Assignment 12

course Mth 174

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Physics II

06-22-2009

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07:32:23

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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17:01:25

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

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RESPONSE -->

This series is geometric. I found this by taking the ratios of each term and found that the ratio is always y. The first term is also y.

** The common ratio is

y^3 / y^2 = y^4 / y^3 = y.

If we factor out y^2 we get

y^2 ( 1 + y + y^2 + … ),

which is in the standard form a ( 1 + r + r^2 + …) with a = y^2 and r = y.

For | y | < 1 this series would converge to sum
y^2 * Sum ( 1 + y + y^2 + . . . =
y^2 * 1 / (1 - y). **

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17:01:42

how do you get the common ratio?

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r = y

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17:03:01

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

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When you factor out y^2 you get y/1. This helps you determine the first term because it shows you that each term is multiplied by y.

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17:08:53

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

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17:23:20

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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a) To show that the ball dropped at a height of h feet reaches the ground in 1/4 sqrt(h) I made an integral:

Integral from 1/4 sqrt(h) to 0 or 32t = 16t^2 =

16( 1/4 sqrt (h) )^2 - 16 (0)^2 = 16( 1/4 sqrt (h) )^2

I wasn't sure how to do part b of this problem where you show the ball stops bouncing after 11 seconds using this equation other than solving the equation.

** 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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17:25:09

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

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time = 1/4 sqrt(h) so each term is less than this amount because the height is constantly divided by 3/4.

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17:27:00

How far does the ball travel on the nth bounce?

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Bn =(a(1 - 1.4 sqrt(h))/ ( 1- 1/4 squrt (h))

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17:27:13

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

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11 seconds

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17:35:27

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

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I could not find this problem in the book but this is how I solved to see if it is convergent or divergent:

This is a harmonius series because the ratios between each term is different and there is no p value in the series. I can compare

S(4) - 1 + 1/5 + 1/9 + 1/13 is greater than the integral from 4 to 1 of 1/x dx

S(4) = 1.38

The integral of 1/x dx from 4 to 1 = 1/4 - 1 = -3/4

This proves that S(5) is greater than the integral of 1/x from 1 and therefore convergent

Good.

&& 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 between 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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17:35:59

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

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The integral of 1/x dx from 4 to 1 = 1/4 - 1 = -3/4

This converges.

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17:37:53

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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I set up this graph f(x)= 1/x by drawing rectangles that overestimate the integral and taking there values, which were equal to each term. This proved that the integral of 1/x is greater than the actual sequence.

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Assignment 12

** The common ratio is y^3 / y^2 = y^4 / y^3 = y. If we factor out y^2 we get y^2 ( 1 + y + y^2 + … ), which is in the standard form a ( 1 + r + r^2 + …) with a = y^2 and r = y. For | y | < 1 this series would converge to sum y^2 * Sum ( 1 + y + y^2 + . . . = y^2 * 1 / (1 - y). **

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

** The common ratio is

y^3 / y^2 = y^4 / y^3 = y.

If we factor out y^2 we get

y^2 ( 1 + y + y^2 + … ),

which is in the standard form a ( 1 + r + r^2 + …) with a = y^2 and r = y.

For | y | < 1 this series would converge to sum
y^2 * Sum ( 1 + y + y^2 + . . . =
y^2 * 1 / (1 - y). **