Assignment 0

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course Mth 279

q1Question: `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:

• Chain Rule: 3(4)cos(4t+2) = 12cos(4t+2)

o -(12*4)sin(4t+2) = -49sin(4t+2)

• Product Rule: 2cos(3t-1)cos(3t-1) 2cos(3t-1)(-3sin(3t-1)) + cos(3t-1)(-6sin(3t-1)) -6cos(3t-1)sin(3t-1) - 6cos(3t-1)sin(3t-1) = -2(6cos(3t-1)sin(3t-1))

o -2[(6cos(3t-1)(3cos(3t-1))+(-18sin(3t-1))(sin(3t-1))] = 2[18cos^2(3t-1)-18sin^2(3t-1)]

• Assuming in terms of t: A(omega)cos(omega*t+phi)

o -A(omega^2)sin(omega*t+phi)

• Exponentials stay the same, only have to worry about chain rule: 3(2t)e^(t^2-1) = (6t)e^(t^2-1)

o (6t)(2t)e^(t^2-1) = (12t^2)e^(t^2-1)

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The second derivative requires application of the product rule as well as the chain rule to the first derivative.

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating: OK

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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 coefficient of 3 is the altitude of the sine curve. The 4 and 2 inside the parenthesis shift the curve.

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Given Solution:

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Self-critique (if necessary):

I do not know exactly how the 4 and 2 shift the graph.

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You will need to review your precalculus to understand how stretches and shifts are used to obtain one graph from another. This will be important later in the course.

Basically the graph of A f( t - h ) is vertically stretched by factor A and horizontally shifted h units.

When the coefficient of t isn't 1, things get a little more complicated, but this is still precalculus-level information and with a quick review it should not give you any trouble.

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Self-critique rating: OK

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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 the amplitude of the graph and omega and theta_0 shift the cosine curve.

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You are right about the amplitude.

You need to review the details related to the other parameters.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating: OK

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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 integral of a function is just the antiderivative. A constant C must be added because it is an indefinite integral.

• -1/3 e^(-3t) + C

• -2cos(4pi t + pi/4) + C (Values from chain rule cancel)

• Integral of 1/x = ln(x)

o (1/3)ln(3t+2) + C

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating: OK

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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:

These solutions will allow us to find the value of C.

• -1/3 e^(-3(0)) + C = 2

o -1/3(1) + C = 2 C = 7/3

• -2cos(4pi t + pi/4) + C (Values from chain rule cancel)

o -2cos(4pi (1/8) + pi/4) + C = 2pi C = 8.3

• Integral of 1/x = ln(x)

o (1/3)ln(3t+2) + C

o

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

Limits are not my strong suit.

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The key question is what happens to ln(3t + 2) as t approaches infinity?

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Self-critique rating: OK

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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-3)+1/(t+1)

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You are expected to show, at least in brief outline, the details of your solution.

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Given Solution:

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Self-critique (if necessary):

I can’t quite remember this formula…

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Self-critique rating: OK

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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:

Would this be linearization?

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This is a linear approximation. The term 'linearization', though, has a different meaning.

If the slope of a straight line through (2, 5) is .5, what is the y value when x = 2.4?

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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 average slope can be found by the rise over run. .5/.4 = 1.25.

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That's an estimate, but it isn't a best estimate. You have information for three points, and you have based your result on only two of them.

To obtain a reasonable answer to this question you need to consider the trend of the slopes.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I am confused about how to find the equation for the slope so that I can find the value.

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&#This looks good. See my notes. Let me know if you have any questions. &#