Assignment 12 Query

course Mth 174

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Physics II

11-05-2007

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21:25:20

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 -->

The first term in this case is y^2.

The ratio is y.

I found this by considering the series to be like the general equation:

a + ar + ar^2

where a is the first term and r is the ratio.

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21:26:34

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 because it has a first term (a = y^2) and a common ratio (r = y).

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21:28:53

how do you get the common ratio?

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RESPONSE -->

To find the common ratio I looked at each term listed and examined how each one differed. The first term was y^2, and according to the general pattern of a geometric series, the first term is equal to a. Each consecutive term changed by a factor of y, so the ratio must be y.

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21:31:02

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

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RESPONSE -->

Ah, that would make a bit more logical sense. When y^2 is factored out, the sequence becomes:

1 + y + y^2 + y^3 + ...

This makes it pretty clear that it is a geometric series and that the first term is equal to what was just factored (y^2) and that the ratio is y.

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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22:56:34

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) A ball dropped from a height h will take t=(1/4) 'sqrt(h).

To prove this, I noticed that this was in reference to time. The previous questions have been in relation to distance, so I needed something that would relate time and distance. I remembered the acceleration, velocity, and distance functions. I used the distance function along with g = -32ft/sec^2 to get:

distance = - (32 / 2)t^2 + v0 + h

where h is the initial position and v0 is the initial velocity. Since initial velocity is 0 at h and the distance to the ground when it hits the ground is 0, this leaves us with the equation:

0 = - (32 / 2)t^2 + h

Rearranging to solve for h in terms of t:

h = 16t^2

h / 16 = t^2

t = 'sqrt( h / 16 )

t = 'sqrt( h ) / 4

t = (1/4) 'sqrt( h )

So the time until the ball hits the floor can be found in terms of height alone.

b) I can't figure this part out. Since the series is a finite series, I think I should somehow use the formula:

a(1 - x^n) / (1 - x)

But I can't get anything out of it. Dividing (1/4) 'sqrt( h ) out of the equation gives a ratio of:

r = 2 * 'sqrt(3/4) * ( 1 / ( 1 - 'sqrt( 3/4) ) )

But this doesn't match anything that I can come up with.

You're on the right track. Here are the details of my solution:

** 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

Total distance = (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

Total time = 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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23:11:10

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 am not quite sure. In 28 I came up with the closed form of the total vertical distance traveled:

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

This looks related, but I am not sure how.

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23:16:22

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

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RESPONSE -->

Hmm. In the series, the first term would be

(1/4)'sqrt(10)

And each successive term would be something like:

(1/4)'sqrt( 10 * (3/4)^n )

But I am not sure where to go from there.

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23:17:04

How far does the ball travel on the nth bounce?

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RESPONSE -->

On the nth bounce, the distance traveled would be:

10(3/4)^n

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23:18:03

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

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RESPONSE -->

Not quite sure. The only thing I can think of is:

(1/4) 'sqrt( 10 (3/4)^n )

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23:31:48

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

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RESPONSE -->

Hmm. I am not sure which one this is. In my book, 9.2.21 is:

sum ( (1/3)^n, n=4 to 20 )

and I am supposed to find the sum.

As for this problem... I don't see a recurring sequence, a number which will divide 1/5, 1/9, and 1/13. In other words, I don't think this is a geometric series.

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23:53:15

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

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RESPONSE -->

Hmm. I guess, then, the idea is something like:

1 / ( 2n + 1 )

The integral of this could be compared with the integral of:

1 / 2n

The antiderivative of this function is ln(2n). Since this antiderivative grows without bound (diverges) and the antiderivative of 1 / ( 2n+1 ) is greater than it, the original function diverges as well.

That is the idea, but the denominators change by 4 with every increase in n so you would be looking at 1 / (4 n) rather than 1 / (2n). However the argument is identical:

*&*& 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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23:59:47

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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RESPONSE -->

Hmm. Again, not quite sure. Thinking of it in terms of a graph would give a descending slope because the first rectangle would be equal to 1, the next equal to 1/5, the next 1/9, and so on. Looking at it like that would seem to indicate that the series would converge on 1, but then the integral is still increasing by however small an amount as the series continues.

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