Revision of QA00

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course Mth 279

09/14/2011 around 5:30PM I am inserting some revisions on the QA00.

QA00#$&*

course Mth 279

09/13/2011 around 9:00AM.Submitting the QA00 assignment.

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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:

d/dt(3 sin(4 t + 2)) = 12cos(4t+2), d/dt(12cos(4t+2) = -48sin(4t+2)

d/dt(2 cos^2(3 t - 1)) = -12cos(3t-1)sin(3t-1),

d/dt(-12cos(3t-1)sin(3t-1)) I will use the product rule here,

let f = -12 cos(3t-1) f’ = 36sin(3t-1), let g = sin(3t-1) g’ = 3cos(3t-1)

fg’ + gf’ = -36cos^2(3t-1) + 36sin^2(3t-1)

d/dt(A sin(omega * t + phi)) = A*omega*cos(omega*t + phi)

d/dt(A*omega*cos(omega*t + phi)) = -A*omega^2sin(omega*t + phi)

d/dt(3 e^(t^2 - 1)) = 6t*e^(t^2-1)

d/dt(6t*e^(t^2-1)) , I will use the product rule here,

let f = 6t, f’ = 6 let g = e^(t^2-1), g’ = 2t*e^(t^2-1)

fg’ + gf’ = 12t^2*e^(t^2-1) + 6*e^(t^2-1)

confidence rating #$&*:8232;

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Given Solution:

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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 has a maximum value that is the amplitude absolute value of 3 and a minimum value that is negative absolute value of 3.

y = 3sin2(2t+1) the positive number 2 is the number of cycles in an interval of 2 pi, and 2 pi/2 =pi is the period of the lenght of cycle.

for y = 3sin2(2t+1) a basic cycle begins at y = 0 the basic cycle of the graph closest to the origin is contained in the interval

-1 less or equal to x less or equal to -1 + pi and the phase shift is -1.

@& 4 t + 2 changes by 2 pi when 4 t changes by 2 pi, which occurs when t changes by (2 pi) / 4 = pi / 2. So the period of the function is pi/2.

4 t + 2 = 4 * ( t + 1/2). The graph of sin(4 t + 2) is therefore shifted -1/2 unit in the t direction from the graph of sin(4 t).*@

confidence rating #$&*:8232;

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Given Solution:

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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 graph has a maximum value of absolute value of A that is the amplitude and a minimum value of negative of the absolute value of A.

@& The amplitude is A but the max and min values are A + k and -A + k.*@

The period for one cycle is 2pi/omega = t1 - t0.

The value theta_0 = - t0

A = (maximum-minimum)/2

I can identify one basic cycle of the graph that begins at the maximum value, decreases to 0, continue to decrease to minimum value, increases to 0 and then increases to maximum value.

The domain of one cycle is t0 less or equal to t and t less or equal to t1.

@& omega t + theta_0 = 0 when t = -theta_0 / omega. At this point the graph will have its maximum value A + k, so a cycle will start at (-theta_0 / omega, A + k)..

confidence rating #$&*:8232;

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Given Solution:

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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:

Integral of (t) = e^(-3 t) = -1/3e^(-3t)

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

@& The derivative of 2 sin( 4 pi t + pi/4) t is not 2 sin(4 pi t + pi/4).

An antiderivative would be -1 / ( 2 pi) cos(4 pi t + pi/4). You should verify that the derivative of this function is the given function, and review the basic integration techniques necessary to find this integral. I don't think this will give you much trouble, but if it does be sure to ask for additional guidance.*@

Integral of y(t) = 1 / (3 x + 2) = (log(3x +2))/3

@& Note that on all integrals you will also have an added integration constant c. This integration constant will be very important in this course.*@

&&&& Integral of (t) = e^(-3 t) = -1/3e^(-3t) + c

Integral of x(t) = 2 sin( 4 pi t + pi/4) = -1/2 cos (4 pi t + pi/4) + c,

the derivative of -1/2 cos (4 pi t + pi/4) + c = -(-1/2*4sin (4 pi t + pi/4)) = 2 sin( 4 pi t + pi/4)

Integral of y(t) = 1 / (3 x + 2) = (log(3x +2))/3 + c

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confidence rating #$&*:8232;

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Given Solution:

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Self-critique (if necessary):

&&&&I will make sure I will insert the constant c of integration all the time.&&&&

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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:

Integral of (t) = e^(-3 t) = -1/3e^(-3t)

@& This is an antiderivative but its value at t = 0 is -1/3, not 2.

This can be remedied by including the integration constant, which gives you the general antiderivative -1/3 e^-3 t) + c.

At t = 0 the value of this antiderivative is -1/3 + c, which must be equal to 2. Setting -1/3 + c = 2 we solve to find that c = 7/3.

Thus the antiderivative that satisfies the given condition is -1/3 e^(-3 t) + 7/3.*@

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

Integral of y(t) = 1 / (3 x + 2) = (log(3x +2))/3

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The antiderivative of (t) = e^(-3 t) = -1/3e^(-3t) + c at t = 0 is -1/3 + c,

-1/3 + c = 2, c = 2 + 1/3 =7/3

The antiderivative of x(t) = 2 sin( 4 pi t + pi/4) = -1/2 cos (4 pi t + pi/4) + c, at t = 2pi, -1/2cos (4pi*(1/8) + pi/4) + c = 2pi, -1/2cos(3pi/4) + c = 2pi, -1/2*(-sqrt(2)/2) + c = 2pi, c = 2pi - sqrt(2)/4 = 5.93

The antiderivative of y(t) = 1 / (3 x + 2) = (log(3x +2))/3 + c, as t approaches infinity is (log (infinity + 2))/3 + c = -1, c = -1 - infinity

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@& Your antiderivatives are not all correct, and none contains the integration constant c.

This constant is needed in order to satisfy the condition at the given value of t.

See my note on the first of the three questions, and see if you can modify your work on the other two.*@

confidence rating #$&*:8232;

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Given Solution:

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Self-critique (if necessary):

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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:

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

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

2t + 4 = At + A + Bt- 3B

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

A+B = 2, A = 2-B

A-3B = 4, 2-B-3B = 4, 2-4B = 4, -4B = 4-2, B = -1/2,

A+(-1/2) = 2, A = 2+1/2 = 5/2

So A = 5/2 and B = -1/2

confidence rating #$&*:8232;

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Given Solution:

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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:

we know that f’(2.4) = 0.5

the slope of the secant line is:

f’(2.4) = (2.4-2)/(x-5) where x is the unknown value of f’(2.4)

so, 0.5 = (2.4-2)/(x-5)

1/2 = 0.4/(x-5)

2(0.4) = x-5

x = 0.8+5

x = 5.8

so the best estimate of f(2.4) = 5.8

confidence rating #$&*:232;

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Given Solution:

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@& Good attempt, but 5 is a y value, and slope = change in y / change in x.

So you would have

1/2 = (y - 5) / (2.4 - 2)

Solving we get y = 5.2.

Another way of reasoning:

x changes by .4 and the slope if 1/2.

So the change in y is slope * change in x = 1/2 * .4 = .2.

y starts at 5 and changes by .2, yielding final value 5.2.*@

&&&& we know that f’(2.4) = 0.5

the slope of the secant line is:

@& terminology note: a secant goes from one point of the graph to another; the line here would be a tangent, not a secant.*@

f’(2.4) = (y-5)/(2.4-2) where x is the unknown value of f’(2.4)

so, 0.5 = (y-5)/(2.4-2)

1/2 =(y-5)/(0.4)

0.4 = 2*(y-5)

0.4 = 2y-10

y = 10.4/2 = 5.2

y = 5.2

so the best estimate of f(2.4) = 5.2

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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:

The derivative represents the slope of the tangent line at a specific x value. The closest x value to x = 3 is x = 3.2, so the best estimate of g’(3) is the slope of the secant line passing through (3,4) and (3.2,4.4).

g’(3) = (4.4-4)/(3.2-30) = 0.4/0.2 = 2

so g’(3) = 2.

confidence rating #$&*:232;

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Given Solution:

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Self-critique (if necessary):

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@& That's a good and reasonable solution.

However the fact that the slope between the next two points is less than 2 might indicate that the derivative is decreasing, which would imply that the derivative at x = 3 is a little greater than 2.*@

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#*&!

@& You clearly have a decent calculus background sufficient to begin this course.

However you did have a few errors, and it will be worth your time to make some modifications and be sure you are clear on all these points. This will save you a significant amount of time in the long run.

&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.

*@"

Revision of QA00

@& terminology note: a secant goes from one point of the graph to another; the line here would be a tangent, not a secant.*@

@& Good. Check my additional notes.*@