Assn 1

course Mth174

Your access code is 05-00-124

01-22-2008......!!!!!!!!...................................

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

10:57:19

Query Section 6.1 #15 f'=1 on (0,2), -1 on (2,3), 2 on (3,4), -2 on (4,6), 1 on (6,7); f(3) = 0

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RESPONSE -->

Graph of f: slope=1 from (0,-1) to (2,1)

concave down

slope=-1from (2,1) to (3,0)

concave up

slope=2 from (3,0) to (4,2)

concave down

slope=-2from (4,2) to (6,-2)

concave up

slope= 1from (6,-2) to (7,-1)

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

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10:57:34

What was your value for the integral of f'?

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RESPONSE -->

Because the graph begins and ends on the same y axis (-1) the positive and negative cancells

out and becomes -1.

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10:57:51

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

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RESPONSE -->

the value is -1 for both of them

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10:58:40

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 -->

Please see first answer, after question was given.

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10:59:19

Was the graph of f(x) continuous?

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RESPONSE -->

yes

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11:00:19

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

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RESPONSE -->

because f' is the slope which does not connect because there is a sharp bend in f. Where as f is continuous because the sharp bends meet.

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11:00:58

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

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RESPONSE -->

a straight line with the slope of f' over that interval.

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11:01:22

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 -->

The areas were porportional the same on either side (+ or -) of the x axis because the graph

begins and ends with on the same Y=-1

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11:02:41

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.

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RESPONSE -->

okay

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11:08:05

Query Section 6.1 #25 outflow concave up Jan 93 -Sept, peaks Oct, down somewhat thru Jan 94; inflow starts lower, peaks May, down until Jan; equal abt March and late July

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RESPONSE -->

moving on to questions

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11:10:23

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

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RESPONSE -->

the quanitity of water is the greatest when the inflow is greater than the outflow, therefore water is being added to the total quantity, this happened in about May.

The quanity was the least when the outflow was greater than the inflow and overall water was being taken away. This happened in October.

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11:13:49

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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RESPONSE -->

correction: the prevous answer would be more suited to this question. The maxima and minima of quanity would not be the maxima and minima of outflow vs. inflow. The quanity would be the greatest at the end of the high inflow period (july) and lowest in Jan when the graph ends.

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13:36:28

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 -->

This is similar to optimization, the graph of outflow vs. inflow is actually more of an q'', the two compare to give you the rate q'. and that will inturn give you the actuall quantity q

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13:38:48

Query Section 6.2 #38

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RESPONSE -->

f'(x)= 'sqrt(x)

= x^.5

x^n = x^n+1 / n+1

= x^1.5 / 1.5

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13:43:49

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

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RESPONSE -->

f'(x)= x^2

x^n= (x^(n+1))/(n+1)

f(x)= (x^3)/ 3

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13:44:32

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

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RESPONSE -->

Mine was (x^3) / 3 but due to the fact that constants don't show up in a derivative there are an infinate number of antiderivatives.

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13:45:26

What in general do you get for an antiderivative of f(x) = x^2?

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RESPONSE -->

x^n = (x^(n+1)) / (n+1)

= x^3 / 3

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13:47:58

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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RESPONSE -->

I understand the formula for finding an antiderivative of this nature, but I am still alittle unsure about why we use this formula (x^n)= x^(n+1)/ (n+1) is you would further explain why we use this it would help further my understanding of the topic.

For example because taking the derivative reduces the power of a power function by 1, you know for example that to find a function whose derivative is x^2 the power you are looking for must be 3.

However if you take the derivative of x^3 you get 3 x^2, not just x^2.

To get just x^2 you therefore need to take the derivative of 1/3 x^3.

For n = 2, 1/3 x^3 is 1 / (n+1) x^(n+1).

This generalizes:

An antiderivative of x^n must be of power n+1. However the derivative of x^(n+1) is (n + 1) x^n, not x^n.

So an antiderivative of x^n would be 1 / (n+1) * x^n.

This rule generalizes to any power n except n = -1. An antiderivative of x^(-1) is ln(x).

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13:52:38

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

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RESPONSE -->

t 'sqrt(t) + 1/ (t 'sqrt(t))

t (t^.5) + 1 / (t (t^.5))

t ( t^1.5 / 1.5) + 1 / (t^1.5 / 1.5)

Right idea but see the details below:

** The function can be written t^(3/2) + t^(-3/2). Both are power functions.

An antiderivative of t^(3/2) would be 1 power higher. 3/2 + 1 = 3/2 + 2/2 = 5/2.

However the derivative of t^(5/2) would be 5/2 t^(3/2).

So an antiderivative would be 2/5 t^(5/2).

When the power -3/2 is increased by 1 you get the -1/2 power, leading to antiderivative

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

-2 t^(-1/2) =

-2 / t^(1/2) =

-2 / sqrt(t).

An 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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13:52:59

What did you get for the indefinite integral?

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RESPONSE -->

please see the past answer

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13:53:41

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

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RESPONSE -->

t 'sqrt(t) = t (t^.5)

antiderivative = t (t^1.5 / 1.5)

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13:54:09

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

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RESPONSE -->

1/ ( t ( t^1.5 / 1.5))

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13:54:24

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

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RESPONSE -->

t^ .5

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13:55:28

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

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RESPONSE -->

1 / t^.5

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13:58:30

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 -->

how did you figure the t to be 2/5?

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13:59:20

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

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RESPONSE -->

moving on to questions

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13:59:25

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

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RESPONSE -->

f=(sin(t) + cos(t))

f'=(-cos(t) + sin(t))

t= pi/4

f'=0

f= 1 + 0

f= 1

Good. For comparison:

** 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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13:59:30

What was your numerical value?

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RESPONSE -->

1

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13:59:38

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

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RESPONSE -->

(-cost + sint) +c

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13:59:47

Why doesn't it matter which antiderivative you use?

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RESPONSE -->

Because the c when set equal to zero will ""leave the formula alone""

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14:01:03

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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RESPONSE -->

okay

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14:04:42

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

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RESPONSE -->

moving on to questions

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14:04:52

What is your value of c?

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RESPONSE -->

c= 6/5

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14:05:53

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

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RESPONSE -->

f'(x)= 6 / x^2

f(x)= -6 / x

since we're evaluating for the difference between 1 and c and it must equal 1 we get.

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

-6/ c+6= 1

6/c=-5

c=6/5

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14:07:40

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.

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RESPONSE -->

okay

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14:08:12

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

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RESPONSE -->

Since the cos and sin formulas worked out similar to there dirvatives I would assume that the e^n function does also

although this may be wrong. So since the e^n is it's own derivative then I guess that this would be it's own antiderivative.

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14:09:30

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 -->

So the derivative is the antiderivative vice versa, simliar to the e^n function.

except for the 1/5 factor on e^(5x), that is the case. This follows directly from the Chain Rule, as I believe you see.

Be sure you see my note at the beginning about your access code.

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