Assignment 3

course PHY 231

Iǎ²_F}{assignment #003

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Physics II

11-01-2007

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19:08:38

query 6.5 #8 Galileo: time for unif accel object to traverse dist is same as if vel was ave of init and final; put into symbols and show why true

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RESPONSE -->

(v0 + vf) / 2 = vAve

If velocity was graphed vs. time, then the definite integral over that time period would be the distance traveled. The fundamental theorem of calculus states(roughly) there must be an average value between two continous points that when multiplied by the domain will give the area. Similarly, Galileo stated that the mean between to velocities, with uniform acceleration, multiplied by time will give the distance traveled.

Good thinking, but your first statement doesn't specify exactly what quantity between the endpoint values gives the same area and does not ensure that the quantity is (v0 + vf) / 2.

It is the linearity of the v vs. t function that ensures this; since the graph is a trapezoid the average value occurs at the midpoint.

However an airtight proof should use the actual integral. One way of proving the result:

** Using s for the distance fallen we can translate Galileo's statement as follows:

t = s / [ (vf + v0)/2 ].

We easily establish that a ball dropped from rest will have velocity function

v = a * t

and position function

s = .5 a t^2.

Both of these results are easily obtained, for the given initial condition that initial velocity is zero, by a straightforward integration.

For given distance of fall s the position function gives us

time of fall t = sqrt(2 s / a).

At this time the velocity function tells us that

Final velocity = a * time of fall = a sqrt(2 s / a) = sqrt( 2 a s).

From this we calculate the average of initial and final velocities and show that at this velocity the time required to fall distance s is the same as above:

The average of initial and final velocities is

Ave of initial and final velocities = (0 + sqrt( 2 a s) ) / 2 = sqrt( a s / 2 ).

If an object travels distance s at this velocity sqrt( a / (2 s) ) then the time required is

Time required at ave of initial and final velocities

= s / sqrt( a s / 2 ) = sqrt(2 s / a),

Which is in agreement with the result obtained from the position function.

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19:08:45

how can you symbolically represent the give statement?

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19:22:26

How can we show that the statement is true?

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RESPONSE -->

A rock falls from 100 ft.

position function:

p(t) = 16t^2 + 100

0 = 16t^2 + 100

t = 2.5 seconds

velocity fuction:

p'(t) = 32t

p'(2.5) = 80

acceleration function:

p''(t) = 32

v0 = 0

vf = 80 ft/s

Galileo's Statement of vAve:

(v0 + vf) / 2 = 40 ft/s

40 ft/s * 2.5 s = 100 ft

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19:22:46

How can we use a graph to show that the statement is true?

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RESPONSE -->

See last response.

In a nutshell, because the v vs. t graph is linear, the average velocity is equal to the average of the initial and final velocities. Since

time of fall = displacement / average velocity,

it follows that

time of fall = displacement / (ave of initial and final vel).

This latter expression is just the time that would be required to fall at a constant velocity equal to the average of initial and final velocities.

More rigorously:

The graph is linear, so the area beneath the graph is the area of a triangle.

The base of the triangle is the time of fall, and its altitude is the final velocity.

By a simple construction we know that the area of the triangle is equal to the area of a rectangle whose length is equal to the base of the triangle, and whose width is equal to half the altitude of the triangle.

It follows that the area of the triangle is equal to half the final velocity multiplied by the time interval.

Since the initial velocity is 0, the average of initial and final velocities is half the final velocity.

So the area of the triangle is the product of the time of fall and the average of initial and final velocities

area beneath graph = time of fall * ave of init and final vel

The area beneath the graph is equal to the integral of the velocity function over the time interval, which we know is equal to the displacement. So we have

displacement = time of fall * ave of init and final vel, so that

time of fall = displacement / ave of init and final vel.

This leads to the same conclusion as above.

Also, more symbolically:

A graph of v vs. t graph is linear. Over any time interval t = t1 to t = t1 + `dt, displacement is represented by the area of the corresponding trapezoid, which is d = (v1 + v2) / 2 * `dt.

Solving for `dt we again obtain `dt = d / [ (v1+v2)/2 ].

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19:55:14

query problem 7.1.22 (3d edition #18) integral of `sqrt(cos(3t) ) * sin(3t)

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RESPONSE -->

integral of `sqrt(cos(3t) ) * sin(3t) =

(2/9)(cos(3t) )^(3/2)

derivative of (2/9)(cos(3t) )^(3/2) =

`sqrt(cos(3t) ) * sin(3t)

You're right except that you are missing a negative sign:

The derivative of cos(3t) is -3 sin(3t), wo the derivative of -2/3 (cos(3t))^(-3/2) is 3 sqrt(cos(3t) sin(3t). This solution didn't account for the Chain Rule.

To perform the integral use substitution:

w = cos (3t) dw = -3 sin (3t)

so that sin(3t) = -dw / 3.

Thus our expression becomes

w^(1/2) * (-dw / 3).

The integral of -1/3 w^(3/2) with respect to w is -1/3 * (2/3) w^(3/2), treating w^(1/2) as a power function.

This simplifies to

-2/9 w^(3/2) or

-2/9 * (cos(3t))^(3/2).

The general antiderivative is

-2/9 * (cos(3t))^(3/2) + c,

where c is an arbitrary constant.

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19:55:23

what did you get for the integral and how did you reason out your result?

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RESPONSE -->

Guess and check

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00:05:13

query problem 7.1.20 (3d edition #21) antiderivative of x^2 e^(x^3+1)

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RESPONSE -->

e^(x^3+1)/3

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00:06:04

what is the antiderivative?

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RESPONSE -->

e^(x^3+1)/3

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00:14:38

What substitution would you use to find this antiderivative?

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RESPONSE -->

The indefinite integral, x^2 * e^(x^3 + 1) dx, has an inside function, (x^3 + 1) we will call w. Then, dw = 3x^2. So, 1/3 dw = x^2, the outside finction.

Then we have:

The indefinite integral e^(w) * 1/3 dw = 1/3 indefinite integral of e^(w) * dw = e^(w) / 3 = e^(x^3 + 1) / 3

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f⛼wnQK

assignment #003

query problem 7.1.37 (3d edition #35) antiderivative of (t+1)^2 / t^2

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17:28:00

what is the antiderivative?

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RESPONSE -->

(2 * ln (abs(x)) + x - 1) / x

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17:32:10

What substitution would you use to find this antiderivative?

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RESPONSE -->

antiderivative of (t + 1)^2 / t^2 =

antiderivative of (t^2 + 2t + 1) / t^2 =

antiderivative of t^2 / t^2 +

antiderivative of 2t / t^2 +

antiderivative of 1 / t^2 =

t + 2 * ln(abs(t)) - 1 / t

I did not use substitution.

Good.

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17:37:57

query 7.1.64 (3d edition #60). int(1/(t+7)^2, t, 1, 3)

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RESPONSE -->

int(1/(t+7)^2, t, 1, 3)

antiderivative of (t + 7)^-2 = -1 / (t + 7)

-1 / (3 + 7) - (-1) / (1 + 7) = .025

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17:38:11

What did you get for the definite integral?

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RESPONSE -->

.025

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17:38:50

What antiderivative did you use?

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RESPONSE -->

-1 / (t + 7)

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17:39:37

What is the value of your antiderivative at t = 1 and at t = 3?

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RESPONSE -->

-1/8 at t = 1

and -1/10 at t = 3

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20:31:35

query 7.1.86. World population P(t) = 5.3 e^(0.014 t).

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RESPONSE -->

(a)

P(t) = 5.3 e^(0.014 t)

In 1990,

P(0) = 5.3 billion

In 2000,

P(10) = 6.096 billion

(b)

int P,t,0,10 = [(5.3 * e^(0.014 * 10) / 0.014) - (5.3 * e^(0.014 * 10) / 0.014)] = 56.889

56.889 / (b - a) = 56.889 / 10 = 5.68894

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20:31:39

What were the populations in 1990 and 2000?

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20:31:42

What is the average population between during the 1990's and how did you find it?

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20:33:54

What is the value of your antiderivative at t = 1 and at t = 3?

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RESPONSE -->

at t = 1

5.3 * e^(0.014(1)) / 0.014 = 383.909

at t = 3

5.3 * e^(0.014(3)) / 0.014 = 394.810

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20:34:08

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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Good work. See my notes.