QA 00

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course Mth 279

9/16 8

Qa 00Most 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)

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Your solution:

y = 3sin(4t+2)

y’ = 12cos(4t+2)

y’’= -48sin(4t+2)

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

y’=12cos(3t-1)sin(3t-1)

y’’=36(cos^2(3t-1)-sin^2(3t-1))

y= Asin(omega*t+phi)

y’=A*omega*cos(omega*t+phi)

y’’= -A*omega^2*sin(omega*t+phi)

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

y’=6t*e^(t^2-1)

y’’=(12t^2-6)(e^(t^2-1))

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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:

The graph of this function will be a sinusoidal wave curve that has an amplitude of 3. When you look at this function you can see that the intervals to use would be pi/4 because if you factor out a two from the inside function you can take 2pi/2 and get pi as the period. Then you can make a table with values of t and plot the points on the graph.

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Good, but the period is pi/2, not pi. If t changes by pi/2, then 4 t changes by 2 pi.

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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:

In this graph, A represents the altitude of the graph. The omega will contribute to the distance between the crests and troughs of the graph. And the theta will dictate the distance that the graph is translated to the left. The k will dictate the distance that the graph is translated up.

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omega * t changes by 2 pi when t changes by 2 pi / omega; this therefore is the period of the function.

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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:

The indefinite integral of f(t) = e^(-3t) is (e^(-3t))/-3 + c

The indefinite integral of x(t) = 2sin(4pi t+ pi/4) is (2cos(4pi t + pi/4))/(-4pi) + c

The indefinite integral of y(t) = 1/(3t+2) is (ln(3t+2))/(3) +c

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Given Solution:

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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:

The antiderivative of f(t) = e^(-3t) is (e^(-3t))/-3 + c

So when we plug in the values we see 2 = (e^(-3*0))/-3+c

2 = (-1/3) + c

So c = 7/3 so the antiderivitive is ((e^(-3t))/-3)) + (7/3)

The antideirvative of x(t) = 2sin(4pi t+ pi/4) is (2cos(4pi t + pi/4))/(-4pi) + c

So when we plug in the values we see 2 pi = (2cos(4pi *(1/8) + pi/4))/(-4pi) + c

2pi = ( sqrt 2/4pi) +c

c = 2pi - ( sqrt 2/4pi) so the antiderivative is (2cos(4pi t + pi/4))/(-4pi) + (2pi - ( sqrt 2/4pi) )

The antiderivative of y(t) = 1/(3t+2) is (ln(3t+2))/(3) +c

So when we plug in the values we see that the as the antiderivative goes to infinity it is equal to negative on therefore c will equal 1.

confidence rating #$&*:

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Given Solution:

Self-critique (if necessary

I don’t think I did the very last one correctly. I wasn’t sure what to do about the limit. Every time I tried to involve a limit I got that the antiderivative would be zero.

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That wasn't exactly a trick question, but it was a little shady. The value of the natural log approaches infinity, as you see, so no added constant can change that.

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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:

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

Confidence rating:

1

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Given Solution:

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Self-critique (if necessary):

I think I could have kept going with this but I wasn’t sure what form you wanted it in.

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To answer the question completely you would need to evaluate A and B. I expect you know how to do this, but if not be sure to ask.

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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 = .5x+4

Y(2) = 5

So y(2.4) = 2.2+4 = 6.2

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You got the approximating linear function, which is the tangent line. Very good.

However note that

y(2.4) = .5 * 2.4 + 4 = 1.2 + 4 = 5.2, not 6.2.

Another way to reason this out:

x changes from 2 to 2.4, a change of 0.4. The slope is 0.5. Rise = slope * run = 0.4 * 0.5 = 0.2.

So the y value changes from 5 to 5 + 0.2 = 5.2.

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Given Solution:

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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 shown by the points you know that the slope will not be constant because the rise over run two different points is not the same. However at the point 3 the derivative would because close to 2.

@&

The average value of the derivative is 2 on the first interval, 0.5 on the second. So it appears that the derivative is decreasing.

It would follow that the derivative of (3, 4) would be somewhat greater than its average value on the first interval.

One fairly simple but reasonable estimate, which takes account of the changing derivative and the rate at which it appears to be changing, might be 2.75.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I wasn’t exactly sure how to get the function and then the derivative. I tried drawing a tangent line but I wasn’t sure how I would be able to find the slope of that line, I think I’m missing something.

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Self-critique rating:

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@&

You appear to be in good shape.

However do check my notes and let me know if you have questions.

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