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course Mth 279
2/2 17
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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 = 3 sin(4 t + 2)
y = 3 cos(4 t + 2) * 4 chain rule
y = 12 cos(4 t + 2)
y = -12 sin(4 t + 2) * 4
y = -48 sin(4 t + 2)
y = 2 cos^2(3 t - 1)
y = [2 cos(3 t - 1)]^2 simplify by algebra
y = 2 * [2 cos(3 t - 1)] * -2 sin(3 t -1) * 3 chain rule
y = 2 * [2 cos(3 t - 1)] * -6 sin(3 t - 1)
y =
??? I am lost at how to proceed for the second derivative
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You could find the derivatives of cos(3t-1) and sin(3t-1).
Then combine them with the functions themselves using the product rule
(fg)' = f ' g + g ' f
and multiply by the constant factor 2 * 2 * (-6) = -24,
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y = sin(omega * t + phi)
y = A cos(omega * t + phi) * omega
y = omega * A cos (omega * t + phi)
y = - omega * A sin (omega * t + phi) * omega
y = -omega^2 * A sin (omega * t + phi)
y = 3 e^(t^2 - 1)
y = 3 e^(t^2 - 1) * 2 t
y = (3e^(t^2 - 1) ) * 2 t + 2 (3e^ (t^2 - 1)) Product Rule
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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:
y = 3 sin(4 t + 2)
Due to the sin function, the graph will oscillate. The max y value of the graph is 3 when the sin function is equal to pi / 2 , as :
sin (pi / 2) = 1
With 4 t + 2 = pi / 2
8 t + 4 = pi
t = (pi - 4) / 8
t = pi / 8 - 1 / 2
So one of the y max values is at ( pi / 8 - 1 / 2 , 3)
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Given Solution:
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Self-critique (if necessary):
??? I was not confident about using degrees vs. radians to find when the value of pi / 2 would occur. I am confident however the max height of the graph would be 3 and the least height would be a y-value of - 3 due to the 3 before the sin function.
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Good start, but more is needed.
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4 t + 2 = pi / 2
can be interpreted as follows, if t is interpreted as time in seconds:
4 has units radians / second
2 has units of radians.
pi/2 also has units of radians.
So the equation is dimensionally consistent, since 4t represents radians / second * seconds = radians.
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To describe the graph you need to also describe its period. Knowing that and one of the points of maximum would allow you to give a formula for all the max points. In this case the period is pi/2 so the max points will be
(pi/8 - 1/2 + (pi/2) * n,
where n can be any integer.
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When we get to applications that involve sine and cosine graphs, and a few weeks, you'll want to be completely up on this.
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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 have a maximum of A + k and a minimum of -A + k. This is because the cos function can be no greater than 1 and no less than -1.
Due to the cos function, the graph will oscillate with between this maximum and minimum with a period of 2 pi.
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Again, good start, but you do have an error, and in any case more is needed.
The period of cos(t + theta_0) would be 2 pi.
The period of cos(omega t + theta_0) depends on omeage.
The graph is shifted along the t axis, and this shift needs to be specified.
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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) = e^(-3 t)
= - 1 / 3 * e^-3 t)
x(t) = 2 sin( 4 pi t + pi / 4) anti-derivative of sin = -cos
= -2 cos( 4 pi t + pi / 4) * (2 pi t^2 + pi t / 4) chain rule
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If you take the derivative of this result, you do not get the original function.
You should integrate this by first letting u = 2 pi t + pi/4.
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y(t) = 1 / ( 3 x + 2 ) = ( 3 x + 2 )^-1
= 1 / 3 log( 3 x + 2)
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Self-critique (if necessary):
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Not bad, but check my notes and be sure to remedy everything that needs to be remedied.
You can probably do this on your own, but if you do have questions they will be welcome.
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