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course Mth 277
09/03/2011 415pm
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= 3 sin(4 t + 2)
y'= 12 cos(4 t + 2)
y""= -48 sin(4 t + 2)
2) y= 2 cos^2(3 t - 1)
y'= 2[cos(3t-1)]*[cos(3t-1)]
= 2{cos(3t-1)*[-3sin(3t-1)] + cos(3t-1) * [-3sin(3t-1)]}
= -12sincos(3t-1)
= -6sin2(3t-1)
= -6sin(6t-2)
y""= -36cos(6t-2)
@& You're pretty much right on y '. However the result is -12 sin(3 t - 1) cos(3 t - 1).
It's probably easier to use the chain rule on squaring function as well. The derivative of cos^2(theta) is -2 sin(theta) cos(theta). This would avoid the use of the product rule.
In any case you do need the product rule in the next step. The derivative of -12 sin(3t-1) cos(3t-1) is 36 ( sin^2(3 t - 1) - cos^2(3 t - 1) ), and if you use the right trig identity this can be put into the form you have reported.*@
3) y=Asin(omega*t + phi)
y'= Acos(omega*t + phi)
y""= -Asin(omega*t + phi)
@& Remember the chain rule.
y'= omega * A cos(omega*t + phi)
y '' = -omega^2 * A sin(omega*t + phi)*@
4)
y= 3 e^(t^2 - 1)
y'= 6t e^(t^2-1)
y""= 6t*[2t*e^(t^2 -1)] + 6e^(t^2-1)
y""= 6t^2 * [2e^(t^2-1) +1]
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Self-critique (if necessary):
I am having a hard to transfering my work from my paper to the computer. I understand typerwriter notation, it's just ALOT of work! and I am unsure if you can see all my steps. I made sure that my work was readible, please tell me if there is anything I can work on, so you can read my work better????????????????????????
@& Until I get a good sense of your work you should continue to provide detail in your responses.
However after a few assignments, ask again and I'll advise you on how much detail you can safely leave out.
I'll also include a related note at the end of this document. *@
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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:
I understand that a sine funcition usually starts at the point (0,1) and because there is a constant 3 in front of the sine function the frequency of the sine function is elongated and goes from 3 to -3. I do not know what the 4t-2 part of the function does. I can assume that this determines the x part of the function. As t increases, so does the inside of the sine function. I am not as strong on my trigonometry as I would like, as I assume for most students, Trig is not their forte. I understand some basic ideas behind creating a graph.
@& Graphing by shifing and stretching transformations, which are generally covered during the first semester of precalculus and reinforced with the trigonometric functions during the second semester, we construct the graph as follows:
Starting with y = sin(t), which has an oscillating graph with period 2 pi and amplitude 1, starting at the point (0, 0) with slope 1, we do the following:
Multiply y by 3, which gives us the graph of y = 3 sin(t) and increases the amplitude to 3 (and also increases the initial slope to 3).
Replace t with 4 t, which decreases the period to 2 pi / 4 = pi / 2 and gives us the graph of y = 3 sin(4 t). This increases the initial slope by factor 4, so that the initial slope is 12.
Replace t once more with t + 1/2, which shifts the graph 1/2 unit to the left and gives us the graph of y = 3 sin(4 ( t + 1/2) ) = 3 sin(4 t + 2).
The resulting graph has amplitude 3, period 4, with a cycle of period pi / 2, originating at (-1/2, 0). The initial slope is 12.*@
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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 constant that multiplies the cosine function. Cosine starts at (0,0) and the frequency is from A to -A. I do not know what the rest of the graph looks like.
@& cos(0) = 1, not 0.
The amplitude of the oscillation is A, the initial slope is 0, and the period is 2 pi / omega.
The graph of A cos(omega * t) thus starts at (0, A), with slope 0. A full cycle occurs between the point (0, A) and (2 pi / omega, A). The y values vary between - A and A.
Adding k raises the graph k units, so that y values now vary between - A + k and A + k. The amplitude is still A, but the axis shifts to the line y = k.
omega t + theta_0 = omega ( t + theta_0 / omega). So the last step is to replace t with t + theta_0 / omega, which shifts the graph -theta_0 / omega units in the x direction.
A cycle therefore starts at (-theta0 / omega, A + k), the amplitude is A and the period is 2 pi / omega.*@
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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:
Integral of e^-3t= (-1/3)(e^-3t) + C
Intergral of 2 sin( 4 pi t + pi/4) = -1/(2*pi) * cos(4*pi*t + pi/4) +C
Integral of 1 / (3 x + 2) = (1/3)ln(3x + 2)
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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:
The anitderivative of f(t) = (-1/3)(e^-3t)+C, set that equal to 2 and t=0, solve for C. Therefore C= 7/3
The antiderivative of x(t) = -1/(2pi)*cos(4*pi*t + pi/4) + C, set that equal to 2pi and t= 1/8, then solve for C. C= (8pi^2 -sqrt(2))/4pi
I don't know how to solve for the third part of the question. The wording is confusing, but if I understand correctly, I want to find C when the limit of the antiderivative approaches -1.
@& The limiting value of ln ( 3 t + 2) as t -> infinity is +infinity, so there is no constant number that could be added to this expression to give you a limit of 1. So the answer to the question is that it is impossible.*@
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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), set t=3, solve for A= 5/2
2t + 4 = A(t+1) + B(t-3), set t=-1, solve for B= -1/2
Subsitute A and B into equation and integrate:
(2t +4)/ (t-3)(t+1) = A / (t - 3) + B / (t + 1), where A= 5/2 and B= -1/2; solve for integration = 5/2ln(t-3) - 1/2ln(t+1)
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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 equation of the graph is y= (1/2)x + 4, substitute x=2.4 into equation and solve. f(2.4)= 5.2
@& Good.
Alternatively 2.4 lies .4 units to the right of 2. Moving .4 units to the right along slope .5 will increase the y value by .2, resulting in y = 5 + .2 = 5.2.
Either way of reasoning is fine, but it's particularly useful to see it both ways.*@
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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 slope between the two points (3.2, 4.4) and (3.4, 4.5) is (1/2). The slope between the two points (3, 4), (3.2, 4.4) is 2. Best guess estimate says that the graph is increasing on a decreasing manner. Therefore, the slope when x=3 will be greater than 2. I would guess g'(3)= 5/2
@& Good.*@
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@& Your calculus is in good shape to start this course.
However be sure to see my notes, especially on the graphing of the trig functions. Everything on which I've given you notes will be important at some point of the course.*@