Assignment 1

course MTH 174

tNRף׻پeassignment #001

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Physics II

06-15-2008

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02:59:30

What was your value for the integral of f'?

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RESPONSE -->

0

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02:59:54

What was the value of f(0), and of f(7)?

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RESPONSE -->

f(0) = f(7) = -1

Two principles can be used to solve this problem?

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.**

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03:03:36

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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RESPONSE -->

f(0) = -1 and is decreasing/concave up to (1,0) where it is increasing/concave up to (2,1). it is increasing/concave down to (3,0) and is increasing/concave up to (4,2), then it is increasing/concave down to (5,0) where it goes decreasing/concave down to (6,-2), then it goes decreasing/concave up to (7,-1) where it ends.

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03:03:42

Was the graph of f(x) continuous?

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RESPONSE -->

Yes

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03:11:14

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

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RESPONSE -->

?

** 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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03:11:23

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

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RESPONSE -->

it overlaps at the points where the graph changes concave up or down except the point on the x-axis.

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03:15:58

What were the areas corresponding to each of the four intervals over which f'(x) was constant? What did each interval contribute to the integral of f'(x)?

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RESPONSE -->

2, -1, 2, -4, 1

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03:17:25

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

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03:17:29

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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03:23:01

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.

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RESPONSE -->

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You did not respond to the question, and should therefore have responded with a very detailed self-critique to the given solution, addressing what you do and do not understand about every of the solution. You should analyze the given solution in this manner, phrase by phrase.

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03:25:16

Query Section 6.2 #38

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03:28:03

antiderivative of f(x) = x^2, F(0) = 0

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RESPONSE -->

f'(x) = (x^3)/3+0, this is the only possibility since C=0

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03:28:15

What was your antiderivative? How many possible answers are there to this question?

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03:30:17

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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03:34:11

Query Section 6.2 #55 (3d edition #56) indef integral of t `sqrt(t) + 1 / (t `sqrt(t))

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03:34:57

What did you get for the indefinite integral?

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RESPONSE -->

(2t^(5/2))/5 - 2t^(-1/2)

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03:35:43

What is an antiderivative of t `sqrt(t)?

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RESPONSE -->

(2t^(5/2))/5

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03:36:06

What is an antiderivative of 1/(t `sqrt(t))?

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RESPONSE -->

-2t^(-1/2)

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03:36:23

What power of t is t `sqrt(t)?

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RESPONSE -->

t^(3/2)

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03:36:38

What power of t is 1/(t `sqrt(t))?

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RESPONSE -->

t^(-3/2)

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03:37:02

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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RESPONSE -->

I left out the constant!!!!

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03:41:02

Query Section 6.2 #68 (3d edition #69) def integral of sin(t) + cos(t), 0 to `pi/4

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03:41:12

What did you get for your exact value of the definite integral?

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RESPONSE -->

(-cos(pi/4) + sin(pi/4)+C)-(-cos(0) + sin(0)+C)=1

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03:41:24

What was your numerical value?

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RESPONSE -->

-cos(pi/4)=-sqrt(2/2)

sin(pi/4)=sqrt(2/2)

(-cos(pi/4) + sin(pi/4))=0

-cos(0)=-1

sin(0)=0

(-cos(0) + sin(0))=-1

(-cos(pi/4) + sin(pi/4))-(-cos(0) + sin(0))=1

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03:41:56

What is an antiderivative of sin(t) + cos(t)?

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RESPONSE -->

-cos(t) + sin(t)

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03:43:14

Why doesn't it matter which antiderivative you use?

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RESPONSE -->

Numericaly you get an exact answer, and using the most general antiderivative you get C's on both sides and they will cancel each other out.

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03:43:53

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.

The 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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03:56:11

Query Section 6.2 #82 (#81 3d edition) v(x) = 6/x^2 on [1,c}; find c

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03:56:22

What is your value of c?

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RESPONSE -->

c = 6/5

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03:56:43

In symbols, what did you get for the integral of 6 / x^2 over the interval [1, c]?

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RESPONSE -->

f(c) - f(1) = (-6/c) - (-6/1) = 1

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03:58:51

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

Evaluating between 1 and c and noting that the result must be 1 we get

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

-6/c+6=1. We solve for c:

-6/c=1-6

6/c=-5

-6=-5c

c=6/5.

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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04:00:06

Extra Problem (formerly from Section 6.2 #44): What is the indefinite integral of e^(5+x) + e^(5x)

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RESPONSE -->

e^(5+x) + e^(5x)

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04:00:42

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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RESPONSE -->

** 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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It looks like you know what you are doing overall. You did make some errors, so be sure to see my notes and let me know if you have questions.