QA_00

#$&*

course Mth 279

850pm 10/6/2011

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.

*********************************************

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:

y= 3 sin(4 t + 2)

y'= 12 cos(4 t + 2)

y''= -48 sin(4 t + 2)

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

y= 2[1+ cos2(3t - 1)/2]

y= 1 + cos2(3t - 1)

y= 1 + cos(6t - 2)

y'= -6sin(6t - 2)

y""= -36cos(6t - 2)

y= A sin(omega * t + phi)

y'= A*omega*cos(omega*t + phi)

y""= -A*omega^2*sin(omega*t + phi)

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]

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating:

*********************************************

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:

The parent function is y = sin(t). Mulitiply it by 3 and it goes from y value -3 to 3. The 4t increases the slope of the graph and shortens the period. The 2 moves the graph 2 units to the left.

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary):

I think I did it right.

------------------------------------------------

Self-critique rating: 3

*********************************************

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:

A is the constant mulitplying the graph and increasing the slope and amplitude. Omega is a coefficient that is multiplying t, which increases/decreases the slope and period. theta_0 is the orgin of the graph.

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary):OK

------------------------------------------------

Self-critique rating:

*********************************************

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:

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)

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating:

*********************************************

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:

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

The antiderivative of y(t) = 1 / (3 t + 2) is ln(3t +2), solve for t=-1, ln(-1) is not a real number

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary):OK

------------------------------------------------

Self-critique rating:

*********************************************

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), 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)

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating:

*********************************************

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:

The equation of the graph is y= (1/2)x + 4, substitute x=2.4 into equation and solve. f(2.4)= 5.2

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary): OK

------------------------------------------------

Self-critique rating:

*********************************************

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:

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

confidence rating #$&*:

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

.............................................

Given Solution:

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

Self-critique (if necessary):OK

------------------------------------------------

Self-critique rating:"

QA_00

@& This looks very good. You appear to be well prepared to start this course.*@