Assignment 12

course MTH 174

֟Rᔰ}ʒassignment #012

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Physics II

12-04-2007

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15:05:01

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 -->

a = 1

ratio = y

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15:05:33

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 1. The common ratio is y.

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15:07:56

how do you get the common ratio?

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RESPONSE -->

I observed that each successive term changes by a factor of y.

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15:17:04

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

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RESPONSE -->

1 + y + y^2 + y^3 + ... + y^(n-1)

a = 1

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15:42:40

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 = 10

ratio = 3/4

10 + 10(3/4) + 10(3/4)^2 + 10(3/4)^3 + ... + 10(3/4)^n

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11:30:47

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 -->

To fall 10 ft. it takes

p(t) = -16t^2 + 10

0 = -16t^2 + 10

10 / 16 = t^2

`sqrt(10) / 4 = t

To fall h ft. it takes

p(t) = -16t^2 + h

0 = -16t^2 + h

h / 16 = t^2

`sqrt(h) / 4 = t

The ball dropped from a height of h feet reaches the ground in 1/4 `sqrt(h) seconds.

The ball falls a total distance of:

S_n = 10 + 10(3/4) + 10(3/4)^2 + 10(3/4)^3 + ... + 10(3/4)^n =

10(1 - (3/4)^n) / (1 - (3/4)).

as n goes to infinity, S_n = 40 ft.

All distances are traveled twice, except for the first ten ft. 2(40 - 10) + 10 = 70 ft.

In this case,

2`ds / `sqrt(2(32)`ds) = `dt

the sum (2 * 10) / `sqrt(64 * 10) + 2(2 * 10(3/4)^n / `sqrt(64 * 10(3/4)^n) as n goes to infinity = approximately 11 seconds.

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Very good. A symbolic solution whose solution matches yours:

** 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) ).**

16:00:08

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

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RESPONSE -->

the sum (2 * 10) / `sqrt(64 * 10) + 2(2 * 10(3/4)^n / `sqrt(64 * 10(3/4)^n) as n goes to infinity = approximately 11 seconds.

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16:05:57

How far does the ball travel on the nth bounce?

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RESPONSE -->

10(3/4)^n ft.

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16:07:19

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

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RESPONSE -->

2(2 * 10(3/4)^n / `sqrt(64 * 10(3/4)^n)

counting rise and fall

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18:18:55

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

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RESPONSE -->

Query 9.2.21

is the sum from 4 to 20 of (1 / 3)^n

a = 1 / 3

x = 1 / 3

(1 / 3)(1 - (1 / 3)^n) / (1 - (1 / 3))

the sum of the first 20 terms - the sum of the first 4 terms = sum from 4 to 20.

(1 / 3)(1 - (1 / 3)^(20)) / (1 - (1 / 3)) = .5

(1 / 3)(1 - (1 / 3)^(40)) / (1 - (1 / 3)) = .493827

.5 - .493827 = .006173

convergence of 1 + 1/5 + 1/9 + 1/13 + ...

equation for the nth term: 1 / (4n - 3)

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19:35:47

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

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RESPONSE -->

convergence of 1 + 1/5 + 1/9 + 1/13 + ...

equation for the nth term: 1 / (4n - 3)

as n goes to infinity, the nth term tends to 0

The sequence converges.

*&*& 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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19:43:06

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 -->

I'm not sure if I need to see the whole problem or if I don't quite understand.

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Your work looks good. See my notes. Let me know if you have any questions. &#