Assignment 12

course MTH 174

kn}Ǧwi`ƩXˮassignment #012

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Physics II

10-23-2008

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19:55:31

Query problem 9.2.8 (3d editin 9.1.6) (was 9.4.6) first term and ratio for y^2 + y^3 + y^4 + ...

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RESPONSE -->

It is a geometric series. The first term is y^2. The ratio is y.

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19:55:49

either explain why the series is not geometric or give its first term and common ratio

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RESPONSE -->

The first term is y^2. The ratio is y.

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20:00:42

how do you get the common ratio?

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RESPONSE -->

Factoring out y^2 makes the series look like:

y^2(1 + y + y^2 + y^3 + y^4 + y^5...)

The reveals that the common ratio is in fact y

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20:01:32

what do you get when you factor out y^2? How does this help you determine the first term?

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RESPONSE -->

By knowing the ratio is y, this means that the first term is y^2

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20:33:27

Query problem 9.2.29 (3d edition 9.1.24) (was 9.4.24) bouncing ball 3/4 ht ratio

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RESPONSE -->

a) g = 32ft/s^2

antiderivative of g is g * t + C. C is 0 because the ball starts at rest. This is initial velocity.

antiderivative of v0 + g * t is v0 * t +g * t^2/2 + C C is zero this time because the ball starts at position 0ft.

h = v0 * (0 ft/s) + g * t^2/2 + 0ft

h = g * t^2/2

2 * h = g * t^2

2 * h/g = t^2

`sqrt( 2 * h/g ) = t

`sqrt( 2 * h / 32 ) = t

`sqrt( h / 16 ) = t

1/4 `sqrt( h ) = t

b) The time required to drop from 10ft is 1/4 * `sqrt(10)

The time must double for the next iteration because it goes up to a height of 10 * (3/4) and then falls that distance.

So 1/4 * `sqrt(10) + 2 * 1/4 * `sqrt(10 * 3/4) + 2 * 1/4 * `sqrt(10 * (3/4)^2) + 2 * 1/4 * `sqrt(10 * (3/4)^3) ... `sqrt(10 * (3/4)^(n-1))

This becomes:

1/4 * `sqrt(10) + 1/2`sqrt(10 * (3/4) *

[(1 - `sqrt(3/4)^n)/(1 - (3/4)]

The limit as n->infinity makes this:

1/4 * `sqrt(10) + 1/2`sqrt(10 * (3/4) *

[(1)/(1 - (3/4)] which is approx 11 seconds.

After the first iteration you have a geometric series with common ratio sqrt(3/4); the ratio isn't 3/4 but you otherwise have the right idea.

** If the ball starts from height h, it falls to the floor then bounces up to 3/4 of its original height then back down, covering distance 3/2 h on its first complete bounce. Then it bounces to 3/4 of that height, covering in the round trip a distance of 3/4 ( 3/2 h). Then it bounces to 3/4 of that height, covering in the round trip a distance of 3/4 [ 3/4 ( 3/2 h) ] = (3/4)^2 * (3/2 h). On the next bounce the round trip will be 3/4 of this, or (3/4)^3 * (3/2 h). All the bounces give us a distance of (3/4) * (3/2 h) + (3/4)^2 * (3/2 h) + (3/4)^3 * (3/2 h) + ... = [ 3/4 ( 3/2 h) ] * ( 1 + 3/4 + (3/4)^2 + ... ) = [9/8 h] * ( 1 + 3/4 + (3/4)^2 + ... ) = (9/8 h) ( 1 / ( 1 - 3/4) ) = 9/2 h. There is also the initial drop h, so the total distance is 11/2 h. But this isn't the question. The question is how long it takes the ball to stop.

The time to fall distance y is ( 2 * y / g ) ^ .5, where g is the acceleration of gravity. This is also the time required to bounce up to height h. The distances of fall for the complete round-trip bounces are 3/4 h, (3/4)^2 h, (3/4)^3 h, etc.. So the times of fall are ( 2 * 3/4 h / g ) ^ .5, ( 2 * (3/4)^2 h / g ) ^ .5, ( 2 * (3/4)^3 h / g ) ^ .5, ( 2 * (3/4)^4 h / g ) ^ .5, etc.. The times for the complete round-trip bounces will be double these times; the total time of fall will also include the time sqrt( 2 h / g) for the first drop.

We can factor the ( 3/4 ) ^ .5 out of each expression, getting times of fall (3/4)^.5 * ( 2 * h / g ) ^ .5 = `sqrt(3)/2 * ( 2 * h / g ) ^ .5, ( sqrt(3)/2 ) ^ 2 * ( 2 * h / g ) ^ .5, ( sqrt(3)/2 ) ^ 3 * ( 2 * h / g ) ^ .5, etc..

Adding these terms up, multiplying by 2 since the ball has to go up and down on each bounce, and factoring out the (2 * h / g ) ^ .5 we get 2 * (2 * h / g ) ^ .5 * [ ( sqrt(3)/2 ) + ( sqrt(3)/2 ) ^ 2 + ( sqrt(3)/2 ) ^ 3 + ( sqrt(3)/2 ) ^ 4 + ... ] .

The expression in brackets is a geometric series with r = ( sqrt(3)/2 ), so the sum of the series is 1 / ( 1 - ( sqrt(3)/2 ) )

h and g are given by the problem so 2 * (2 * h / g ) ^ .5 is just a number, easily calculated for any given h and g.

We also add on the time to fall to the floor after the drop, obtaining total time sqrt(2 h / g) + 2 * (2 h / g)^.5 (1 / (1 sqrt(3)/2) ). Rationalizing the last fraction and factoring out sqrt(2 h / g) we have

sqrt(2 h / g) * ( 1 + 2 * (4 + 2 sqrt(3) ) = sqrt(2 h / g ) ( 9 + 4 sqrt( 3) ).**

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20:33:39

how do you verify that the ball stops bouncing after 1/4 `sqrt(10) + 1/2 `sqrt(10) `sqrt(3/2) (1 / (1-`sqrt(3/4)) sec?

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RESPONSE -->

I demonstrated in my last answer.

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20:33:48

What geometric series gives the time and how does this geometric series yield the above result?

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20:35:02

How far does the ball travel on the nth bounce?

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RESPONSE -->

10 * (3/4)^n up and 10 * (3/4)^n down so

2 * 10 * (3/4)^n

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20:36:15

How long does it takes a ball to complete the nth bounce?

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RESPONSE -->

2 * 1/4 * `sqrt(10 * (3/4)^n) for up and down.

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20:52:07

Query 9.2.21 (was p 481 #6) convergence of 1 + 1/5 + 1/9 + 1/13 + ...

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RESPONSE -->

This series appears to be of the formula 1/(4x - 3)

meaning that it is similar to 1/x which is the p-series

The integral from 1 to infinity of 1/(4x - 3)

let u = 4x -3

du/dx = 4

1/4 du = dx

The integral from 1 to b of 1/4 * 1/u du

Limit as b->infinity of 1 to b of 1/4 * ln |u|

1/4 * ln |4b - 3|

As b gets larger so does ln |4b -3|. It diverges.

Comparing to 1/x^p, this is exactly the same result for p >= 1

*&*& In this case you can actually integrate the expression. In many you cannot and have to rely on comparisons. The solution here is equivalent to yours, but does not require integration of 1 / (4x - 3).

The denominators are 1, 5, 9, 13, ... . These numbers increase by 4 each time, which means that we must include 4 n in the expression for the general term. For n = 1, 2, 3, ..., we would have 4 n = 4, 8, 12, ... . To get 1, 5, 9, ... we just subtract 3 from these numbers, so the general term is 4 n - 3. For n = 1, 2, 3, ... this gives us 1, 5, 9, ... .

So the sum is sum( 1 / ( 4n - 3), n, 1, infinity).

Knowing that the integral of 1 / (4 x) diverges we set up the rectangles so they all lie above the y = 1 / (4x) curve. Since 1 / (4n - 3) > 1 / (4 n) we see that positioning the n = 1 rectangle between x = 1 and x = 2, then the n = 2 rectangle bewteen x = 2 and x = 3, etc., gives us a series of rectangles that lie above the y = 1 / (4x) curve for x > 1.

Since an antiderivative of 1 / (4x) is 1/4 ln | x |, which as x -> infinity approaches infinity, the region beneath the curve has divergent area. Since the rectangles lie above the curve their area also diverges. Since the area of the rectangles represents the sum of the series, the series diverges. *&*&

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20:52:12

with what integral need you compare the sequence and did it converged or diverge?

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21:08:58

Explain in terms of a graph how you set up rectangles to represent the series, and how you oriented these rectangles with respect to the graph of your function in order to prove your result.

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RESPONSE -->

Using a left-hand sum with width 1 for each rectangle shows that the series is larger than the integral.

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Very good, but be sure to see my notes. You had the wrong ratio for your geometric series in the first problem (it's the square root of 3/4, not 3/4). And the more general approach on that last problem would use a comparison.