qa_00

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course Mth 279

6/4 4:15

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:

a) y=3sin(4t+2)

y’= 3cos(4t+2)*4 = 12cos(4t+2)

y’’= -12sin(4t+2)*4 = -36sin(4t+2)

b)y=2cos^2(3t-1)

chain rule y’= 2*2(cos(3t-1))*3-sin(3t-1)*3) = -12cos(3t-1)sin(3t-1)

product rule y’’= -12(cos(3t-1)*3*cos(3t-1))+(sin(3t-1)*(-3)sin(3t-1))

y’’= 36(cos^2(3t-1)+sin^2(3t-1)

c)y = Asin(omega(t) + phi)

y’ = A(omega)cos(omega(t) + phi)

y’’ = -A(omega)^2sin(omega(t) + phi)

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

y’= 3e^(t^2-1)*2t = 6te^(t^2-1)

y’’= 6t*e^(t^2-1)*2t +(e^(t^2-1)*6)

= 12t^2e^(t^2-1) + 6e^(t^2-1)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

Remember some of the rules took longer than others but overall 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:

From the original f(x) function, sin(x) the graph will move according to the function y.

1. stretch graph vertically by factor of 3

2. Shrink graph horizontally by factor of 4

3. Shift graph 2 units to left

@&

Your left shift isn't correct.

*@

confidence rating #$&*:

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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:

`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:

1. stretch graph vertically by factor of A

2. Shrink graph horizontally by factor of omega

3. Shift graph theta units to left

@&

Your left shift isn't correct.

You'll want to review that rule and be sure you understand the reason for it.

*@

4. Shift k units upward

confidence rating #$&*:

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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:

`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:

a. f(t) = e^(-3t)

int(e^-3t) = e^(-3t) + C

b. x(t) = 2sin(4pi*t + pi/4)

int(2sin(4pi*t + pi/4)

= 2 * (1/4pi) int(sin(u)) used u-substitution

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

c. y(t) = 1/(3x + 2) dx

(1/3)int(1/u) du where u= 3x+2 so du = 3 dx

ln|u|/3 = (ln|3x+2|)/3 + C

confidence rating #$&*:

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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:

`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:

a. f(t) = e^(-3t)

F(t) = e^(-3t) + C

2 = e^(-3(1)) +C

C = 2- e^-3

Therefore F(t) = e^(-3t) + 2 -e^(-3)

@&

Your integration isn't correct, as is obvious if you check it by differentiating.

*@

b. x(t) = 2sin(4pi(t) + pi/4)

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

2pi = (-1/(2pi))*(-sqrt(2)/2) + C

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

Therefore X(t) = - 1/(2pi) cos(u) = -1/(2pi) cos(4pi(t) + pi/4) + (2pi - sqrt(2)/(4pi))

c.y(t) = 1/(3t +2)

Y(t) = ln|3(-1) +2|/3 + C = 0

Y(t) = ln|3(-1) +2|/3

@&

Your approach is good.

There is no value of C than works for this last situation.

*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

Unsure if what I did was even what you are asking, especially with part c.

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Self-critique rating:3

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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)

when t =-1; 2(-1) + 4 = A(-1+1) + B(-1-3)

2 = -4B; B= -1/2

when t = 3; 2(3) + 4 = A(3+1) + B(3-3)

10 = 4A; A = 5/2

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

confidence rating #$&*:

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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:

`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=mx + b

5=(.5)(2) +b

5=1 + b

b = 4

Therefore

y=(.5)(2.4) + 4

y= (6/5)+4 = 26/5

@&

Good.

More direct reasoning:

A change in x value of .4 results in a change of .2 in the y value.

*@

confidence rating #$&*:

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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:

`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:

points (3,4) and (3.2, 4.4)

(y2-y1)/ (x2 - x1) = (4.4-4)/(3.2-3) = .4/.2 = 2 (slope)

y=mx + b

4 = (2)(3) +b

b = -2

y=2x -2

y’ = 2

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

First I found the slope of the points then used y=mx+b to find the equation of the line and then took derivative.

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Self-critique rating:2"

@&

Good. But if you consider all the information available, you'll see that a slope somewhat greater than 2 is more appropriate.

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Self-critique (if necessary):

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Self-critique (if necessary):

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#*&!

&#Good responses. See my notes and let me know if you have questions. &#