QA 00

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course MTH 279

1/19 12

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:

y = 3 sin(4 t + 2)

y' = 12cos(4t +2) Chain rule and simplify.

y"" = -48sin(4t + 2) Chain rule and simplify.

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

y' = -12cos(3t - 1)sin(3t - 1) Chain, chain, simplify.

y"" = -12[(cos(3t -1)cos(3t - 1)*3) + (sin(3t - 1)* -sin(3t - 1)*3)] product rule

y"" = -36[cos^2(3t - 1) - sin^2(3t - 1)] simplify

y = A sin(omega * t + phi)

y' = A(omega)cos(omega*t + phi) chain rule and simplify

y'' = -A(omega^2)sin(omega*t + phi) chain rule and simplify

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

y' = 6t*e^(t^2 - 1) chain rule and simplify

y'' =6 [(t*e^(t^2 - 1)*2t + (e^(t^2 - 1) )] product rule with chain rule inside

y'' = [6e^(t^2 - 1)] * (2t^2 +1) simplify

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

This graph oscillates between y = -3 and 3 (it has an amplitude of 3) and reaches this ampltude at approximately t = -0.1 ( (pi - 4)/8 exactly) for +3 and t = 0.68 ( (3pi - 4)/8 exactly) for -3 and has a period of 1/2 pi in which this pattern repeats. It is also shifted horizontally from where a regular sine wave is (it doesn't cross at 0,0), but I don't know by exactly how much. I know the phase shift has something to do with the +2 inside of the equation, but I don't know how that affects it exactly when the period is also manipulated. I can say that it crosses the y axis when t = -1/2 and (pi - 2)/4. I plotted a few key points along the way to see the general behavior of the graph. Points that would make sin = 0, 1, and -1 particularly. I also used my knoweldge of graphical transformations of sinusoids to see if my plotted points made sense.

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We start with sin(t), which has period 2 pi.

The argument of the function is not t but 4 t + 2, which can be expressed as 4 ( t + 1/2).

Replacing t by t + 1/2 has the effect of shifting the graph 1/2 unit to the left (since in order to get a given value, the value of t in t + 1/2 has to be 1/2 unit less than the value ot t).

The factor 4 then causes the argument 4 ( t + 1/2) to go through a complete cycle 4 times as fast as would t + 1/2 alone. Thus the function goes through a cycle as t changes by 2 pi / 4 = pi/2.

The period of the function is therefore pi/2, as you say, and the graph is shifted 1/2 unit to the left so that the graph point at the origin shifts to t = -1/2.

Note that your (pi - 2) / 4 is the same as pi/4 - 1/2. This is the point that would correspond to t = pi on the graph of sin(t). Having 1/4 the period of the sine function, the zeros of the transformed function will be separated by pi/4 rather than pi. So when the graph is shifted 1/2 unit to the left and the period compressed by factor 4, the point t = pi transforms to t = pi / 4 - 1/2, consistent with your calculation.

Sine and cosine graphs will be very important later in the course. You obviously have a good background and have approached this problem correctly; hopefully the preceding paragraphs of this note will amplify the transformation rules (almost all of which you applied correctly) and the reasons for them..

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): I feel that my descriptions of graphs aren't concise enough.

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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:

The value of A is the amplitude of the function. It determines the maximum and minimum y values of the graph.

Omega determines the period and frequency of the graph. The period of the graph would be 2pi/omega and the frequency, which is the reciprocal of the period, would be omega/2pi. Therefore, a larger omega woud give you a smaller period (larger frequency) or more complete cycles in a unit t.

theta_0 is the initial phase shift of the graph. Cosine waves normally start at a peak at t = 0, but if you have a phase shift theta_0, they may start at + or - that normal location. A positive theta_0 shifts the graph to the left that many units, while a negative theta_0 shifts your graph to the right that many units.

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Good, but your horizontal shift isn't quite right.

If you see the cosine function can be written as

cos(omega (t + theta_0 / omega) )

it should become clear that the peak of the function cos(t) which occurs on the positive y axis is transformed to the point t = -theta_0 / omega. This makes sense, because this is the value that makes the argument of the cosine function zero (i.e., when t = -theta_0 / omega you get cos(0) = 1 ).

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

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

F(t) = -(1/3)e^(-3t) + C

I used u sub and let u = -3t, then du = -3 dt

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

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

let u = 4pi*t + pi/4, then du = 4pi dt. made sure to put a 4pi inside and a 1/(4pi) on the outside.

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

Y(t) = 1/3 ln(abs(3x +2) ) + C

let u = 3x + 2, then du = 3 dx

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

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

F(t) = -(1/3)e^(-3t) + C

2 = -(1/3)e^(-3*0) + C

2 = -(1/3) + C

7/3 = C

F(t) = -(1/3)e^(-3t) + 7/3

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

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

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

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

2pi = -1/(2pi) cos(3pi/4) + C

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

(8*pi^2)/sqrt(2) = C

X(t) = -1/(2pi) cos(4pi t + pi/4) + (8*pi^2)/sqrt(2)

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

Y(t) = 1/3 ln(abs(3t +2) ) + C

No idea.

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The problem was improperly posed.

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Since the log approaches infinity, there is no fixed value of C that can satisfy the given condition. Any fixed value of C would still lead to a limit of infinity.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I'm not sure what the third part was asking for. I interpreted it as saying as t -> infinity, the antiderivative is -1. I knew that at t went to infinity, ln would go to infinity....so I had no idea how I could get a value for C, since -1 = infinity + C was what I was coming up with. I must be reading the intent of the problem wrong, but I can't find another way to interpret it.

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Self-critique rating: OK

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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:

A / (t - 3) + B / (t + 1)

[A(t + 1)] / [(t - 3)(t + 1)] + [B(t + 3)] / [(t - 3)(t + 1)]

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

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

A + B = 2

B = 2 - A

A - 3B = 4

A = 4 + 3B

B = 2 -(4 + 3B)

B = -1/2

A = 4 + 3(-1/2)

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: OK

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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 - y_1 = m(x - x_1)

y - 5 = .5(x - 2)

y = .5x - 1

f(2.4) = .5(2.4) - 1

f(2.4) = 5.2

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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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:

Since I had only points, I decided to find the slope from point a to point b and from point a to point c. I then took the average of these two slopes. Connecting the points in a linear fashion like this would result in their slopes being the derivatives of their respective equations.

m_1 = 2

m_2 = 1.25

m_avg = 1.625

g'(3) = 1.6 (to my best guess)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

There may be a more efficient way of doing this, but I couldn't think of anything else off the top of my head.

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At first glance yours would seem to be a very reasonable conjecture. But on closer inspection, it doesn't hold up.

Since the slope appears to be decreasing, it would be reasonable to expect that the slope at x = 3 would be greater than either of the slopes between the points. A slope estimate of 2.5 or 3 might be more appropriate.

There are more sophisticated ways to answer the question (e.g., find the quadratic function that fits the three points), but the idea here is that the trend being toward decreasing slopes, the first slope should be the greatest.

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Self-critique rating: 3"

Self-critique (if necessary):

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Self-critique rating:

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You've done very well on this exercise. I do not expect that you will have any problem with the calculus and precalculus aspects of this course.

However do check out my notes, especially related to the horizontal shift of a transformation.

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