QA_00

#$&*

course Mth 279

6/4 4:14

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)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

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:

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

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

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

@&
Pretty good, but one of your answers is incorrect.

The graph would shift 1/2 unit to the left.

You should review both the rules and the reasoning behind the rules so you understand them deeply.
*@

confidence rating #$&*:

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Given Solution:

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

Self-critique (if necessary):

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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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

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

4. Shift k units upward

@&
The left shift isn't theta units. Otherwise good.
*@

confidence rating #$&*:

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Given Solution:

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

Self-critique (if necessary):

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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)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

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

@&
The derivative of this expression is not equal to the original expression.

You need to always check your integrations by differentiating your results.
*@

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:

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

Self-critique (if necessary):

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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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

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 quite right.
*@

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

confidence rating #$&*:

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Given Solution:

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

Self-critique (if necessary):

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

@&
You have the right approach.

There is no value of c that fits the condition in 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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

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:

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

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)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

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:

From x = 2 to x = 2.4 is an interval `dx = .4.

If the slope remains .5, as it does on the tangent line, the change in y will be

`dy = slope * `dx = .5 * .4 = .2.

Thus the new y value will be 5 + .2 = 5.2, which is of course the same as your 26/5.
*@

confidence rating #$&*:

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

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Given Solution:

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

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

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:

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

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"

@&
Not bad, but there is a trend to the slopes that indicates a slope at x = 3 which is a bit greater than 2.

You have data for three points, so your best estimate would consider all the information.
*@

Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

#*&!

QA_00

@& Pretty good, but one of your answers is incorrect. The graph would shift 1/2 unit to the left. You should review both the rules and the reasoning behind the rules so you understand them deeply. *@

@& The left shift isn't theta units. Otherwise good. *@

@& The derivative of this expression is not equal to the original expression. You need to always check your integrations by differentiating your results. *@

@& Your integration isn't quite right. *@

@& You have the right approach. There is no value of c that fits the condition in part c. *@

@& Good. *@

@& More direct reasoning: From x = 2 to x = 2.4 is an interval `dx = .4. If the slope remains .5, as it does on the tangent line, the change in y will be `dy = slope * `dx = .5 * .4 = .2. Thus the new y value will be 5 + .2 = 5.2, which is of course the same as your 26/5. *@

@& Not bad, but there is a trend to the slopes that indicates a slope at x = 3 which is a bit greater than 2. You have data for three points, so your best estimate would consider all the information. *@

@& You're clearly in good shape to start this course, but check the notes I've inserted for a few missing pieces. *@

@&
Pretty good, but one of your answers is incorrect.

The graph would shift 1/2 unit to the left.

You should review both the rules and the reasoning behind the rules so you understand them deeply.
*@

@&
The left shift isn't theta units. Otherwise good.
*@

@&
The derivative of this expression is not equal to the original expression.

You need to always check your integrations by differentiating your results.
*@

@&
Your integration isn't quite right.
*@

@&
You have the right approach.

There is no value of c that fits the condition in part c.

*@

@&
Good.
*@

@&
More direct reasoning:

From x = 2 to x = 2.4 is an interval `dx = .4.

If the slope remains .5, as it does on the tangent line, the change in y will be

`dy = slope * `dx = .5 * .4 = .2.

Thus the new y value will be 5 + .2 = 5.2, which is of course the same as your 26/5.
*@

@&
Not bad, but there is a trend to the slopes that indicates a slope at x = 3 which is a bit greater than 2.

You have data for three points, so your best estimate would consider all the information.
*@

@&
You're clearly in good shape to start this course, but check the notes I've inserted for a few missing pieces.
*@