Assignment 0

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course Mth 279

2/5Not to worry, this is a good start, and I have the next few days to work on much of the rest of it. Will be caught up soon, possibly by next week and be ready for the first test.

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:

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

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:

Its easier to describe, rather than sketch, so I shall paint a mental picture.

The function: y = 3 sin(4t + 2) is easiest to drescribe when we talk about the original function that this one is based off.

y = sin(t)

now each element of this graph can be broken down, and describe how it looks different from its similar parent funtion.

The 3 in front increases the amplitude of the sine wave which was originally 1. Sine goes from y = -1 to y = 1. This funtion goes from -3 to 3.

The 4 in front of the t, makes the wavelengths of the sine function shorter, giving more sine curves in the same space.

The 2 added to 4t shifts the graph 2 to the right.

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2 added to 4 t doesn't shift the graph 2 units to the right.

sin(4 t + 2) = sin(4 ( t + 1/2) ),

which shifts the graph 1/2 unit to the left.

The function f(t - k) will encounter the same argument at f(t), provided t is k units greater. So the graph of f(t - k) is shifted k units to the right of the graph of f(t).

Note that sin(4 (t + 1/2) ) is of the form f(t - (-1/2) ), for the function f(t) = sin(4 t).

This should provide you with the information you need to think through this part, but if not be sure to let me know and I'll be glad to clarify further.

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Other than those changes it resembles a normal sine curve.

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Other than your horizontal shift, your analysis and your explanations are very good.

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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:

y = A cos(omega * t + theta_0) + k

This graph also can be described by its parts.

A is the amplitude of this cosine curve function. Whatever A is, will be the height, or depth of the crests and troughs.

Depending on what omega is, will change the wavelengths of the cosine curves.

Giving theta_0 a value could shift the graph to the left or right, depending on if the value was positive(right) or negative(left).

K would do something similar to theta_0. It could shift the graph up or down if given a value.

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The horizontal shift would be -theta_0 / omega. Let me know if you're not completely sure why.

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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 / (3t + 2)

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Your solution:

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

Integration of (e^(-3 t) dt)

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

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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

confidence rating #$&*:

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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 approximatly + 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 approximatly 6.396 in order to get an antiderivative of 2 pi.

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

The antiderivative from question 4 was:

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

It appears that no antiderivatives that meet those qualities exist.

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

It seems impossible to get it to equal negative one, everytime.

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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:

Using partial fractions you could Integrate the function -> (2 t + 4) / ( (t - 3) ( t + 1) ).

But for this question I beleive that just the form is wanted, so...

we need to find A, and B.

First we use a little bit of cross multiplication, following along with the partial fractions integration method.

(2t + 4) = A(t+1) + B(t - 3)

= At + A + Bt -3B

Setting up 2 equations

A + B = 2, and A - 3B = 4

Subtracting the second equation from the first cancels out A, so that B can be solved

A + B = 2

-(A - 3B = 4)

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4B = -2

B = (-1/2)

Then pluging B into the first equation, can be solved for A

A + (-1/2) = 2

A = 2.5

Therefore, (2 t + 4) / ( (t - 3) ( t + 1) ) = 2.5 / (t - 3) + (-1/2) / (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:

To have both of these conditions, I first consider y = x as a function, to see what slope it has.

Its slope is well known to be 1 rise/run, which is 1/1.

The slope of (1/2)x gives a better slope and points that start to look like they would work, but does not contain (2,5).

Finally the function f(x) = (1/2)x+4 contains the point (2,5), and if you plug in the number it works!

(1/2)x+4 -> by adding 4 we shifted the original function up 4 units.

(1/2)(2)+4

1+4

5

which shows that f(2) = 5, with this function and slope.

f(2.4) = (1/2)(2.4)+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:

I have tried a few things in an attempt to solve this, but alas cannot remember anything to help with it.

What I did figure out though,

The slope between the first 2 points is 2, but the slope between the last 2 is (1/2).

apon doing a chart and listing these points, I saw that there were 0.2 gaps in the x, and a 0.4, then a 0.1 gap in the y, which is frustrating!

from that I gathered the slopes

I tried some functions to do with (1/2)x and 2x, but can't seem to find that one equation that makes all the numbers work.

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The trend seems to be toward decreasing slope. So you would infer that g ' (3) would be somewhat greater than the average slope between t = 3 and t = 3.5, which you correctly calculate as 2.

So the slope will be somewhat greater than 2.

If you assume that the average slope between two points is equal to the actual slope about halfway between those two points, then you expect that slope 2 will occur roughly at t = 3.1, while slope 1/2 will occur roughly at t = 3.3. So the slope appears to decrease by 3/2 over an interval of length .2.

Between t = 3 and t = 3.1, then, the slope might be expected to decrease by about half this amount, by about 3/4.

This would give us a slope of 2 3/4 at t = 3.

This is a pretty crude set of inferences. More sophisticated models could improve on this conclusion, but without additional information any model will depend on a fairly arbitrary set of assumptions.

One assumption would be that the graph is approximately quadratic in the vicinity of these three points. A parabola through the three points has equation -3.75 t^2 + 25.25 t - 38. The derivative of this expression, evaluated at t = 3, is 7.5 t + 25.25 = 7.5 * 3 + 25.25 = 2.75, or 2 3/4. Coincidentally, this agrees with the result obtained above.

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Given Solution:

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Good work. You clearly have a solid background, and you explain your thinking very well.

You did have a couple of errors, so be sure to see my notes, especially on a couple of the earlier problems. If you have questions be sure to let me know.

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