course Mth 174

Resubmission. Originally submitted on 3/11/07.

w»’ö¹™ Ï§ÂѦõ§~w|ŸyºÆ†©‚óassignment #012

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Physics II

03-11-2007

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10:51:07

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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10:54:42

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. It has the form a + ax + ax^2 + ... + ax^(n-2) + ax^(n-1) + ax^n + ...

The first term a is y^2 and the ratio (constant multiple) is y.

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10:55:42

how do you get the common ratio?

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The common ratio is found by dividing each term by the previous term. For example,

y^3/y^2 = y

y^4/y^3 = y

etc.

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11:01:30

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

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I understood the series to begin with y^2. However, if the series is understood to begin with a constant, then factoring out y^2 results in the first term becoming 1. Now the series is 1 + y + y^2 + ...

Upon rereading the section, I couldn't find anything that says the first term a must be a constant. Must it be?

The sum of a geometric series can be expressed in a number of different ways.

Your text says that a + a x + a x^2 + ... sums to a / (1 - x). If you use this form then the series must start with constant number a.

An alternative statement is evey more restrictive:

1 + x + x^2 + ... = 1 / (1 - x).

Of course this is the same as the preceding form, just dividing through by the non-zero constant a.

You could express this series as

y^2 ( 1 + y + y^2 + ...), in which case you would use a = y^2 and get the result

a / (1 - y) = y^2 / (1 - y).

Alternatively you could just say that since the expression 1 + y + y^2 + ... is 1 / 1 - y, then y^2 ( 1 + y + y^2 + ...) = y^2 / (1 - y).

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11:01:51

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

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11:28:27

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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Using the result of part (a), t = (1/4)sqrt h, the times for each bounce and return to 0 height form a geometric series with

a = sqrt (10*(3/4)) * (1/2) (the 1/2 comes from 1/4 mulitiplied by 2, since the time to bounce up and return to 0 is double the time of the bounce up)

ratio = sqrt (3/4) Since each successive bounce is 3/4 the height of the previous and t = 1/4 sqrt h, the multiplier is sqrt (3/4). The constant 1/4 in the t equation is the same for all terms in the series, so it is part of a.

The sum of the times for the ball to be moving is the time for the first drop ((sqrt 10)/4) plus the sum of the geometric series.

Using the sum of a finite series formula:

Total time = (sqrt 10)/4 + (a(1-x^n))/(1-x)

= (sqrt 10)/4 + {[(1/2) sqrt (10*(3/4))]*[1-(sqrt(3/4))^n]}/[1-sqrt(3/4)]

Taking the limit as n>>infinity

Total time = (sqrt 10)/4 + [(1/2)*(sqrt 10)*(sqrt (3/4))*(1-0)]/[1-sqrt (3/4)]

= (sqrt 10)/4 + [(1/2)*(sqrt 10)*(sqrt (3/4))]/[1-sqrt(3/4)]

= (approx.) 11.8 sec.

I believe your result is correct if timing starts from the instant the ball first strikes the floor. The solution below includes the initial time to fall to the floor:

** 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) ).**

For h = 10 feet and g = 32 ft / sec this gives us about 12.59 sec.

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11:28:47

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

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See previous response.

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11:51:49

How far does the ball travel on the nth bounce?

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In part (a) I used the velocity of the ball = -gt + C = -32t + v(sub 0).

The distance traveled then becomes the integral of the velocity function, or

S = -16t^2 + v(sub 0)*t + S(sub 0)

Since S(sub 0) = h and v(sub 0) = 0

We can solve for t

0 = -16t^2 + h

h = 16*t^2

If we look at the bounce from the top of the bounce to the return to h = 0, then

t(sub n) = [sqrt (10*(3/4)^n)]/4

sustituting in the equation for h above (which assumes velocity initially 0, which it is at the top of the bounce)

h = 16*{[sqrt (10*(3/4)^n)]/4}^2

h = 10*[(3/4)^n] feet (double this if counting both the bounce up and return to h=0)

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11:52:38

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

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See previous response. (Note: Time is for one way, double for both ways).

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11:56:53

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

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12:35:25

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

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The formula for generating these terms is 1/(4n-3), so the integral to compare with is:

Integral of 1/(4x-3) dx, evaluated from 1 to infinity

Using integration by subsitution u = 4x-3, du = 4dx

Integral of (1/u) (du/4) = (1/4) ln|u|, evaluated from 1 to infinity

= (1/4) ln|4x-3|, evaluated from 1 to infinity,

Since as x goes to infinity, this expression grows without bound, the integral diverges and therefore the series diverges

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12:56:47

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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Graphing y = 1/(4x-3), then setting up a left hand Riemann sum, with rectangles beginning at (1,1), (2,1/5), (3,1/9), ... gives partial sums 1 + 1/5 + 1/9 + ... which are always greater than the area under the curve. Thus the partial sums of the series are greater than the integral of the curve. Since the integral gets arbitrarily large as n >> infinity, and the partial sums exceed the integral, the series must diverge.

By the way, I could not find this problem in the current edition of the text.

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Good work, including that last problem.