Q 00

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course Mth 279

6/9/13, 9:42pm

Most students coming out of most calculus sequences won't do very well on these questions, and this is particularly so if it's been awhile since your last calculus-related course.So give it your best shot, but don't worry if you don't get everything.

I'm trying to identify areas on which you might need a refresher, as well as familiarize you with terminology and ideas that might not have been covered in your prerequisite courses.
Most of this is these questions are related to things you don't want to get distracted by when they pop up in your assignments.
Give me your best thinking, and I'll give you feedback, including a lot of additional explanation should you need it.
 

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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: 

Assuming radians

1st derivative: 3 sin(4 t + 2) = 12*cos(4t+2)

2nd derivative: -48*sin(4t+2)

1st derivative: 2cos^2(3 t - 1) = -12sin(3t-1)*cos(3t-1)

2nd derivative: 36 - 72*cos^2(3t-1)

1st derivative: 2cos^2(3 t - 1) = -12sin(3t-1)*cos(3t-1)

 1st derivative: A sin(omega * t + phi) = A*omega*cos(omega*t+phi)

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

1st derivative: 3 e^(t^2 - 1) = 6*t*e^(t^2 - 1)

2nd derivative: (e^t^2)*(12*e^-1*t^2 + 6*e^-1)

 

confidence rating #$&*:

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Not completely sure about all the answers. 

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Given Solution: 

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Self-critique (if necessary):

 OK

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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: 

 The graph looks like a sine wave. The amplitude is 3, the period is pi/2. The y intercept is shifted to the right by a value of 2.

I am basing this on my knowledge of graphing. I may be wrong about which way the graph gets shifted.

 

@&

The argument of the sine function is 4t + 2, which is equal to 4 ( t + 1/2 ). This shifts the graph left 1/2 unit.

Otherwise very good.

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 Confidence rating:

OK

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Given Solution: 

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

Self-critique (if necessary):

 I may be wrong about which direction the wave shifts.

 

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Self-critique rating:

3

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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: Gives magnitude of the amplitude.

Omega: changes the period

theta_0: shifts the wave left or right depending on the sign in front of it.

 

confidence rating #$&*:232; 

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OK

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Given Solution: 

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

Self-critique (if necessary):

 OK

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Self-critique rating:

3

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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^(-3 t) --> -(1/3)*e^(-3t) + c

x(t) = 2 sin( 4 pi t + pi/4) --> (1/(2*pi)) * (-cos(4 pi t + pi/4)) + c

 y(t) = 1 / (3 x + 2) --> t / (3x + 2) + c

 Confidence rating:  

OK

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Given Solution: 

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

Self-critique (if necessary):

 OK

 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: 

I am not sure what to do with this problem. I know how to take the antiderivatives (aka integrals)… do I just set the value of the variable equal to the point given (e.g. t = 0) and then set the whole antiderivative equal to value given (e.g. value of integral is 2)?

@&

Your antiderivative for the first function is

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

When t = 0 this expression has to give you 2.

So set

-1/3 e^(-3t) + c = 2

and solve for c.

The others are done similarly.

*@

confidence rating #$&*:232; 

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

1

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Given Solution: 

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

Self-critique (if necessary):

 I don't know how to answer this.

 Self-critique rating:

1

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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: 

 I don't remember how partial fractions work. I think I remember that the numerator needs to be rewritten, but I don't remember how to accomplish this.

@&

You need to review partial fractions, which occur frequently (and soon) in this course. You can find numerous sources online, or review your precalculus and first-year calculus.

*@

  

confidence rating #$&*:232; 

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

0

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Given Solution: 

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

Self-critique (if necessary):

 I don't know how to do partial fractions. I tried to google how, but I just got confused.

  

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Self-critique rating:

@&

Try Khan Academy. Very good, I highly recommend it.

Keep me posted on your progress.

*@

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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: 

 Given that a line tangent to our function has a slope of 1/2, we can determine that the slope of our function is -2. The slope allows us to calculate the y intercept giving a final function of f(x) = -2x + 9.

f(2.4) = -2*2.4 + 9 = 4.2. f(2.4) = 4.2

@&

The slope of a function at at point is the slope of the tangent line, not its negative reciprocal.

*@

  

confidence rating #$&*:232; OK

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: 

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

Self-critique (if necessary):

 OK

 

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Self-critique rating:

OK

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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 have no idea how to do this.

 Confidence rating:  1

@&

Try starting with a sketch, which I hope you thought to do, then find the slopes of the line segments between the points.

From there make a reasonable conjecture, assuming a smooth function g(t).

*@

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Given Solution: 

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Self-critique (if necessary):

 I'm bad.

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Self-critique rating:

1"

Self-critique (if necessary):

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Self-critique rating:

@&

You'll want to do a revision of this document, to be sure you have some of the basics down. And you'll want to review certain topics.

Do include the process you used to find the integrals and derivatives on the first couple of problems.

&#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.

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