and 6 the value is -2*2=-4

Between 6 and 7 the value is 1*1=1 ??????? &&&&

Between 2 and 3 the area is -1*1=-1 so the slope is -1

Between 3 and 4 the area is 2*1=2 so the slope is 2

Between 4 and 6 the area is -2*2=-4 so the slope is -4/2=-2???

Between 6 and 7 the area is 1*1=1 so the slope is 1

Constructing the graph of F. If F(3) =0 then that point would be plotted at (0,-3). The slope between 0 and 2 is 1 so the next point would be (2, -1). The slope between 2 and 3 is -1 so the next point would be (3,-2). The slope between 3 and 4 is -1 so the next point would be (4,0). The slope between 4 and 6 is -2 so the next point would be (6,-4). The slope between 6 and 7 is 1 so the next point would be (7, -3).

X Y Change in Y F(x)

0 -3 F(0)= 1-2=-1

2 -1 2 F(2)=0-(-1)=1

3 -2 -1 F(3)=0 (given)

4 0 2 F(4) =0+2=2

6 -4 -4 F(6) =2-4=-2

7 -3 1 F(7)=-2+1=-1

So F(0)=-1 and F(7)=-1

So the intregal would be the changes which are 2-1+2-4+1=0 or you could subtract F(0) from F(7) which is -1-(-1) or -1+1=0

confidence rating #$&*: 3

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Given Solution:

the change in f from x=0 to x=2 is 2 (area beneath line segment from x=0 to x=1), then from x=2 to x=3 is -1, then from x=3 to x=4 is +2, then from x=4 to x=6 is -4, then from x=6 to x=7 is +1.

If f(3) = 0 then f(4) = 0 + 2 = 2, f(6) = 2 - 4 = -2 and f(7) = f(6) + 1 = -1.

Working back from x=3, f(2) = 0 - (-1) = 1 and f(0) = 1 - 2 = -1.

The integral is the sum of the changes in f ' which is 2 - 1 + 2 - 4 + 1 = 0.

Alternatively since f(0) = -1 and f(7) = -1 the integral is the difference f(7) - f(0) = -1 - (-1) = 0.

Let me know if you disagree with or don't understand any of this and I will explain further. Let me know specifically what you do and don't understand.

Alternative solution:

Two principles will solve this problem for you:

1. The definite integral of f' between two points gives you the change in f between those points.

2. The definite integral of f' between two points is represented by the area beneath the graph of f' between the two points, provided area is understood as positive when the graph is above the x axis and negative when the graph is below.

We apply these two principles to determine the change in f over each of the given intervals.

Answer the following questions:

What is the area beneath the graph of f' between x = 0 and x = 2?

What is the area beneath the graph of f' between x = 3 and x = 4?

What is the area beneath the graph of f' between x = 4 and x = 6?

What is the area beneath the graph of f' between x = 6 and x = 7?

What is the change in the value of f between x = 3 and x = 4? Since f(3) = 0, what therefore is the value of f at x = 4?

Now that you know the value of f at x = 4, what is the change in f between x = 4 and x = 6, and what therefore is the value of f at x = 6?

Using similar reasoning, what is the value of f at x = 7?

Then using similar reasoning, see if you can determine the value of f at x = 2 and at x = 0.**

STUDENT QUESTION: I did not understand how to obtain the value of f(0), but I found that f(7)

was 10 by adding all the integrals together

INSTRUCTOR RESPONSE:

The total area is indeed 10, so you're very nearly correct; however the integral is like a 'signed' area--areas beneath the x axis make negative contributions to the integral--and you added the 'absolute' areas

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Self-critique (if necessary):

### It took me a long time to wrap my head around this problem. I think I have got the solution and steps down now. On a test, will we have a graph illustration as in the book or will we have any points to plot the f’ graph??? If the instructions only say, “f '(x) =1 for x on the interval (0,2), -1 on (2,3), 2 on (3,4), -2 on (4,6), 1 on (6,7)” with no illustration or listed points, I have a difficult time envisioning the f’graph in order to determine the area underneath the points to I can determine the slope of the f graph????

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Self-critique Rating:3

See also the notes I inserted into your solution.

You would be expected to be able to construct the graph from that description. Piecewise-defined functions are a standard precalculus topic and are usually encountered in first-semester calculus.

However a description of such a function on a test would probably be more detailed, and might include a graph.

If you want more information on piecewise-defined functions, let me know. You could ask using the Submit Work Form, though the Submit Question Form at http://vhcc2.vhcc.edu/dsmith/forms/question_form.htm would be preferable as it helps you pose a properly detailed question.

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Question: (Note that no problem reference is given, meaning that this question applies to the current problem. Any question that is not preceded by a problem number is likely to be in reference to the current problem.)

Describe your graph of f(x), indicating where it is increasing and decreasing and where it is concave up, where it is straight and where it is concave down.

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Your solution:

For my graph of F(x).

Between X=0 and X=2, the graph is increasing at a slope of 2/1 =1

The slope is 1, but the calculation 2/1 is not relevant to determining the slope. The slope is 1 because the value of the f ' (x) graph on this interval is 1. If the interval ran from x = 0 to x = 50, with the same y value, the slope would still be 1.

&&&&I understand now that the value of the slope is equal to the value of the interval on the f’(x) graph. &&&&

Between X=2 and X=3 the graph is decreasing at a slope of -1.

Between X=3 and X=4 the graph is increasing at a slope of 2.

Between X=4 and X=6 the graph is decreasing at a slope of -4/2=-2

Between X=6 and X=7, the graph is increasing at a slope of -1.

The graph is straight because the slope is constant throughout each interval.

What is the slope of the f graph between x = 0 and x = 2? 1

What is the slope of the f graph between x = 3 and x = 4? 2

What is the slope of the f graph between x = 4 and x = 6? -2

What is the slope of the f graph between x = 6 and x = 7? 1

Given that f(3) = 0 and using the value of the slope of the f graph between x = 3 and x = 4, describe the f graph between these two points.

Slope for (3 4) is 2 so on the f graph, between points X=3 and X=4 the graph is rising at a constant rate of increasing y by 2 and x by 1.

Using similar information describe the graph for each of the other given intervals.

(0,2) slope is 2/1=1 so the graph is rising at a constant rate of increasing y by 1 and x by 1 for each increment

(2,3) slope is -1 so the graph is decreasing at constant rate of decreasing y by 1 and increasing x by 1

(4,6) slope is -4/2 increments = -2 so the graph is decreasing y by 2 and increasing x by 1 for each increment

(6,7) the slope is 1 so the graph is increasing y by 1 and x by 1.

Also answer the following:

What would have to be true of the f' graph for the f graph to be concave up? Same question for concave down. **

In order for f to be concave up, F’ would have to be rise at an inconsistent rate or slope???

In order for f to be concave down, F’ would have to fall at an inconsistent rate or slope???

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18:37:09

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confidence rating #$&*: 2

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Given Solution:

** The graph of f(x) is

increasing, with slope 1, on the interval (0,2), since f'(x) = 1 on that interval,

decreasing, with slope -1, on the interval (2,3), where f'(x) = -1,

increasing, with slope +2, on the interval (3,4), where f'(x) = +2,

decreasing, with slope -2, on the interval (4,6), where f'(x) = -2, and

increasing, with slope +1, on the interval (6,7), where f'(x) = +1.

The concavity on every interval is zero, since the slope is constant on every interval.

Since f(3) = 0, f(4) = 2 (slope 2 from x=3 to x=4), f(6) = -2 (slope -2 from x = 4 to x = 6), f(7) = -1 (slope +1 from x=6 to x=7).

Also, since slope is -1 from x=2 to x=3, f(2) = +1; and similar reasoning shows that f(0) = -1. **

** The definite integral of f'(x) from x=0 to x=7 is therefore f(7) - f(0) = -1 - (-1) = 0. **

** Basic principles:

1. The slope of the graph of f(x) is f'(x). So the slope of your f graph will be the value taken by your f' graph.

2. Note that if the slope of the f graph is constant for an interval that means that the graph is a straight line on the interval.

Using these principles answer the following questions:

What is the slope of the f graph between x = 0 and x = 2?

What is the slope of the f graph between x = 3 and x = 4?

What is the slope of the f graph between x = 4 and x = 6?

What is the slope of the f graph between x = 6 and x = 7?

Given that f(3) = 0 and using the value of the slope of the f graph between x = 3 and x = 4, describe the f graph between these two points.

Using similar information describe the graph for each of the other given intervals.

Also answer the following:

What would have to be true of the f' graph for the f graph to be concave up? Same question for concave down. **

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18:37:09

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Self-critique (if necessary):

### What would have to be true of the f' graph for the f graph to be concave up? Same question for concave down. **

In order for f to be concave up, F’ would have to be rise at an inconsistent rate or slope???

In order for f to be concave down, F’ would have to fall at an inconsistent rate or slope???

I am not sure if the above answer is correct????

If the graph of f(x) is concave up, what is happening to its slope?

The slopes of the f(x) graph are the values of the f ' (x) graph. Whatever your description of the trend of the slopes of the f(x) graph, that will be the description of the values of the f ' (x) graph.

What therefore is the trend of the values of f ' (x)?

What would the graph of f ' (x) therefore have to look like? There are many possible descriptions of f ' (x) graphs that correspond to f(x) graphs which are concave up, but all such f ' (x) graphs have one thing in common. What is that one thing?

Having answer this, you should be able to answer the question for f(x) being concave down.

&&&&if the graph of F(x) is concave up, then its slope is increasing so values of f’(x) integral would have to be positive values. Trends for f’(x) for concave up would be positive values. The graph of f’(x) would have to be all positive values above the x axis. The one thing the description of all f’(x) graphs that correspond to f(x) graphs which are concave up are that the values of f’(x) are positive numbers.

For f(x) graph to be concave down, the slope would have to be decreasing which would mean that the values of f’(x) integral would have to be negative values???? &&&&

@&
To be concave up, as you say, slopes have to be increasing. They don't have to be positive. For example the graph of y = x^2 is concave up everywhere, both to the left of the y axis (where the slopes are negative) and to the right.

So the one thing you can say about the f ' (x) graph where f(x) is concave up is that it is increasing. The slopes of the f ' (x) graph will therefore be positive.

Similarly the slopes of f ' (x) are decreasing where f(x) is concave down.
*@

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Self-critique Rating:3

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Question: Was the graph of f(x) continuous?

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Your solution:

The graph of F(x) does not appear to be continuous because it had different slopes for different integrals of x. The graph had jumps throughout and jagged appearance. However, for each intergral of x, there is a consistent slope. For example (0 2) slope is 1, (2 3) slope is -1 etc. So each integral of x is continuous although the entire graph as a whole is not???

** A function f(x) is continuous at x = a if the limit of the f(x), as x approaches a, exists and is equal to f(a).

I don’t follow??? I know that the function is continuous between each integral as it has a consistent slope for each integral. Does this mean that the graph is consistent???

Is this condition fulfilled at every point of the f(x) graph? **

At every point of f(x) graph, there is a constant and continuous slope, although overall there is not consistency for the entire graph???

confidence rating #$&*: 2

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Given Solution:

** A function f(x) is continuous at x = a if the limit of the f(x), as x approaches a, exists and is equal to f(a).

Is this condition fulfilled at every point of the f(x) graph? **

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18:37:15

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Self-critique (if necessary):

### ** A function f(x) is continuous at x = a if the limit of the f(x), as x approaches a, exists and is equal to f(a).

I don’t follow??? I know that the function is continuous between each integral as it has a consistent slope for each integral. Does this mean that the graph is consistent???

Is this condition fulfilled at every point of the f(x) graph? **

At every point of f(x) graph, there is a constant and continuous slope, although overall there is not consistency for the entire graph???

??? is the above correct or am I misunderstanding???

You don't seem to be familiar with the definition or concept of continuity.

In your test continuity has not really been defined at this point, but it has been discussed in more than one place. You'll need to review the idea, which isn't that difficult (applying the definition can be challenging, but as I say that comes later in the text). I expect that just be reading the text discussions and descriptions you'll get the idea.

You should go to the table of contents and look at every discussion of continuity up that precedes Chapter 6. I think you can get the idea within an hour.

Of course if you still have questions or want further guidance you're welcome to ask.

Having reviewed that material, and if necessary asked for my additional assistance, you should be able to answer this question.

I will pose one very relevant question for you at this point. Would it be possible for you to graph the f(x) function without lifting your pencil from the paper?

&&&&I understand now from reading back through the previous chapters of the text, that for a function to be continuous at an interval that the graph should have no gaps or breaks, so I understand your comment about being able to draw the F(x) function without lifting the pencil from the paper. So yes the graph of f(x) was continuous throughout, I could draw the entire graph without lifting my pencil and there are no holes, gaps, or jumps. &&&&

@&
Good.
*@

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Self-critique Rating:3

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Question:

How can the graph of f(x) be continuous when the graph of f ' (x) is not continuous?

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Your solution:

The graph of F(x) is continuous between each integral of x because each integral has a consistent and continuous slope??? I’m not sure why F(x) would be continuous and not F’(x) except for the fact that the above question said, “A function f(x) is continuous at x = a if the limit of the f(x), as x approaches a, exists and is equal to f(a).” Which is a statement I don’t fully understand???

confidence rating #$&*: 2

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Given Solution:

18:38:11

** f' is the slope of the f graph; f' has 'jumps', which imply sudden changes in the slope of the f graph, causing the graph of f to have a jagged shape as opposed to a smooth shape. However this does not cause the graph of f itself to have discontinuous 'jumps'. **

** f ' determines the slope of f; the slope of f can change instantaneously without causing a 'jump' in the values of f. Continuity is, roughly speaking, a lack of 'jumps' in a graph. **

** Basically, if f ' is finite and does exceed some fixed bound over a small interval about x = a, then the change `dx in x has to be small. More specifically:

f(x) is continuous at x = a if the limit of f(x) as x -> a is equal to f(a).

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Self-critique (if necessary):

###I’m not sure I follow. I understand that the graph of f has continuous slopes in each integral. So this would imply continuity. Even though the graph of f looks jagged, it still has continuity with its slopes. Is this all I need to understand or am I missing anything???

You need to take a little time to nail down the basic idea of continuity, per my previous note.

&&&&after reviewing continuity I believe I understand the concept now that continuity means that you can draw the graph without lifting your pencil. So to answer the question, “How can the graph of f(x) be continuous when the graph of f ' (x) is not continuous? “ The answer is that although the values of the integrals of f’(x) change and cause changing slopes in the f(x) graph, these changes in values and corresponding values in slopes do not cause a discontinuous f(x) graph because the slopes change creating a jagged graph but the graph is still continuous because there are no holes or breaks in the graph??? &&&&

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Self-critique Rating:3

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Question:

What does the graph of f(x) look like over an interval where f ' (x) is constant?

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Your solution:

IF f’(x) has a value that is constant such as y value always being 3 and the graph of f’(x) ran from point (0,3) to point (10,3) then the area under this would be 10*3=30/10 intervals for a slope of 3. Therefore, the graph of f(x) would have a constant rising slope of 3. So the graph of f(x) would have a consistent slope and would rise consistently from left to right.

confidence rating #$&*: 3

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Given Solution:

** If f ' is constant then the slope of the f(x) graph is constant, so the graph of f(x) must be linear **

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Self-critique (if necessary):

### I didn’t use the word linear, but I think my description is correct???? I will remember the term linear for the future.

Your answer was very good.

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Self-critique Rating:3

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Question: Section 6.1, Problem 10: The graph of outflow vs. time is concave up Jan 1993 -Sept, peaks ub October,

then decreases somewhat thru Jan 1994; the inflow starts lower than the outflow, peaks in May, then decreases until January; inflow is equal to outflow around the middle of March and again in late July.

**** When was the quantity of water greatest and when least? Describe in terms of the behavior of the two curves.

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Your solution:

For this problem I created a table with the labeled months and the outflow and inflow based on a scale of 1 I drew in on the y axis of 1 which represents 1 million gallons of water.

Date Inflow Outflow Water Qty

Jan93 2 3 2-3=-1

Feb 2.5 3 -1+2.5-3=-1.5

March 3 3 -1.5+3-3=-1.5

April 5.5 3.25 -1.5+5.5-3.25=.75

May 6 3.5 .75+6-3.5=3.25

June 5.5 3.75 3.25+5.5-3.75=5

July 4 4 5+4-4=5----- Highest months (june and july)

August 3.75 4.25 5+3.75-4.25=4.5

Sept 3.5 5 4.5+3.5-5=3

Oct. 3 6 3+3-6=0

Nov 2.75 5.5 0+2.75-5.5=-2.75

Dec 2.5 4.5 -2.75+2.5-4.5=-4.75

Jan94 2 4 -4.75+2-4=-6.75----Lowest month Jan

I created a table as listed above by penciling in values by increments of 1 on the y axis which represent 1 million gallons of water. I began with the given values of inflow and outflow for January 1993 and entered inflow and outflow values on the table and began for January by adding water inflow and subtracting water outflow. Each month I took the net water balance from the previous month and added water inflow and subtracted water outflow. By completing the chart and observing the graph in the book, I found that the max water value was in July and the minimum water value was in January of 1994.

Over what time interval(s) is the amount of water increasing?

The water increases from March to July

Over time interval(s) is the amount of water decreasing? **

Water decreases from Jan-Feb of 93 and again decreases from July - Jan 1994

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**** When was the quantity of water increasing fastest, and when most slowly? Describe in terms of the behavior of the two curves.

The water increased the fastest in the month of May because the inflow curve is much more than the outflow curve.

The water increased the slowest in October (when it is actually the most water lost) because the outflow curve is much more than the inflow curve.

What aspect of which graph gives you the rate at which water is flowing out of the reservoir?

The water is flowing out when the outflow curve is higher than the inflow curve. So when the outflow graph is at its largest y value over the y value of the inflow graph, water is flowing out at a higher rate.

What has to be true of the two graphs in order for the amount of water in the reservoir to the increasing at an increasing rate?

For water to be increasing at an increasing rate, the y value of the inflow curve has to be higher and higher than the corresponding y value of the outflow curve as we move left to right.

What has to be true of the two graphs in order for the amount of water in the reservoir to the increasing at a decreasing rate?

For water to be increasing at a decreasing rate, the inflow rate has to be greater than the outflow rate but the gap between the inflow and outflow has to be closing in. So as we move left to right the y value of the inflow curve is less and less higher than the y value of the outflow curve.

What has to be true of the two graphs in order for the amount of water in the reservoir to the decreasing at an increasing rate?

If the water is to be decreasing at an increasing rate, the water has to be going out faster and faster than the water is coming in. So the outflow graph has to have a higher y value than the inflow graph does and this space between the outflow and the inflow has to be getting greater and greater. So as we move left to right the y value of the outflow is increasing and the y value of the inflow is decreasing.

What has to be true of the two graphs in order for the amount of water in the reservoir to the decreasing at a decreasing rate? **

For the water in the lake to be decreasing at a decreasing rate, the outflow graph has to be more than the inflow graph, so the outflow must have a higher y value than the inflow, but the gap between the 2 has to be getting less and less (closing in). As we move left to right the y value of the outflow is decreasing and the y value of the inflow is increasing, but the outflow curve is still above the inflow curve.

confidence rating #$&*: 3

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Given Solution:

** Between any two dates the corresponding outflow is represented by the area under the outflow curve, and the inflow by the area under the inflow curve.

When inflow is greater than outflow the quantity of water in the reservoir will be increasing and when the outflow is greater than the inflow quantity of water will be decreasing.

We see that the quantity is therefore decreasing from January 93 through sometime in late February, increasing from late February through the beginning of July, then again decreasing through the end of the year.

The reservoir will reach a relative maximum at the beginning of July, when the outflow rate overtakes the inflow rate.

The amount of water lost between January and late February is represented by the difference between the area under the outflow curve and the area under the inflow curve. This area corresponds to the area between the two graphs. The amount of water gained between late February and early July is similarly represented by the area between the two curves. The latter area is clearly greater than the former, so the quantity of water in the reservoir will be greater in early July than on Jan 1.

The loss between July 93 and Jan 94, represented by the area between the two graphs over this period, is greater than the gain between late February and early July, so the minimum quantity will occur in Jan 94.

The rate at which the water quantity is changing is the difference between outflow and inflow rates. Specifically the net rate at which water quantity is changing is

net rate = inflow rate - outflow rate.

This quantity is represented by the difference between the vertical coordinate so the graphs, and is maximized around late April or early May, when the inflow rate most greatly exceeds the outflow rate. The net rate is minimized around early October, when the outflow rate most greatly exceeds the inflow rate. At this point the rate of decrease will be maximized. **

** When inflow is > outflow the amount of water in the reservoir will be increasing. If outflow is < inflow the amount of water will be decreasing.

Over what time interval(s) is the amount of water increasing?

Over time interval(s) is the amount of water decreasing? **

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**** When was the quantity of water increasing fastest, and when most slowly? Describe in terms of the behavior of the two curves.

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18:47:03

The curve increases most between Jan and Apr and it decreases most between

July and October

** What aspect of which graph gives you the rate at which water is flowing into the reservoir?

What aspect of which graph gives you the rate at which water is flowing out of the reservoir?

What has to be true of the two graphs in order for the amount of water in the reservoir to the increasing at an increasing rate?

What has to be true of the two graphs in order for the amount of water in the reservoir to the increasing at a decreasing rate?

What has to be true of the two graphs in order for the amount of water in the reservoir to the decreasing at an increasing rate?

What has to be true of the two graphs in order for the amount of water in the reservoir to the decreasing at a decreasing rate? **

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18:47:04

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Self-critique (if necessary):

### I hope I answered all of the components of this question. I feel pretty confident but let me know if I am wrong or missed anything???

Your answers are good.

Note, as I believe you understand, that the greater rate of increase occurs when the inflow is higher than the outflow, and the gap between the two is greatest, which appears to happen around late April or early May.

Think through the other questions, just to be sure, and let me know if there's anything you don't think you understand.

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Self-critique Rating:

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Question:

Section 6.2, Problem 5 [[6.2.5 (previously 6.2 #26)]] antiderivative of f(x) = x^2, F(0) = 0

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Your solution:

The antiderivative of f(x) =x^2 would be 1/3x^3 + C which can be checked to confirm that the derivative of 1/3x^3 +C is x^2. Evaluated at F(0) is 1/3(0)^3 +C= 0 so C=0 so the final answer is F(x)= 1/3x^3+0

confidence rating #$&*: 3

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Given Solution:

** An antiderivative of x^2 is x^3/3.

The general antiderivative of x^2 is F(x) = x^3/3 + c, where c can be anything. There are infinitely many possible specific antiderivative.

However only one of them satisfied F(0) = 0. We have

F(0) = 0 so 0^3/3 + c = 0, or just c = 0.

The antiderivative that satisfies the conditions of this problem is therefore F(x) = x^3/3 + 0, or just F(x) = x^3/3. **

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18:47:58

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Self-critique (if necessary):

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Self-critique Rating:3

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Question:

Section 6.2, Problem 8 [[(previously 6.2 #56)]] indef integral of t `sqrt(t) + 1 / (t `sqrt(t)) **** What did you get for the indefinite integral?

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Your solution:

I began by re-writing the square roots as powers

Indef intergral of t*t^1/2 + t^-1*t^-1/2

Indef intergral of t^3/2 + t^-3/2

I then thought about what the integral would be and found the answer below

2/5t^5/2 -2t^-1/2 +c

Which can be confirmed by taking the derivative which would give you t^3/2 + t^-3/2.

confidence rating #$&*: 3

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Given Solution:

** The function can be written t^(3/2) + t^(-3/2). Both are power functions of the form t^n. Antiderivative is

2/5 * t^(5/2) - 2 t^(-1/2) + c or

2/5 t^(5/2) - 2 / `sqrt(t) + c. **

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11:39:51

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Self-critique (if necessary):

Ok except I didn’t put parenthesis around by fraction powers.

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Self-critique Rating:

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Question:

6.2.9 (previously 6.2 #50) definite integral of sin(t) + cos(t), 0 to `pi/4

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Your solution:

When taking a definite integral you have to first find the antiderivative and then evaluate at the given points. The antiderivative of sin(t) would be -cos(t) which we can verify by taking the derivative of -cos(t) which is -(-sin(t)) which is sin(t). The antiderivative of cos(t) is sin(t) which we can verify by taking the derivative of sin(t) which is cos(t)

So the antiderivative of the function is -cos(t)+sin(t)+c

We now evaluate the antiderivative at pi/4 -cos(pi/4) + sin(pi/4)=0

We now evaluate the antiderivative at 0 -cos(0) + sin(0)= -1

So to find the definite integral we say 0-(-1) which is 0+1=1

So the answer is 1 +c

confidence rating #$&*: 3

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Given Solution:

** An antiderivative is -cos(t) + sin(t), as you can see by taking the derivative.

Evaluating this expression at `pi/4 gives -`sqrt(2)/2 + `sqrt(2)/2 = 0. Evaluating at 0 gives -1 + 0 or -1. The antiderivative is therefore 0 - (-1) = 1. **

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Self-critique (if necessary):

## ok except I put c at the end of the antiderivative which I think is correct???

The c isn't correct, for the following reason:

It is true that an antiderivative of sin(t) + cos(t) has the general form

-sin(t) + cos(t) + c.

So when evaluated at pi/4 you would get

-sin(pi/4) + cos(pi/4) + c

and when evaluated at 0 you would get

-sin(0) + cos(0) + c.

However when you find the change in the antiderivative, which is the definite integral, you subtract these values. That gives you

(0 + c) - (-1 + c) = 0 + c + 1 - c = 1.

The c's subtract out.

This will always be the case with the integration constant in a definite integral. So when doing a definite integral it's always safe to leave the integration constant out.

This is also the reason that any antiderivative will work when calculating the definite integral, as long as you evaluate the same antiderivative at both points.

You do need of course to include an integration constant when doing an indefinite integral.

&&&&I understand what you are saying, the c integration constant cancels out on definite integrals but is always included in indefinite integrals. &&&&

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Self-critique Rating:

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Question:

Why doesn't it matter which antiderivative you use?

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Your solution:

In this problem it doesn’t matter which value you use for C because you are finding the -cos T and the Sin t for a specific number in this case pi/4 and 0. So whichever value you use for C would be the same for pi/4 and 0.

confidence rating #$&*: 2

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Given Solution:

** General antiderivative is -cos(t) + sin(t) + c, where c can be any number. You would probably use c = 0, but you could use any fixed value of c.

Since c is the same at both limits of the integral, it subtracts out and has no effect on the value of the definite integral. **

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Self-critique (if necessary):

### Is my answer correct, I am a little confused and want to make sure I am understanding correctly???

I'm not sure you completely understood this when you answered. However I believe you'll clarify the point from my preceding note.

&&&&I do understand this concept now. &&&&

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Self-critique Rating:3

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Question:

6.2.13 (previously 6.2 #60) The average of v(x) = 6/x^2 on the interval [1,c} is 1. Find the value of c.

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Your solution:

We first have to find the antiderivative of 6/X^2. The first thing I do is rewrite that as 6x^-2. To find the antiderivative I figured -6x^-1 which I then verified by taking the derivative which would give me 6x^-2 so I confirm the antiderivative is correct. -6X^-1 can be written as -6/x

So now we want to know the average on interval (1,C) for -6/x.

The interval at 1 is -6/1=-6

The interval at c is -6/C

The average interval at 1- interval at C

-6 - (-6/C) = 1-C

-6+6/C=1-C now we multiply by c to get the c off the denominator

-6C + 6= c - c^2 now move everything to the right

C^2-7c+6=0 now factor

(c-1)(c-6) = 0

Solve each one

C=1 c= 6

We don’t know which of these C is so we start substituting in to see if one doesn’t work

Int. 1 - int C

-6-6=-12

-6-1=-7

At this point I’m not sure what to do since it appears either value of c will work.

confidence rating #$&*:

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Given Solution:

An antiderivative of 6 / x^2 is F(x) = -6 / x.

The definite integral is equal to the product of the average value and the length of the interval. In this case average value is 1 and the interval from x = 1 to x = c has length c - 1. So the definite integral must be 1 * ( c - 1).

Evaluating between 1 and c and using the above fact that the result must be 1 we get

F(c) - F(1) = -6/c- (-6/1) = c - 1 so that

-6/c+6=c - 1. We solve for c, first getting all terms on one side:

c - 7 + 6/c = 0. Multiplying both sides by c to get

c^2 - 7 c + 6 = 0. Either be factoring or the quadratic formula we get

c = 6 or c = 1.

If c = 1 the interval has length 0 and the definite integral is not defined. This leaves the solution

c= 6.

STUDENT QUESTION

I had some trouble with this problem

I got -6/x for antideri. So I thought that at F(1) = -6 and F(1.5) = -4

Then I got really confused for some reason used the logic

F(b)- F(a) = -4 - (-6) = 2 when divided by 2 = 1.

I see what you did but not so sure about the logic.

INSTRUCTOR RESPONSE

Note that the length of the interval between x = 1 and x = 1.5 is .5. The integral is 2, but the average value between x=1 and x=1.5 is (integral) / (length of interval) = 2 / .5 = 4, not 2.

The average value of the integral must be 1.

The integral of a function over an interval is equal to its average value over that interval, multiplied by the length of the

interval:

• ave value = definite integral / length of interval

It follows immediately that

• definite integral = ave value * length of interval

In this case the interval has length (c - 1) and the average value must be 1.

The integral must therefore be 1 * (c - 1).

The integral is from x = 1 to x = c. So

The integral of 6/x^2 from x = 1 to x = c must equal 1 * (c - 1).

The integral of 6/x^2 from x = 1 to x = c is -6 / c - (-6 / 1) = 6 - 6/c.

Thus 6 - 6/c = 1 * (c - 1).

We solve to get c, and we obtain c = 6.

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Self-critique (if necessary):

## I got all of the work to where c could either be 1 or 6 But I don’t understand the part below

“The definite integral is equal to the product of the average value and the length of the interval. In this case average value is 1 and the interval from x = 1 to x = c has length c - 1. So the definite integral must be 1 * ( c - 1)”

I thought you subtracted the top interval from the bottom interval which would be int c - int 1 which would be either 6-(-6) or 1-(-6) I don’t understand what I am doing wrong????

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Self-critique Rating:

On the interval [1, c), c is greater than 1. The length of the interval is therefore c - 1, not 1 - c.

The definite integral from 1 to c is -6 / C - 6, not 6 - (-6 / C).

So the definite integral must be 1 * (c - 1), and this must be equal to 6 - (-6 / c). Your equation would then read

6 - (-6 / C) = 1 ( c - 1).

The two sides of this equation are negatives of the two sides of your equations, so the solutions are identical.

It was just the reversal of signs that appears to have confused your interpretation.

&&&&I understand now thank you for clarifying. &&&&

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Question:

6.2.14 (previously 6.2 #44) What is the indefinite integral of e^(5+x) + e^(5x)

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Your solution:

When taking the indefinite integral of e^(5+x) + d^(5x) you begin by taking the intergral of each separate piece.

Integral of e^(5+x) Integral of E^5x

Use formula e^u * du =e^u +c use formula e^u * du =e^u +c

Let u = 5+x let u = 5x

Du = 1 du=5x^0 which is 5

E^(5+x)*1=e^(5+x) e^5x*5=E^5x

E^(5+x) + c is antiderivative divide by 5

1/5*e^5x + c is antiderivative

Now put the two together to find the indefinite integral which is

E^(5+x) + 1/5e^5x + C

confidence rating #$&*: 3

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Given Solution:

** The derivative of e^(5+x) is, by the Chain Rule, (5+x)' * e^(5+x) = 1 * e^(5 + x) = e^(5 + x) so this function is its own antiderivative.

The derivative of e^(5x) is (5x) ' * e^(5x) = 5 * e^(5x). So to get an antiderivative of e^(5x) you would have to use 1/5 e^(5x), whose derivative is e^(5x). **

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Self-critique (if necessary):

### I believe I got the correct answer, please let me know if I did not I didn’t see a combined answer with a + C#####

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Self-critique Rating:3

Your answer is correct.

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

#*&!

You've done very well here. You clearly have excellent ability and your background appears solid overall.

As is the case just about any time when you transition from a first-semester course at one institution to a second-semester course at another, you will have covered some topics we didn't during our first-semester course, we will have covered some you didn't.

The concept of continuity is one such concept.

Our course might also have had more emphasis on graphical representations, though you appear to have done pretty well on problems that involved such representations.

Be sure you review continuity, as I suggested, and revise your answers to the questions that rely on that concept.

I've inserted a number of notes, many of which don't necessarily require a response on your part. However responses would be appropriate to my direct questions, and to anything on which you still have a question.

I don't know if your first-semester course included an introduction to integration comparable to Chapter 5 in this text. Some courses do and some concentrate on differentiation through the entire first semester. You should thoroughly review Chapter 5 and work through anything you aren't sure you understand. If you think it necessary, you're more than welcome to submit any parts of any Chapter 5 assignments of which you aren't confident. You can find those assignments at the Calculus I homepage, at

http://vhcc2.vhcc.edu/cal1fall/frames%20pages/Mth%20173%20Homepage.htm

On the whole, I believe you're off to a very good start in this course and I'm looking forward to working with you throughout.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

&&&&Thank you for all of your input and help. I will work very hard to master the concepts of this course so that I can make a good grade. It has been a while since my calculus 1 class so I am a little rusty but I am willing to put in the time and effort to do well in this course. I see we have access to the test bank. Are we allowed to work through some of those questions and submit to you for suggestions on improvement before the test??? &&&&

@&
Absolutely.
*@

@&
Very good.

Let me know if I've missed anything.

As I've already said you appear to be off to a very good start in this course. You appear to be doing all the right things to brush up, when that is necessary, and to master the concepts.

I do look forward to continuing to work with you.
*@

Query 1 revised

@& Good. *@

@& Absolutely. *@

@& Very good. Let me know if I've missed anything. As I've already said you appear to be off to a very good start in this course. You appear to be doing all the right things to brush up, when that is necessary, and to master the concepts. I do look forward to continuing to work with you. *@

@&
Good.
*@

@&
Absolutely.
*@

@&
Very good.

Let me know if I've missed anything.

As I've already said you appear to be off to a very good start in this course. You appear to be doing all the right things to brush up, when that is necessary, and to master the concepts.

I do look forward to continuing to work with you.
*@