Assignment 0

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course MTH 279

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:

1. First : 12cos(4t+2)

Second: -48sin(4t+2)

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

Second: -36(cos^2(3t-1)) + 12(sin^2(3t-1))

3. First: A(omega)cos (Omega *t + phi)

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

4. First: 6t e^ (t^2 -1)

Second: 12t^2 (e^ (t^2 -1))

@&
t is implicitly the variable, as it is in the other expressions.

Of course you could have taken the derivative with respect to A or omega or phi.

In any case your result is correct, as is the case on the other expressions.

However note that you were asked for a solution, which in this course would include at least an outline of your steps and reasoning, and not just an answer (which could, for example, have been obtain
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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I did not understand the form of the 3rd question, I have in fact not seen this denotion before, so I took it to be that “t” was the assigned variable of which to take derivation.

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Self-critique rating:

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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:

This is a sinusoidal graph with a magnitude of 3, its intercepts are moved back 2 units on the x axis and the frequency of the wavelength takes Ľ the time.

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You communicated the nature of the graph just fine. You can always review the terminology.
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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I don’t know all the correct word usage all the time.

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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 is a cosinusoidal wave with an magnitude A values larger, placed negative theta units on the x axis and the frequency (length of wavelength) is 1/omega units

@&
The wavelength of the graph would be the period if t is interpreted as time, and is 2 pi / omega. The frequency is omega / 2 pi.

The horizontal shift occurs on the t axis and is affected by both theta_0 and omega; it would be -theta_0 / omega, not -theta_0.
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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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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:

1. D = e^u * du, thus integral e^u / du

-1/3 (e^(-3t)) + C

2. D = cos (u) *du thus Integral = -cos (u) / du

- ˝ pi (cos (4pi (t) + pi/4)) + C

3. Using u substitution u = 3t +2 du = 3 integral of du/u = ln (u)

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

@&
Good, and it's particularly good that you showed your reasoning.
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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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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:

1. -1/3 (e^(-3t)) + C

t = 0 : -1/3 (e^(-3(0))) + C = 2

C = 7/3

-1/3 (e^(-3t)) + (7/3)

2. - ˝ pi (cos (4pi (t) + pi/4)) + C

t = 1/8 : - ˝ pi (cos (4pi (1/8) + pi/4)) + C = 2 pi

4 C = 8 pi - (sqrt (2)*pi)

C = (8 pi - (sqrt (2)*pi)) / 4

- ˝ pi (cos (4pi (1/8) + pi/4)) + (8 pi - (sqrt (2)*pi)) / 4

3. 1/3 ( ln (3t+2)) + C

limit as t approaches infinity: 1/3 ( ln (3(infinity)+2)) + C = -1

1/3 (ln (3(infinity)+2)) + C = -1

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I did not understand the 3rd question I believe, I am not sure how to find the constant when the limit is truly approaching infinity.

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Self-critique rating:

@&
There's good reason why you were unable to find that value of C.

The log function does not approach a finite limit, so there is no value of C that satisfies the specified condition.

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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:

At + A + Bt - 3B = 2t + 4

A + B = 2

A - 3B = 4

3A + 3B = 6

4A = 10

A = 5/2

B = -1/2

(2 t + 4) / ( (t - 3) ( t + 1) ) = (5/(2(t-3))) - ( 1/ (2(t+1)))

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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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:

Y - Intercept = (0,4)

Y = .5x+4

Y(2) = 5

So y(2.4) = 2.4 (.5) +4 = 1.2+4 = 5.2

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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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 function seems to be a square root function, thus the derivative, the function goes from negative infinity increases at an increasing rate as approaches x=0, where it would be undefined.

@&
You can find the average slope from each point to the next. The progression of these slopes will give you the basis for an estimate of the desired derivative.
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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Assignment 0

@& You communicated the nature of the graph just fine. You can always review the terminology. *@

@& The wavelength of the graph would be the period if t is interpreted as time, and is 2 pi / omega. The frequency is omega / 2 pi. The horizontal shift occurs on the t axis and is affected by both theta_0 and omega; it would be -theta_0 / omega, not -theta_0. *@

@& Good, and it's particularly good that you showed your reasoning. *@

@& There's good reason why you were unable to find that value of C. The log function does not approach a finite limit, so there is no value of C that satisfies the specified condition. *@

@& You can find the average slope from each point to the next. The progression of these slopes will give you the basis for an estimate of the desired derivative. *@

@& Good work. I've inserted a few notes, which you should review. Let me know if you have questions. *@

@&
You communicated the nature of the graph just fine. You can always review the terminology.
*@

@&
The wavelength of the graph would be the period if t is interpreted as time, and is 2 pi / omega. The frequency is omega / 2 pi.

The horizontal shift occurs on the t axis and is affected by both theta_0 and omega; it would be -theta_0 / omega, not -theta_0.
*@

@&
Good, and it's particularly good that you showed your reasoning.
*@

@&
There's good reason why you were unable to find that value of C.

The log function does not approach a finite limit, so there is no value of C that satisfies the specified condition.

*@

@&
You can find the average slope from each point to the next. The progression of these slopes will give you the basis for an estimate of the desired derivative.
*@

@&
Good work.

I've inserted a few notes, which you should review.

Let me know if you have questions.
*@