Assignment 0

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course Mth 279

Sorry if this assignment is late. I have a full course load here at Virginia Tech as well as the mix-up in the mail with my course DVD's has me a little behind in the course material.

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 derivative of 3sin(4t+2): 12cos(4t+2)

=3[d/dt sin(4t+)]=d/dt(sin(4t+2))=(dsin(u)/du)/(du/dt)

u=4t+2 dsin(u)/du=cos(u)

=3(cos(4t+2)(d/dt(4t+2))) = 3cos(4t+2)[4(d/dt(t))+(d/dt(2))]

=3cos(4t+2)[4(1)+0]

=3cos(4t+2)[4]

Second derivative of 3 sin(4t+2): 48(-sin(4t+2))

=12[d/dt cos(4t+2)]=d/dt(cos(4t+2))=(dcos(u)/du)/(du/dt)

u=4t+2 dcos(u)/du= -sin(u)

=12(-sin(4t+2))[4]

First derivative of 2cos^2(3t-1): -12cos(3t-1)sin(3t-1)

=2[d/dt(cos^2(3t-1))]=(du^2/du)(du/dt)

u=cos(3t-1) du^2/du=2u

=2[2cos(3t-1)(d/dt(cos(3t-1))]

u=3t-1 dcos(u)/du= -sin(u)

=4cos(3t-1)[-sin(3t-1)(d/dt(3t-1))]

= -4cos(3t-1)sin(3t-1)[3]

Second derivative of 2cos^2(3t-1): -36cos(2-6t)

First derivative of Asin(omega*t+phi): omega/[sqrt(1-(omega*t+phi)^2)]

=sin-1(omega*t+phi) [dsin^-1(u)/du](du/dt)

u=omega*t+phi [dsin^-1(u)/du]=1/(sqrt(1-(omega*t+phi)^2))

=(d/dt(omega*t+phi))/[(sqrt(1-(omega*t+phi)^2))]

Second derivative of Asin(omega*t+phi):

[(-omega*t*phi)-(phi^2)+1]/[(1-((omega*t)+(phi^2))^(3/2)]

@& The derivative of A sin(omega t + phi) is

A * ( (omega t + phi) ' ) * cos(omega t phi) ) = omega A cos(omega t + phi).

This is an application of the chain rule, with 'inner function' z = omega t + phi and 'outer function' sin(z).

The second derivative is -omega^2 A sin(omega t + phi).*@

First derivative of 3e^(t^2-1): 6e^((t^2)-1)(2t)

=3[d/dt(e^((t^2)-1))

u=t^2-1 du=e^u

=3[e^((2t)^2-1)(d/dt(t^2-1))

=3e^((2t)^2-1)(2t)

Second derivative of 3e^(t^2-1): ?unsure how to complete this 2nd order derivative

@& Apply the product rule (f g) ' = f ' g + f g ' with f(t) = 6 e^(t^2 - 1) and g(t) = t.*@

confidence rating #$&*: out of 3

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Given Solution:

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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:

The graph is in the shape of a standard sin curve however it repeatedly extends between 3 and -3. I used a chart of t values and plugged them into the function to find corresponding y values. I think that the portion of the equation within the sin function will affect the width of each repetition of the sin curve but I’m not sure on this fact.

@& You've got some good insights but should understand the following, based on knowledge of precaclulus (or trig-analysis):

Using shifting and stretching transformations, as covered in a standard precalculus class, we find that

A cos(omega t + phi)

has amplitude a, period 2 pi / omega and is shifted -phi / omega in the t direction from the graph of A cos(omega t).*@

confidence rating #$&*:ut of 3

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Given Solution:

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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:

A will affect the height of the cos curve, and I think that omega and theta_0 have an effect on the width of the curve repetitions. If so, omega will effect it to a greater extent as it is a multiplier of t while theta_0 is just in addition.

@& You've got some good insights but should understand the following, based on knowledge of precaclulus (or trig-analysis):

Using shifting and stretching transformations, as covered in a standard precalculus class, we find that

A cos(omega t + phi)

has amplitude a, period 2 pi / omega and is shifted -phi / omega in the t direction from the graph of A cos(omega t).*@

confidence rating #$&*: out of 3

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Given Solution:

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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:

f(t)=e^(-3t) = (e^(-3t))/(-3loge)+Constant

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

u=4pit+pi/4

du=4pi dt

=1/2piintegral(sin(u))du = [-cos(u)/2pi]+Constant

= -(cos(pi(4t+1/4)))/(2pi)+Constant

y(t)=1/(3x+2) = (log3x+2)/3

@& That would be the natural log of e in the denominator. Since the natural log is the inverse of the exponential function, ln(e) = 1. So your result is -1/3 e^(-3 t) + c, which is correct.*@

confidence rating #$&*:ut of 3

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Given Solution:

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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:

I completely forget how to approach these types of problems. Amount of time since my last Calculus course is causing me issues…

@& You need a brush-up on some things but don't appear to be in bad shape, assuming you do some reviewing.

For example:

The general antiderivative of e^(-3 t) is -1/3 e^(-3 t) + c.

When t = 0 this is -1/3 e^0 + c = -1/3 * 1 + c = -1/3 + c.

If the value is 2, then we solve -1/3 + c = 2 to get c = 7/3.

Our function is therefore -1/3 e^(-3 t) + 7/3.

This sort of process is very important in your course. You should complete the next two yourself, using similar reasoning.*@

confidence rating #$&*:ut of 3

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Given Solution:

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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:

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

Substituting t=-1 B=-1/2

= [(5/2)/(t-3)] + [(-1/2)/(t+1)]

confidence rating #$&*:ut of 3

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Given Solution:

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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:

Best estimate = 5.2

confidence rating #$&*: out of 3

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Given Solution:

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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:

I graphed the given points and found a curve and a slopes on this curve, however I am not sure how to go about finding the derivative of this curve without knowing the equation of the curve.

confidence rating #$&*:ut of 3

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Given Solution:

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Self-critique rating:

@& You weren't asked for an exact value but an estimate.

Based on your curve and other information what is your best estimate? A good starting point would be to find the slopes between the points, and see what your curve tells you about how the requested derivative is related to those slopes.*@

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#*&!

@& You're not in bad shape to start out, but probably do need some review. Be sure to check out the recommended resources on the Assignments page for your course.

I recommend that you submit a revision to your work on this exercise. You got most of the problems but you will be served well by spending a little more time on some of them, guided by my notes.

&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions 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.

If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.

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