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course Mth 279
6/8 11:03 pm
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:
1. y' = 12cos(4t+2)
y'' = -48sin(4t+2)
2. y' = -12sin(3t-1)
y'' = -36cos(3t-1)
3. I Only think I know this because it's a physics formula
y' = A*omega*cos(omega*t+phi)
y'' = -A*omega^2*sin(omega*t+phi)
4. y' = 6te^(t^2-1)
y'' = 12t^2e^(t^2-1)
confidence rating #$&*:
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Given Solution: none
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Self-critique (if necessary):
I feel like I got the first and second one right. I’m not very good at differentiating and integrating and frequently get them confused.
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You've got some errors in there, and some are correct.
I'm going to ask you to resubmit your work on the second and third expressions and include the details of your thinking, without skipping any steps. Not all of your answers on those expressions are wrong, but I need to give you more feedback on the process.
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Self-critique rating:0
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Question:
`q002. Sketch a graph of the function y = 3 sin(4 t + 2). Don't use a graphing calculator, use what you know about graphing. Make your best
attempt, and describe both your thinking and your graph.
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Your solution:
The graph goes between 3 and -3. It is a sine curve shifted 2 units to the left. At t = -2, the graph is on the x-axis.
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You didn't mention the period of the function.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
I know the 4t does something but I don't know what
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Self-critique rating:
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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 is the amplitude, theta_0 is the direction the graph is shifted (left if positive, right if negative). I don't know what omega does.
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This is covered in the videos on the Introductory Disk (Disk 0).
You really need to understand these functions for the work we'll be doing during the middle third of the course.
If after viewing the introductory series of videos you still aren't comfortable with trigonometric functions and questions of this nature, I can recommend additional materials.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Question:
`q004. Find the indefinite integral of each of the following:
• f(t) = e^(-3 t)
• x(t) = 2 sin( 4 pi t + pi/4)
• y(t) = 1 / (3 x + 2)
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Your solution:
1. -1/3 e^(-3t) +C
2. 1/(2pi)*cos(4 pi t +pi/4) +C
3. ln (3x+2) +C
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You should be indicating how you arrive at your results, at least in a brief outline that describes the main operations.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
I think the first one is right
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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) +2
2. -1/(2pi) cos(4 pi t +pi/4) + pi/2
3. ln(3t+2) -2
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
I think the first one is right. I completely guessed on the other two
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Self-critique rating:
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Question:
`q006. Use partial fractions to express (2 t + 4) / ( (t - 3) ( t + 1) ) in the form A / (t - 3) + B / (t + 1).
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Your solution:
I know the first step is to multiply the two parts of the denominator (not sure why)
(t-3)(t+1) = t^2-2t-3
Since there are no t^2 terms, we ignore that. A will be the t terms and B the constants
A = -2, B = -3
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
I used to know this really well. That seems way too easy to be the answer though.
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What you have done isn't correct.
Partial fractions is a standard Algebra II topic, and their application to integration is standard in second-semester calculus courses.
However students often manage to come out of those courses not understanding the process. I had a couple of exceptional students in my differential equations class this spring, and they were a little fuzzy on the topic.
You will be using partial fractions a lot in this course, and fairly soon, so review it now. You don't want to be dealing with the differential equations part of the course and stumbling on partial fractions.
There are numerous explanations of partial fractions online.
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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:
I think Euler's method is how you solve this. But I don't remember how to do that.
I think the equation was
y_x = y_0 +h* f(y_(x-1), t_(x-1))
But I don't know what h is, or really what this equation means.
Or maybe something with calculating using trapezoids? But I feel like there isn't enough information given to do that.
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Your last idea, about trapezoids, is good.
Construct a trapezoid starting at (2, 5), with a side going straight down from that point to the x axis.
Then construct a line with slope .5, part of which will form the top of the trapezoid.
The bottom of the trapezoid runs .4 units from the point (2, 0) to the point (2, 2.4).
What is the length of the fourth side?
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
I really don't have any idea. That's not to say that I wasn't taught it, because I'm sure that I was. I just can't remember
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Self-critique rating:
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Question:
`q008. The graph of a function g(t) contains the points (3, 4), (3.2, 4.4) and (3.4, 4.5). What is your best estimate of the value of g ' (3), where the ' represents the derivative with respect to t?
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Your solution:
I think this is the trapezoid thing.
Using the points, I make 2 trapezoids, and find the area of both
Area of first one = (b1+b2)/2*h = (4+4.4)/2*.2 = 0.62
Area of second = (4.4+4.5)/2*.2 = 0.89
.89+.62 = 1.51
g'(3) = 1.51
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You have estimated the area underneath the graph. This would be an estimate of the integral of g(t) from t = 3 to t = 3.4.
What aspect of the graph of a function is its derivative related to?
What information would your trapezoids give you about that aspect of this graph?
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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#*&!
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You're doing some good things.
There are also some things you'll want to do more work on, and some things you'll want to review.
&#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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