query 12
Question:
`qProblem 9.2.10 (4th edition 9.2.8 3d edition 9.1.6) (formerly 9.4.6) first term and ratio for y^2 + y^3 + y^4 + ...
.either explain why the series is not geometric or give its first term and common ratio
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Your solution:
Confidence Assessment:
Given Solution:
`a** 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). **
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 even 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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Question:
`q problem 9.2.16 (4th edition 9.2.28) (3d edition 9.1.23) (formerly 9.4.24) bouncing ball 3/4 ht ratio
Your solution:
Confidence Assessment:
Given Solution:
`a** 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 the total distance 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 ).
Factoring out sqrt(3) / 2 that expression becomes
sqrt(3)/2 * (1 + (sqrt(3) / 2) + (sqrt(3) / 2)^2 + ... )
so the sum of the series is
sum of series = sqrt(3) / 2 * (1 / ( 1 - ( sqrt(3)/2 ) ) ).
The total time of fall would therefore be
sqrt(2 h / g) + 2 * (2 h / g)^.5 * sqrt(3) / 2 * (1 / (1 - sqrt(3)/2) )
= sqrt(2 h / g) + (6 h / g)^.5 * (1 / (1 - sqrt(3) / 2).
where we have added in the time sqrt( 2 h / g) for the initial drop to the floor.
h and g are given by the problem so sqrt( 2 h / g) and (6 * h / g ) ^ .5 is just a number, easily calculated for any given h and g.
Rationalizing 1 / ( 1 - sqrt(3) / 2) we get
(1 + sqrt(3) / 2) / (1/4) = 4 + 4 sqrt(3) / 2 = 4 + 2 sqrt(3)
so the total time is
total time = sqrt( 2 h / g) + (6 h / g)^.5 * (4 + 2 sqrt(3)).
An alternative expression is obtained by factoring out sqrt( 2 h / g) to get
total time = sqrt( 2 h / g) * ( 1 + 4 sqrt(3) + 6) = sqrt( 2 h / g) * (4 sqrt(3) + 7).
22:45:05
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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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22:48:05
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What geometric series gives the time and how does this geometric series
yield the above result?
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22:49:09
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How far does the ball travel on the nth bounce?
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22:49:32
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How long does it takes a ball to complete the nth bounce?
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Question:
`q Query Work this problem (Omitted from 4th and 5th editions so work from given information) (3d edition 9.2.21) (was p 481 #6) convergence of 1 + 1/5 + 1/9 + 1/13 + ...
With what integral need you compare the sequence and did it converge or diverge?
Your solution:
Confidence Assessment:
Given Solution:
`a **The integral was: int(1/x, x, 1, 4n-3). The integral does not converge. **
`q 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.
`a ** STUDENT ANSWER:
I didn't really set up a graph, but if I had to it would be a series of rectangles which get smaller with each successive one. The area of each rectangle would be equal to the respective terms in the series. So, the area of the first would be 1, the second would be 1/5, the third would be 1/9, and so on. The series of rectangles would go on forever and thus the series would not converge.**
INSTRUCTOR SOLUTION
*&*& 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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