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course MTH 279

7/21 7

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.

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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:

3sin(4t +2)

1st. Derivative = 3cos(4t + 2)*4 = 12cos(4t + 2)

2nd.Derivative = -12sin(4t + 2)*4 = -48sin(4t + 2)

2cos^2(3t - 1)

1st = 2(cos(3t - 1))^2 = 4(cos(3t -1)sin(3t - 1) * 3) = -12cos(3t - 1)sin(3t - 1)

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2(cos(3t - 1))^2 = 4(cos(3t -1)sin(3t - 1) * 3)

is clearly a false statement, the right-hand side being the derivative of the left.

Don't make false statements. They have a tendency to confuse everybody, but mostly you.

The = sign means just what it means, 'is equal to', not 'is related to' or 'follows from'.

Don't feel badly. Almost everyone gets a note to this effect near the beginning of this course.

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Your derivative is fine.

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

= 12(3cos^2(3t - 1) - 3sin^2(3t - 1)) = 36(cos^2(3t - 1) - sin^2(3t - 1))

A sin(omega * t + phi)

This problem is giving me trouble. I do not know what to treat some of the figures, A, omega, and phi, as. I don’t know if they should be variables or contants.

3 e^(t^2 - 1)

1st = 3e^(t^2 -1) * (2t) = 6te^(t^2 - 1)

The derivative of e raised to an exponent is always the same, multiplied by the derivative of the exponent.

2nd = 6te^(t^2 - 1) * (2t) = 12t^2 * e^(t^2 - 1)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I remember most of the ideas from calculus, however I need more work to become used to knowing what is and isn’t constants. Also, I need further practice with chain rule derivatives.

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Self-critique rating:3

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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:

I wrote out a table with polar x coordinates and Cartesian y coordinates. I always use this method when graphing functions with trigonometric functions because it makes it easier to see patterns and repetitions in the data. From the table and graph I drew using it, I determined that the graph is oscillating from 2.73 at (pi/2), pi, (3pi/2), 2pi to -2.73 at (pi/4), (3pi/4), (5pi/4), (7pi/4), which is to be expected with a sin graph.

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Good, but the maximum and minimum values of this function do not occur at multiples of pi/2. They occur at some of the points where 4 t + 2 is a multiple of pi/2.

You will want to review graphing of trigonometric functions, which become extremely important about halfway through the course.

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The maximum and minimum values of a sine or cosine are +1 and -1. This makes the max and min of the given function +3 and -3.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I created a table and drew the graph and saw a repetitive wave that would be expected with a sin graph. There is no given solution to compare to.

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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:

A is a number that dictates the amplitude of the graph. Whatever value comes out of the cosine will be multiplied by A. If A is greater than 1, there will be a vertical stretch, whereas if it is less than one, there will be a vertical compression.

Omega indirectly dictates horizontal compression. I am unsure due to the trigonometric function, however in most other functions, the value before the variable, t, in a situation like this would create a horizontal stretch if greater than one and a horizontal compression if less than one. This may be incorrect here though because the value is in a cosine.

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You have that backwards. The coefficient of t is the horizontal compression, so a large number indicates a lot of compression. A number less than 1 indicates a 'decompression' or horizontal stretch.

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Theta_0 is an angle which in this case would dictate a horizontal shift. If positive, the graph will shift left and if negative, the graph will shift right.

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theta_0 is not the horizontal shift, but it is an important part of finding the horizontal shift.

You are on the right track here. A good review and a little practice, and you'll be fine with this.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I wrote my answer according to how I have seen graphs react to their formulas in the past. However, I do not have as much experience with the graphs of trigonometric functions. I wrote my doubts in my answer. There is no given solution to compare to.

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Self-critique rating:3

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

Here, the derivative of e raised to a power is always the same multiplied by the derivative of the power. I simply worked backwards and since it is indefinite, added a constant.

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

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

Here, I pulled the constant outside of the integral to begin with. Then, I took the integral of the trig function with consideration to the variables inside. I also added a constant, C, because the integral is indefinite.

y(t) = 1/(3x + 2) = int(1/(3x + 2)) = (1/3)ln(3x + 2) + C

Here, I know the derivative of a natural logarithm is 1 over whatever the natural log is being taken of. I was able to work backwards with respect to the variable inside the natural log and add the constant.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I wrote out my steps on paper and I believe I got the correct answers. There is no given solution to compare to.

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You are doing the right things here. Very good.

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Self-critique rating:3

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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^(-3t), when t = 0, antid. = 2

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

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

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

Ans.: (-1/3)e^(-3t) + 7/3

x(t) = 2sin(4 pi t + pi/4), when t = 1/8, antid. = 2pi

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

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

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

(-1/(2pi))(-sqrt(2)/2) + C = 2pi, C = 2pi - sqrt(2)/(4pi)

Ans.: (-1/(2pi))cos(4 pi t + pi/4) + 2pi - sqrt(2)/(4pi)

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

I wrote out the limit as t approaches infinity of the function plus C and set it equal to negative 1. However, since the limit of the function as t approaches infinity is infinity, I do not believe it is possible to add any constant that will make the function approach -1.

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You are correct.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

For the first two functions, I wrote out the antiderivative, or integral, and set it equal to the given value. With this, I was able to find the constant, C. However, I was not able to solve the third problem and I believe it may be impossible. I am not confident about this and that is why I gave a confidence rating of 2. There is no given solution to compare to.

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Self-critique rating:3

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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), Split into partial fractions

2t + 4 = A(t + 1) + B(t - 3), multiply both sides by the denominator

2t + 4 = At + A + Bt - 3B, expand the multiplication

2t + 4 = (A + B)t + A - 3B, group like terms

2 = A + B and 4 = A - 3B, create two equations

-2 = 4B, B = -1/2, A =5/2, solve both equations simultaneously

Final Form: 5/(2(t - 3)) - 1/(2(t + 1))

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I used the method I remembered from Calculus to split the equation, factor, and solve for the variables. There is no given solution to compare to, but I am confident about my work.

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Self-critique rating:3

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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:

Assuming the graph is linear or near linear, I drew the graph and generated the function, f(x).

If f(2) = 5, where the slope is .5, I can write the formula f(x) = .5x + 4, because I know 4 is the y-intercept. Using this function, the value of f(2.4) = 5.2

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

The approximation is difficult in this problem because it does not say whether the graph is linear or quadratic. Assuming linear, I believe I took the correct steps to generate the function, draw the graph, and use that information to find the value at x = 2.4.

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Self-critique rating:3

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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:

I took the slopes between the three points. From (3,4) to (3.2,4.4), the slope is 2. From (3.2,4.4) to (3.4,4.5), the slope is .5. This makes the graph appear to be quadratic. Assuming the graph continues changing slope at this rate, the slope between x = 2.8 and x = 3 would be 4.

For the slope at x = 3, I took the average of the slope on either side (2 and 4), which would give a slope of 3.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I have no idea what the proper way to solve a problem like this is. However, my method allowed me to visualize the graph and make what I believe is a reasonable approximation for the amount of information given. No solution is given to compare to.

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You made a very reasonable conjecture about the slope.

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Self-critique rating:3

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Self-critique (if necessary):

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Self-critique rating:

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It took you awhile to get to this point, but your work here is very good. You've got a couple of rough edges, so be sure to see my notes, but you appear to be very well-prepared to begin this course. Very encouraging.

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