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course MTH 279

6/3 11

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:

y=3sin(4t+2)

y’=3cos(4t+2)*(4)=12cos(4t+2)

y’’=-12sin(4t+2)(4)=-48sin(4t+2)

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

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

Asin(omega*t+phi)

y’=Acos(omega*t+phi)*(omega)

y’’=A*omega^2*cos(omega*t+phi)

3e^(t^2-1)

y’=3e^(t^2-1)*(2t)

y’’=6e^(t^2-1)+12t^2*e^(t^2-1)

confidence rating #$&*:

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I think I may have made an error for the second equation. I tried to use the chain rule, but the second derivative gets rather long and tricky with my solution for the first derivative, which leads me to believe I made a mistake somewhere.

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The derivative of cos(3t-1) sin(3t-1) is a fairly straightforward application of the product rule. The chain rule gives you a factor of 3 with each derivative, and you end up with multiples of cos^2(3t-1) and sin^2(3t-1).

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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:

My graph is sinusoidal, but is shifted to the right as a result of the +2 in the parenthesis. The amplitude is 3 because of the 3 at the beginning of the equation. Because t is multiplied by 4 the period is shorter than a normal sin function.

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Specifically it's 4 times shorter, so the period is 2 pi / 4 = pi/2.

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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 will be shifted in the y direction by the value k. The amplitude will be A. The graph will be shifted in the x direction by theta not. Omega will stretch or shrink the graph in the x direction.

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The t shift is -theta_0 / omega.

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

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

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

confidence rating #$&*:

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If I remember these steps correctly my answers should be fine. I am a little rusty on some of the rules however so some (mainly Y(t)) might be incorrect.

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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)=-1/3(e^-3t)+2

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F(0) would be -1/3 + 2, since e^0 = 1.

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

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t = 1/8 gives you -pi/2 cos(pi/2), which is zero. So you would add 2 pi.

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Y(t)=1/3ln(3t+2)+ (the natural log function approaches infinity, so it is impossible for the limit as t approaches infinity to be -1 by adding a constant. I either integrated incorrectly or this is a trick question)

confidence rating #$&*:

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I was unable to determine the answers for the second two problems. I am relatively confident with the theory behind them so my guess is I either made an arithmetic mistake or I am rusty on some of the integration rules.

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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+1)/((t-3)(t+1))+B(t-3)/((t-3)(t+1))=(2t+4)/((t-3)(t+1)) so numerators must be equal.

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

at t=-1 A zeroes out giving B=-5/2 and at t=3 B zeroes out giving A=5/2

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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:

f’(x)=.5

f(x)=.5x+C

f(2)=.5(2)+C

C=4

f(2.4)=.5(2.4)+4=5.2

confidence rating #$&*:

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I feel confident in my calculations, but I have a feeling that I am missing something in this question.

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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:

my estimate for g’(3) is 3

confidence rating #$&*:

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The graph seemed to be increasing at a decreasing rate, so from the points given the slope (the derivative) should be highest at the first point. The line between the first two points has a slope of 2, so I chose 3 for the tangent line at the first point, but I feel as though there is a more “mathematical” way of reaching a conclusion.

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Good work.

One purpose of this assignement is to provide a review, the other to anticipate some procedures and concepts that will be useful in the course.

I've inserted some comments.

You appear to be in good shape at this point.

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