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course Mth 279
5/27 9
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:
1) y’ = 12cos(4t+2)
y” = -36sin(4t-+2)
2) y’ = -12cos(3t-1)sin(3t-1)
y” = -36cos(2-6t)
3) y’=omega*Acos(omega*t-phi)
y” = -omega^2*Acos(omega*t-phi)
4) y’ = 6te^(t^2-1)
y” = 12t^2e^(t^2-1)
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These are correct, but you haven't shown how you obtained these results.
In general you need to show, at least in outline, how your results are obtained.
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confidence rating #$&*:
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Question:
`q002. Sketch a graph of the function y = 3 sin(4 t + 2). Don't use a graphing calculator, use what you know about graphing. Make your best
attempt, and describe both your thinking and your graph.
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Your solution:
The graph has a range of (-3,3) which is 3 times the size of a typical sine curve. A period occurs every pi/2. That is denoted by the 4t. The sine curve is also shifted to the left by 2. This is denoted by the 2. In short, this graph has the general shape of a sine curve but has many more peaks and valleys, 3 times taller, and it is shifted to the left.
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Both this graph and that of sin(t) have a countably infinite number of peaks and valleys. It would be more appropriate to say that the frequency of peaks and valleys is 4 times greaer for the given function.
I believe the graph would be shifted 1/2 unit to the left of that of sin(t).
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Confidence rating: 3
Question:
`q003. Describe, in terms of A, omega and theta_0, the characteristics of the graph of y = A cos(omega * t + theta_0) + k.
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Your solution:
This graph is a cosine curve that has been shifted up k and left theta_0).
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The shift would not be -theta_0. Be sure to review how these graphs transform (you've got most of it, but you tend to miss the shift on the t axis).
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The graph has also been stretched/shrunk to a height of A. This means the graph ranges from (A+k, A-k). A period occurs every 2pi/omega.
confidence rating #$&*:
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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:
1) F(t) = -1/3 * e^(-3t) + C
2) X(t) = (-cos(4pi*t) - sin(4pi*t))/(2sqrt(2)pi) +C
3) Y(t) = -ln(3x+2) + C
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Again you would want to show your steps.
The derivative of -ln(3x+2) is not equal to 1 / (3x + 2), so your last answer is clearly 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:
1) F(t) = -1/3 * e^(-3t) + C
2 = -1/3 + C, C = 7/3
F(t) = -1/3 * e^(-3t) +7/3
2) X(t) = (-cos(4pi*t) - sin(4pi*t))/(2sqrt(2)pi) +C
2pi = 1/(2sqrt(2)pi) + C, C = 4sq(2)*pi^2
X(t) = (-cos(4pi*t) - sin(4pi*t))/(2sqrt(2)pi) + 4sq(2)*pi^2
3) Y(t) = -ln(3x+2) + C
-1 = 1/infinity + C, C = -infinity
Y(t) = -ln(3x+2) -infinity
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This is not a valid function. Infinity isn't a number.
The bottom line is that in this case, a solution satisfying the given conditions does not exist.
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Confidence rating: 3
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:
y’ = 2t/(t-3) + 4/(t+1)
confidence rating #$&*:
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I do not remember how to do the partial fraction process.
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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(2.4) = f(2) + f’(2)*(2.04-2)
f(2.4) = 5 + .5*.04 = 5.02
Confidence rating: 3
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:
First slope = (4.5-4)/(3.4-3.2) = 2.5
Second slope = (4.4-4)/(3.2-3) = 2
g’(3) can be approximated to be 1.5
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Good conjecture.
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Confidence rating: 2
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Despite a couple of errors, it's obvious that you are in good shape to start this course.
However it would be much to your advantage to reconcile your thinking with my inserted notes.
If you have questions I'll be glad to answer them.
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