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course Mth 279
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:
**3 sin(4 t + 2)
12 cos(4t + 2)
-48 sin(4t + 2)
**2 cos^2(3 t - 1)
-12 cos(3t - 1) sin(3t - 1)
-36 sin^2(3t - 1) + 36cos^2(3t - 1)
**A sin(omega * t + phi)
A*omega cos(omega*t + phi)
A*omega^2 cos(omega*t + phi)
**3 e^(t^2 - 1)
6t e^(t^2 - 1)
12t^2 e^(t^2 - 1)
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You have at least one error in sign.
You also need to show your work, not just your final results.
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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:
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You at least need to indicate how you would approach this question.
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With that information I could provide some help, in the event it is needed.
Graphs and functions of this sort will be very important in this course.
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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 change from a positive slope to a negative slope (or vice versa) where the function (A*omega sin(omega*t +theta_0)) is equal to zero, and will change concavity when the function (-A*omega^2 cos(omega*t +theta_0)) is equal to zero.
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That's not a bad start.
However there are many more details to these graphs.
Graphs of this nature are addressed on the introductory disk.
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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^(-3 t)
-1/3 e^(-3t) + C
** x(t) = 2 sin( 4 pi t + pi/4)
(-1/2pi) cos(4pi t + pi/4) + C
** y(t) = 1 / (3 x + 2)
3 ln(3x + 2) + C
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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:
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For example if you set
-1/3 e^(-3t) + C
equal to 2, with t = 0, what is the value of C?
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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:
After working this problem out on paper, I got it to be expressed as the following:
(2 t + 4) / ( (t - 3) ( t + 1) ) = 2.5/(t-3) - .5/(t+1)
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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:
If f(2) = 5 and f’(2) = .5, then we can infer from the slope that f(2.4) = 5.8 because .4/.8 equals
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Right idea, but slope * run = rise, so rise is .5 * .4 = .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:
If g(3) = 4 and g(3.2) = 4.4 then by estimating slope we find that g’(3) = .4/.2 = 2
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That would be close, but it could be improved upon.
There is a trend to the slope, established by the 3 points.
You haven't taken account of that trend.
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&#This looks good. See my notes. Let me know if you have any questions. &#