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course MTH-279

06/22 around 8:15PM.

`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:

f(t) = 3 sin(4 t + 2)

f '(t) = d/dt(3 sin(4 t + 2))

= 3(d/dt(sin(4 t + 2))) -> factor out constants

= 3(cos(4 t + 2)(d/dt(4t + 2))) -> using the chain rule

= (3cos(4t + 2))(4) -> power rule, d/dt(4t + 2) = 4

= 12cos(4t + 2) -> simplify, 4 times 3 = 12

f ''(t) = d/dt(12cos(4t + 2))

= 12(d/dt(cos(4t + 2))) -> factor out constants

= 12(-sin(4t + 2)(d/dt(4t + 2))) -> using chain rule

= (-12sin(4t + 2))(4) -> power rule, d/dt(4t + 2) = 4

= -48sin(4t + 2) -> simplify, 4 times 12 = 48

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

cos^2(u) = 1/2 + 1/2cos(2u), u = 3t -1

f(t) = cos(6t -2) + 1/2 -> by using the above trig identity

@&

This is good, but unless you remember those trig identities it's simpler to use the chain rule.

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f '(t) = d/dt(cos(6t -2) + 1/2)

= -sin(6t -2)(d/dt(6t - 2)) -> using chain rule

= (-sin(6t - 2))(6) -> power rule, d/dt(6t - 2) = 6

= 6sin(2 - 6t) -> simplify, and distribute negative sign.

f ''(t) = d/dt(6sin(2 - 6t))

= 6(d/dt(sin(2 - 6t))) -> factor out constants

= 6(cos(2 - 6t)(d/dt(2 - 6t))) -> using chain rule

= 6(cos(2 - 6t)(-6) -> power rule, d/dt(2 - 6t) = -6

= -36(cos(2 - 6t) -> simplify, 6 times -6 = -36

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f(t) = A sin(omega * t + phi)

f '(t) = d/dt(A sin(omega * t + phi))

= A(d/dt(sin(omega * t + phi))) -> factor out constants

= A(cos(omega * t + phi)(d/dt(omega * t + phi))) -> using chain rule

= (A(cos(omega * t + phi))(omega) -> power rule, d/dt(omega * t + phi) = omega

= A(omega)cos(omega * t + phi) -> simplify

f ''(t) = d/dt(A(omega)cos(omega * t + phi))

= A(omega)(d/dt(cos(omega * t + phi))) -> factor out constants

= A(omega)(-sin(omega * t + phi)(d/dt(omega * t + phi))) -> using chain rule

= (-A(omega)(sin(omega * t + phi))(omega) -> power rule, d/dt(omega * t + phi) = omega

= -A(omega)^2sin(omega * t + phi) -> simplify

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f(t) = 3 e^(t^2 - 1)

f '(t) = d/dt(3 e^(t^2 - 1))

= 3 (d/dt(e^(t^2 - 1))) -> factor out constants

= 3 (e^(t^2 - 1)(d/dt(t^2 - 1))) -> using chain rule

= (3 (e^(t^2 - 1))(2t) -> power rule, d/dt(t^2 - 1) = 2t

= 6te^(t^2 - 1) -> simplify

f ''(t) = d/dt(6te^(t^2 - 1))

= 6(d/dt(te^(t^2 - 1))) -> factor out constants

= 6(t(d/dt(e^(t^2 - 1))) + (d/dt(t))(e^(t^2 - 1)) -> product rule, d/dt(uv) = v(du/dt)+u(dv/dt)

= 6te^(t^2 - 1)(d/dt(t^2-1)) + e^(t^2 - 1) -> using chain rule, and d/dt(t) = 1

= (6te^(t^2 - 1))(2t) + e^(t^2 - 1) -> power rule, d/dt(t^2-1) = 2t

= 12t^2e^(t^2 - 1) + e^(t^2 - 1) -> combine 6t and 2t

= e^(t^2 - 1)(12t^2 + 1) -> simplify by factoring out e^(t^2 - 1)

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Given Solution:

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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 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 4 times shorter than a normal sin function.

@&

You don't say how far this function shifts, and the shift is 1/2 unit to the left.

Be sure you understand not only the rules for the shift but the reasons for those rules.

Everything but the shift is correct.

*@

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Given Solution:

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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 realized as A. The graph will be shifted in the x direction by theta_0. Omega will stretch or shrink the graph in the x direction.

@&

The shift is not theta_0. See also my preceding note.

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Given Solution:

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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

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x(t) = 2 sin( 4 pi t + pi/4)

Integration of (2 sin( 4 pi t + pi/4) dt)

= 2 (Integration of (sin( 4 pi t + pi/4) dt) -> factor out constants

= (1/2)(Integration of (sin(u)du)) -> substitute u=( 4 pi t + pi/4), du=(4 pi )dt

= (-cos(u))/(2 pi) + constant -> integration of sin(u) = -cos(u)

= (-cos( 4 pi t + pi/4))/(2 pi) + constant -> substitute u with original function

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y(t) = 1 / (3t + 2)

Integration of (1 / (3t + 2) dt)

= (1/3) Integration of ((1/u) du) -> substitute u=(3t + 2), du= 3 dt

= (1/3)(ln(u)/3) + constant -> (1/u) = ln(u)

= (1/3)(ln(3t + 2)) + constant -> substitute u with original function

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Given Solution:

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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) = e^(-3 t) = 2 when, t = 0

The antiderivative from question 4 was:

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

when t=0 we can find a solution for the constant

-(1/3) + constant = 2 -> substitution of values from f(t) = e^(-3 t) = 2, when t = 0

From this the constant must equal (7/3) in order to get an antiderivative of 2.

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x(t) = 2 sin( 4 pi t + pi/4) = 2 pi, when t = (1/8)

The antiderivative from question 4 was:

(-cos( 4 pi t + pi/4))/(2 pi) + constant

When t = (1/8), we can find a solution for the constant

0.113 approximately + constant = 2 pi -> substitution of values from x(t) = 2 sin( 4 pi t + pi/4) = 2 pi, when t = (1/8)

From this the constant must equal approximately 6.396 in order to get an antiderivative of 2 pi.

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y(t) = 1 / (3 t + 2) = -1, as t approaches infinity

The antiderivative from question 4 was:

(1/3)(ln(3t + 2)) + constant

It appears that no antiderivatives that meet those qualities exist.

@&

This is correct. There is no constant number that will make the limit -1.

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As t increases, the whole function increases and remains positive.

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Given Solution:

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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:

2t + 4 = A(t+1) + B(t-3), set t=3, solve for A= 5/2

2t + 4 = A(t+1) + B(t-3), set t=-1, solve for B= -1/2

Substitute A and B into equation and integrate:

(2t +4)/ (t-3)(t+1) = A / (t - 3) + B / (t + 1), where A= 5/2 and B= -1/2; solve for integration = 5/2ln(t-3) - 1/2ln(t+1)

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Given Solution:

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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:

The equation of the graph is y= (1/2)x + 4

substitute x=2.4 into equation and solve

f(2.4)= 5.2

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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 slope between the two points (3.2, 4.4) and (3.4, 4.5) is (1/2). The slope between the two points (3, 4), (3.2, 4.4) is 2. Best guess estimate says that the graph is increasing on a decreasing manner. Therefore, the slope when x=3 will be greater than 2. I would guess g'(3)= 5/2

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Given Solution:

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@&

Very well done, but give the horizontal shift some careful thought (per my inserted notes).

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