qa_00_Resubmission

#$&*

course Mth 279

6/14 around 1:30 am. Resubmitted upon request with instructor notes and corrections.

qa_00#$&*

course Mth 279

6/9 around 9:15pm

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Question: `q001. Find the first and second derivatives of the following functions:

3 sin(4 t + 2)

2 cos^2(3 t - 1)

A sin(omega * t + phi)

3 e^(t^2 - 1)

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Your solution:

#### First derivatives:

12*cos(4t - 2)

-12*cos(3t-1)*sin(3t-1) = -6* cos (6t-2)

A*t*cos(omega*t + phi)

6t * e^(t^2-1)

Second derivatives:

-48*sin(4t - 2)

36*sin (6t - 2)

-A*t^2*sin(omega*t + phi)

12t^2 * e^(t^2 - 1)

You've got errors on a number of your derivatives, and you've made some incorrect steps in simplification.

You need to indicate your steps and your thinking.

&&&& First Derivatives:

1.) 12 cos(4 t + 2)

2.)-12 cos(3 t - 1) * sin(3 t - 1)

3.) A*omega cos( omega * t + phi)

4.) 6t * e^(t^2 - 1)

Second Derivative:

1.) -48 sin(4 t + 2)

2.) -12[ cos(3t- 1) * 3 cos(3t - 1) + sin(3t - 1) * -3 sin(3t - 1)] using chain rule,

which simplifies to = -36 [ cos^2 (3t - 1) - sin^2 (3t - 1) ]

3.) -A * omega^2 * sin( omega * t + phi)

4.) 6 * e^(t^2 - 1) + 12 * t^2 * e^(t^2 - 1), using chain rule

which simplifies to = (6 e^(t^2 - 1)) * (1 + 2 t^2)&&&&&

confidence rating #$&*:#2

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

####I am not one-hundred percent sure on some of those, since it has been a while since I have done this and am a little rusty, but I understand most of the concepts.

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Self-critique rating:####3

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Question:

`q002. Sketch a graph of the function y = 3 sin(4 t + 2). Don't use a graphing calculator, use what you know about graphing. Make your best

attempt, and describe both your thinking and your graph.

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Your solution:

####Since the graph is being multiplied by 3 on the outside it will oscillate completely through three times between 0 and 2*pi. As well it will start at around y = 2, because two is being added to it on the inside.

confidence rating #$&*:#1

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

####I am confused on what the (4t + 2) does to alter the graph. I cannot recall what it does to change it??????????

You need to view the Introductory Disk, which summarizes the behavior of functions of this nature. Surprisingly few students understand these functions at the beginning of this course.

&&&& The amplitude of the graph is 3 units, meaning that it will have a maximum of 3 and a minimum of -3, because it is being multiplied by 3 on the outside.

The period of the function will be one-fourth of an original sine graph, because the variable ,t, is being multiplied by 4. Therefore the function will cycle through 4 times as fast as a normal sine graph, meaning it will cycle through every (pi/2) since a normal sine graph cycles through every 2*pi.

The graph will also be shifted two units to the left because 2 is being added inside the parentheses. It works in reverse of what is expected, and any number added will cause the graph of the function to be shifted that many units to the left, and any number subtracted would shift it to the right

@&
Close, but the function is of the form f(t + 1/2), not f(t+2), with f(t) = 3 sin(4t+2).

f(t+2) would be 3 sin(4t + 10).

This is based on the principal that f(t+c) is shifted -c units in the t direction, relative to the function f(t).
*@

&&&&

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Self-critique rating:####1

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Question:

`q003. Describe, in terms of A, omega and theta_0, the characteristics of the graph of y = A cos(omega * t + theta_0) + k.

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Your solution:

####The graph will oscillate completely through A times, while it will start at y = theta_0 and the peaks will be omega apart.

confidence rating #$&*:#1

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

####This is the same as the previous problem. I am having difficulty recalling what each does to alter the graph.

&&&& The graph will have an amplitude of A, and if the value of A is negative, the graph will be flipped.

The graph will have a period of (2*pi)/omega.

The graph will also be shifted theta_0 units, to the left if theta_0 is positive and to the right if it is negative.

@&
If f(t) = A cos(omega * t), this function is f(t + theta_0 / omega), not f(t + theta_0).

Be sure to ask if you don't understand this.
*@

The graph will be shifted vertically k units, upward if k is positive and downward if k is negative.&&&&

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Self-critique rating:####1

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Question:

`q004. Find the indefinite integral of each of the following:

f(t) = e^(-3 t)

x(t) = 2 sin( 4 pi t + pi/4)

y(t) = 1 / (3 x + 2)

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Your solution:

#### The integrals are

(e^-3t)/(-3) + C

-8*pi cos( 4*pi t + 4/pi) + c

1/3 ln(3x + 2) + C

You need to take the derivatives of your results to check that your integration is correct.

On two of the three problems, it is.

&&&& Integral of x(t) = -1/(2*pi) cos( 4 pi t + pi/4) + C&&&&

confidence rating #$&*:#3

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

####Ok

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Self-critique rating:####3

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Question:

`q005. Find an antiderivative of each of the following, subject to the given conditions:

f(t) = e^(-3 t), subject to the condition that when t = 0 the value of the antiderivative is 2.

x(t) = 2 sin( 4 pi t + pi/4), subject to the condition that when t = 1/8 the value of the antiderivative is 2 pi.

y(t) = 1 / (3 t + 2), subject to the condition that the limiting value of the antiderivative, as t approaches infinity, is -1.

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Your solution:

#####

For f(t), the integral is (e^-3t)/(-3) + C. At t=0, 1 + C = 2, so C = 1. Therefore the antiderivative is (e^-3t)/(-3) + 1.

For x(t), the integral is -8*pi cos( 4*pi t + 4/pi) + c. At t= 1/8, -8*pi cos( pi/2 + 4/pi) + c = 2pi. So the antiderivative is -8*pi cos( 4*pi t + 4/pi) - 1/[4(cos( pi/2 + 4/pi))].

For y(t), the integral is 1/3 ln(3t +2) + C. As t approaches infinity, the limiting value is -1 so c = -1. The antiderivative is 1/3 ln(3t + 2) -1.

confidence rating #$&*:#3

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

####I think I pretty much understand how to do the first two, but I am confused on how to do the third one. Could you explain it to me???????

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Self-critique rating:####2

You're applying the conditions correctly, except on one question. That is the one on which your general anti-derivative is also incorrect.

&&&& for x(t), the integral is -1/(2*pi) cos( 4 pi t + pi/4) + C. At t= 1/8, -1/(2*pi) cos((3 pi)/4) + C = 2*pi.

From there you multiply over by -2 and divide by pi to get, cos ((3 pi)/4) + C = -4

Using the unit circle, you can conclude that cos((3 pi) / 4) is equal to -1/sqrt(2). Divide both sides by that to get, C = 4 * sqt(2).

Therefore the anti-derivative is, -1/(2*pi) cos( 4 pi t + pi/4) + 4 * sqrt(2). &&&&

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Question:

`q006. Use partial fractions to express (2 t + 4) / ( (t - 3) ( t + 1) ) in the form A / (t - 3) + B / (t + 1).

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Your solution:

#### To solve for A and B, you should multiply everything by the denominator ((t-3)(t-1)).

When you do that, you get (2t-4) = A(t+1) + B(t-3). From here you solve it at both x = -1 and x = 3.

At x = -1:

(2*(-1) + 4) = A*(0) + B * (-4), which becomes 2 = -4B, and then solve for B to get B = -1/2

At x = 3

(2*(3) + 4) = A*(4) + B*(0), which becomes 10 = 4A, and then solve for A to get A = 5/2

Therefore, the partial fraction becomes 5/(2 * (t-3)) - 1/(2 * (t + 1)).

confidence rating #$&*:#3

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

####Ok

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Self-critique rating:####3

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Question:

`q007. The graph of a function f(x) contains the point (2, 5). So the value of f(2) is 5.

At the point (2, 5) the slope of the tangent line to the graph is .5.

What is your best estimate, based on only this information, of the value of f(2.4)?

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Your solution:

#### The tangent line at the point is y-5 = .5(x-2).

After multiplying the .5 out and bringing over the 5, it becomes y = .5x + 4.

Then plug in x = 2.4 to get .5(2.4) + 4, which equals 5.2.

confidence rating #$&*:#3

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

####I am pretty sure I did this one right, assuming I found the equation in the appropriate manner.

Good. You could also have multiplied the slope by the 'run' to get .5 * .4 = .2, which is the change in the y value.

Self-critique ####3

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Question:

`q008. The graph of a function g(t) contains the points (3, 4), (3.2, 4.4) and (3.4, 4.5). What is your best estimate of the value of g ' (3), where the ' represents the derivative with respect to t?

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Your solution:

####The average slope of the graph is (4.5-4) / (3.4 - 3), which goes to .5/.4 = 1.25

Since the basic form of a line is y = mx + b, with m being the slope, the derivative of that line is y' = m. Therefore g'(3) will equal 1.25.

confidence rating #$&*:#2

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

####I am not sure that assuming it was a straight line and finding the slope was the correct way to solve this problem.

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Self-critique rating:####2.""

The slope is clearly changing. Your calculation gives the slope of a secant line, without considering the fact that the slope changes.

&&&&The slope between the two points (3,4) and (3.2, 4.4) on the graph is (4.4 - 4) / (3.2 - 3) = .4/.2, which comes out to be a slope of 2. Therefore my best estimate for the value of g'(3) would be 2.&&&&

Not bad overall, but you did make some errors that you don't want to be making later on. So for your future good, I'm asking for a revision.

&#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.

*@"

Self-critique (if necessary):

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Self-critique rating:

The slope is clearly changing. Your calculation gives the slope of a secant line, without considering the fact that the slope changes.

Not bad overall, but you did make some errors that you don't want to be making later on. So for your future good, I'm asking for a revision.

&#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.

*@"

Self-critique (if necessary):

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Self-critique rating:

#*&!

qa_00_Resubmission

@& Close, but the function is of the form f(t + 1/2), not f(t+2), with f(t) = 3 sin(4t+2). f(t+2) would be 3 sin(4t + 10). This is based on the principal that f(t+c) is shifted -c units in the t direction, relative to the function f(t). *@

@& If f(t) = A cos(omega * t), this function is f(t + theta_0 / omega), not f(t + theta_0). Be sure to ask if you don't understand this. *@

@& Good work, but you still have one misconception, so be sure to see my notes. *@

@&
Close, but the function is of the form f(t + 1/2), not f(t+2), with f(t) = 3 sin(4t+2).

f(t+2) would be 3 sin(4t + 10).

This is based on the principal that f(t+c) is shifted -c units in the t direction, relative to the function f(t).
*@

@&
If f(t) = A cos(omega * t), this function is f(t + theta_0 / omega), not f(t + theta_0).

Be sure to ask if you don't understand this.
*@

@&
Good work, but you still have one misconception, so be sure to see my notes.
*@