qa 00

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course Mth 279

1/22 10pm

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’=12cos(4t+2)

y’’=-48sin(4t+2)

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

y’=-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 #$&*:8232; 

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Given Solution: 

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Self-critique (if necessary):

 

 

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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 I obtained is sinusoidal, shifted to the right. The amplitude is 3. The period is shortened as well.

 

 

confidence rating #$&*:232; 

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Given Solution: 

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Self-critique (if necessary):

 

 

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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: 

 graph is shifted in the y direction by k. Amplitude is A. T shift is -theta 0/ omega

 

 

confidence rating #$&*:232; 

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Given Solution: 

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Self-critique (if necessary):

 

 

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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 #$&*:232; 

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Given Solution: 

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Self-critique (if necessary):

 

 

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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

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

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

confidence rating #$&*:232; 

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Given Solution: 

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Self-critique (if necessary):

 

 

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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 —>B=-5/2

t=3 B —> A=5/2

 

 

confidence rating #$&*:232; 

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Given Solution: 

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Self-critique (if necessary):

 

 

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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 #$&*:232; 

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Given Solution: 

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Self-critique (if necessary):

 

 

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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 best estimate of g’(3) would be 3 for a tangent line

 

 

confidence rating #$&*:232; 

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Given Solution: 

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Self-critique (if necessary):

 

 

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Self-critique rating:"

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Some of your answers included sufficient detail, while some were short on detail. However your results are good, and I assume you obtained them in the appropriate manner.

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