course Mth 174
???????????
Physics II
07-19-2008
......!!!!!!!!...................................
15:17:35
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 + ...
......!!!!!!!!...................................
RESPONSE -->
The first term is y^2
It is geometric
The ratio is n*y
y^2*y = y^3
y^3*y = y^4...
Good.
** 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). **
.................................................
......!!!!!!!!...................................
15:18:15
either explain why the series is not geometric or give its first term and common ratio
......!!!!!!!!...................................
RESPONSE -->
Its first term is y^2
The ratio is a * y
.................................................
......!!!!!!!!...................................
15:20:13
how do you get the common ratio?
......!!!!!!!!...................................
RESPONSE -->
I am not sure on a proven system, maybe i missed it in the text. But I just looked at the pattern and found a ratio.
.................................................
......!!!!!!!!...................................
15:21:44
what do you get when you factor out y^2? How does this help you determine the first term?
......!!!!!!!!...................................
RESPONSE -->
When I factor out y^2 I get y^2(1 + y + y^2 + y^3...)
I just though the first term was the term to the far left of the series (implying there is not a ... preceding the first term).
.................................................
......!!!!!!!!...................................
15:24:36
Query problem 9.2.29 (3d edition 9.1.24) (was 9.4.24) bouncing ball 3/4 ht ratio
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
15:43:33
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?
......!!!!!!!!...................................
RESPONSE -->
** 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) ).**
.................................................
......!!!!!!!!...................................
15:43:35
What geometric series gives the time and how does this geometric series yield the above result?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
15:43:37
How far does the ball travel on the nth bounce?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
15:43:53
How long does it takes a ball to complete the nth bounce?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
15:43:56
Query 9.2.21 (was p 481 #6) convergence of 1 + 1/5 + 1/9 + 1/13 + ...
......!!!!!!!!...................................
RESPONSE -->
*&*& 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. *&*&
.................................................
......!!!!!!!!...................................
15:43:58
with what integral need you compare the sequence and did it converged or diverge?
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
15:44:00
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.
......!!!!!!!!...................................
RESPONSE -->
.................................................
&#Let me know if you have questions. &#