qa 00

#$&*

course Mth 279

7/23 3am

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:

3 sin(4 t + 2).

f’(t) = 3cos(4t+2)(4) = 12cos(4t+2)

f’’(t) = -12sin(4t+2) = -48sin(4t+2)

2cos^2(3t-1)

f’(t) = 4cos(3t-1)(-3sin(3t-1)) = -12cos(3t-1)sin(3t-1)

f’’(t) = -12(-3sin(3t-1)sin(3t-1)+ cos(3t-1)3cos(3t-1))

= -12(-3sin^2(3t-1) + 3cos^2(3t-1))

= 36(cos^2(3t-1)-sin^2(3t-1))

Asin(omega*t + phi)

f’(t) = Acos(omega*t + phi)*omega = A*omega*cos(omega*t + phi)

f’’(t) = -A*omega*sin(omega*t + phi)*omega = -A*omega^2*sin(omega*t + phi)

3e^(t^2-1)

f’(t) = 3(2t)e^(t^2-1) = 6te^(t^2-1)

f’’(t) = 6e^(t^2-1) + 6te^(t^2-1)*2t = 6e^(t^2-1)+12t^2e^(t^2-1)

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

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:

Oh boy it’s been a while since I’ve graphed trig functions. Well. I’m thinking that 3 is the amplitude, so y is going to oscillate between -3 and 3. Now normally I know sin(0) is 1, so the peak of a wave would normally be at the origin, but I know something inside those parentheses is going to shift my graph left or right.

@&

Think of the unit-circle definition of the trig functions. sin(0) is 0. cos(0) is 1.

*@

I think when it’s positive it shifts it to the left?

@&

Right. You should also know why this is. If you add something to the argument of sin(t), say, then values of t that are that much less will give you the same result. This shifts the graph to the left.

*@

And the period of a sine wave is normally 2 pi I think, and there’s something I have to do with dividing. Maybe I divide 2pi by 4 to get a period of pi/2?

@&

The period of sin(t) is 2 pi.

The period of sin(4 t) is the change in t necessary to change 4 t by 2 pi. The required change in t would be 2 pi / 4 = pi / 2.

*@

So from peak to peak is a change in t of pi/2? And then shift everything left by 2 because of the +2 inside the parentheses? I think some reviewing of graphing trig functions might be in order!! 

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

I know I need to review graphing trig functions! Which is pretty awful since I know I had to use that in my physics lab fall semester at VT. But hey, the more times I have to review it the better it’ll stick, I suppose.

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

Self-critique rating: 1. I’m not sure I understand how to find the period of the sine function, or how to tell how far left or right it should shift.

@&

You're on the right general track with the rules.

You'll be much better off if you can reason out the rules rather than just quoting them. Check my notes.

*@

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

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:

The graph has an amplitude of A and has been shifted upwards k units. y(0) would normally equal zero for a cosine graph, but the graph has also been shifted to the left, though I’m not sure by how much. The period has also been changed, possibly from 2pi to omega/2pi? What’s the formula for shifting sine graphs, and for calculating the period????

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

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:

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

Well I know the derivative of e^u is e^u*du, so working backwards, to obtain e^(-3t) as the derivative, my integral must be -1/3e^(-3t) + C.

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

The derivative of cos(u) is -sin(u)du, so in order to get 2 sin( 4 pi t + pi/4) when I take the derivative of something, my integral must be -1/(2pi)*cos(4pi*t + pi/4) + C.

y(t) = 1 / (3 x + 2)

The derivative of lnx is 1/x, so the indefinite integral of y(t) = 1 / (3 x + 2) must be 1/3ln(3x+2)

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

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 antiderivative means to integrate, and now that I am given initial conditions I can find out what C is.

When t=0, -1/3e^(-3t) + C = 2.

-1/3e^0 + C = 2.

-1/3 + C = 2.

C = 7/3.

So my antiderivative is -1/3e^(-3t) + 7/3.

When t = 1/8, -1/(2pi)*cos(4pi*t + pi/4) + C = 2pi.

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

-cos(6pi/8)/(2pi) + C = 2pi.

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

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

So my antiderivative is -1/(2pi)*cos(4pi*t + pi/4) + 2pi - sqrt(2)/(4pi)

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

I must be doing something wrong here???? I got before that the indefinite integral for that was 1/3ln(3t+2) + C. I don’t get how, when t approaches infinity, 1/3ln(3t+2) + C = -1. I thought ln(infinity) was infinity. So how is infinity + C = -1?

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

I don’t get the last question with the antiderivative of 1/(3x+2). Perhaps my understanding of limits is not where it should be?

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

Self-critique rating:

@&

That wasn't exactly a trick question. The point is that the given limit is not possible, as you pretty much demonstrated.

*@

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

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:

Multiply both sides by (t-3)(t+1).

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

When t = -1,

2(-1)+4 = B(-1-3).

-2+4 = -4B.

2 = -4B.

B = -1/2.

When t = 3,

2(3)+4 = A(3+1).

6+4 = 4A.

A = 5/2.

(2t+4)/((t-3)(t+1)) = 5/(2(t-3)) - 1/(2(t+1))

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

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:

With a point and a slope, we can approximate the function f(x) linearly.

f(x)-5 = 0.5(x-2)

f(x)-5 = 0.5x-1

f(x) = 0.5x + 4

f(2.4) = 0.5(2.4)+4 = 5.2

My best estimate of f(2.4) is 5.2

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

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:

g’(3) is the slope of g(t) at t = 3. While we don’t know the slope at t = 3, we can find the average slope from t= 3 to t=3.2.

(4.5-4.4)/(3.4-3.2) = 0.1/0.2 = 0.5.

0.5 is my best estimate of the value of g’(3), although it is likely higher than that, as the slope at t = 3.2 is visibly lower than the slope at t = 3.

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

Not sure if I answered what the question was getting at for that one.

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

Self-critique rating:

@&

Good.

You are correct about the trend of the slope, and the best estimate would make a reasonable assumption about how that would affect the result.

*@

"

Self-critique (if necessary):

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

Self-critique rating:

"

Self-critique (if necessary):

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

Self-critique rating:

#*&!

@&

You are in pretty good shape here.

Do see my notes, especially on the graphs of the trig functions (that is the most common area of weakness for students beginning this course).

*@