Assignment1

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course Mth 174

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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

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 aninterval 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 thisentire 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 withrespect 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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18:28:51

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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 [[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: For 0 to 2, the height is 1…so 2*1 is 2. For (2,3), the height is -1, so the area is -1. For (3,4), the height is 2, so the area is 2. For (4,6), the height is -2, so the area is -4. For (6,7), the height is 1, so the area is 1. If you add all those together: 2+-1+2+-4+1=0.

confidence rating #$&*

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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): 3

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Self-critique rating #$&*

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Question: 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: The graph of f(x) would be increasing at (0,2), decreasing at (2,3), increasing at (3,4), decreasing at (4,6) and increasing at (6,7)

confidence rating #$&* 3

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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): ok

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Self-critique rating #$&* ok

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

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Your solution: yes

confidence rating #$&* 3

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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): ok

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Self-critique rating #$&* ok

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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 a step function, so it is broken, therefore, NOT continuous.

confidence rating #$&* 3

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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).

If f ‘ (x) is bounded in some vicinity of x = a, then this condition must be satisfied. Specifically if for | x – a | < epsilon we have | f ‘ | < L, it follows that on this same interval | f(x) – f(a) | < epsilon * L. So as x -> a, f(x) -> f(a) and the function f is continuous at a.

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Self-critique (if necessary): ok

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Self-critique rating #$&* ok

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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) was constant, then the graph of f(x) would be constantly increasing or decreasing, which means it would be linear.

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): ok

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Self-critique rating #$&* ok

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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: I am not sure how to solve this problem.

confidence rating #$&0
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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): ok

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Self-critique rating #$& ok
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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 is (x^3)/3+C, you just add one to the exponent and divide by that number. If you plug in zero for the x value, then the value of c would be 0, so the full formula is (x^3)/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): ok

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Self-critique rating #$&* ok
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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: So I would first change the square roots to ½, so it would be t^3/2 + t^-3/2. Then I would take the integral of each and add them. I end up with:
(t^5/2)/(5/2)+(t^-1/2)/(-1/2)+C

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

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Self-critique rating #$&* ok
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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: the integral for sin(t) + cos(t)= -cos(t) + sin(t). when its evaluated at pi/4 and 0 its:
-cos(`pi/4) + sin(`pi/4) – (-cos(0) + sin(0))
= (-sqrt 2/2)+(sqrt 2/2) – (-1+0)
=0+1
=1

confidence rating #$&*
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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

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Self-critique rating #$&* OK
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Question:
Why doesn't it matter which antiderivative you use?

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Your solution: Both of those terms evaluated at the numbers will end up equaling the same thing either way.

confidence rating #$&* 3
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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): ok

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Self-critique rating #$&* ok
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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: I know I have to get the antiderivative of the function which is: 18/x^2. I am not really sure on what to do after this.

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.

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Self-critique (if necessary): My answer was different than the solution, mainly the antiderivative part. After looking at the solution, I really don’t know why the answer is how it is.

The derivative of 18 / x^2 is -36 / x^3. May be easier to see if you express 18 / x^2 as 18 x^(-2) and the derivative as -2 * 18 x^(-3).

So 18 / x^2 is not an antiderivative of 6 / x^2.

-6 / x is an antiderivative of 6 / x^2. Possibly easier to see if you write -6 / x as -6 x ^ (-1), with derivative -1 * (-6) x^(-2).

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Self-critique rating #$&* ok

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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: I can split these up and just do the integral of each and add them together. The integral of e^(5+x) is e^(5+x). The integral of e^(5x) is e^(5x)/5. Therefore the integral is e^(5+x) + e^(5x)/5

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): ok

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Self-critique rating #$&* ok

&#Be sure to see my note(s), inserted at various places in this document, and let me know if you have questions. &#

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