course Mth 174
I was just wonderin about how well am i doin in this class right now because all i need to graduate is a C or better and i am hoping for that out of this class.
assignment #011ʣ֗ngymӮg]~w
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Physics II
10-23-2007
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21:10:25
what is the mean daily catch?
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RESPONSE -->
The mean is calculated by using the equation
sum of all the tons of fish / the total number of times fish were caught
so we will get 28 ton / 5 fish catchings = 5.6 tons of fish is the mean
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21:11:21
What integral(s) did you perform to compute a mean daily catch?
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the integral that i performed here to compute the mean was from 2 to 8 and used N as the total number of catches that were recorded.
** You are asked here to find the mean value of a probability density function.
The linear functions that fit between the two points are y = .04 x and y = .6 - .06 x.
You should check to be sure that the integral of the probability density function is indeed 1, which is the case here.
The mean value of a distribution is the integral of x * p(x). In this case this gives us the integral of x * .04 x from x = 2 to x = 6, and x = x(.6 - .06 x) from x = 6 to x = 8.
int( x*.04 x, x, 2, 6) = .04 / 3 * (6^3-2^3) = .04*208/3=8.32/3 = 2.77 approx.
int(-.06x^2 + .6x, x, 6, 8) = [-.02 x^3 + .3 x^2 ] eval at limits = -.02 * (8^3 - 6^3) + .3 ( 8^2 - 6^2) = -.02 * 296 + .3 * 28= -5.92 + 8.4 = 2.48.
2.77 + 2.48 = 5.25.
The first moment of the probability function p(x) is the integral of x * p(x), which is identical to the integral used here. The mean value of a probability distribution is therefore its first moment. **
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21:11:33
What does this integral have to do with the moment integrals calculated in Section 8.3?
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do not know
You use the formula to obtain the mean that you use to find the first moment. We used the same mean formula to get the first moment in the application to physics section.
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21:13:16
which function might best represent the probability for the time the next customer walks in?
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the best function that might best represent the probability for the time the next customer walks in is
p(t) = 3e^(-3t) for t greater or equal to 0
** Our function must be a probabiity density function, which is the case for most but not all of the functions.
It must also fit the situation.
If we choose the 1/4 function then the probability of the next customer walking in sometime during the next 4 minutes is 100%. That's not the case--it might be 5 or 10 minutes before the next customer shows up. Nothing can guarantee a customer in the next 4 minutes.
The cosine function fluctuates between positive and negative, decreasing and increasing. A probability density function cannot be negative, which eliminates this choice.
This leaves us with the choice between the two exponential functions.
If we integrate e^(-3t) from t = 0 to t = 4 we get -1/3 e^-12 - (-1/3 e^0 ) = 1/3 (1 - e^-12), which is slightly less than 1/3. This integral from 0 to infinity will in fact converge to 1/3, not to 1 as a probability distribution must do.
We have therefore eliminated three of the possibilities.
If we integrate 3 e^(-3t) from 0 to 4 we get 1 - e^-12, which is almost 1. This makes the function a probability density function. Furthermore it is a decreasing function. **
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21:13:48
for each of the given functions, explain why it is either appropriate or inappropriate to the situation?
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dont know how to explain this answer
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21:14:15
Query Add comments on any surprises or insights you experienced as a result of this assignment.
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i didnt know how to answer one question from each of the sets of problems here so any feed back here would be greatly appreciated.
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FpmҐ
assignment #012
ʣ֗ngymӮg]~w
Physics II
10-23-2007
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21:18:54
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 is givne the first term of y^1 and the common ration here is y^n+1 so it would go from y^1 to y^2 just adding one every time.
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21:19:31
how do you get the common ratio?
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i just looked at the series given and it just goes up by 1 every time so i just thought that the n+1 factor best suits this series of numbers.
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21:20:42
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^3, y^4 and so on and this helps determine the first term because you just substract the highest number from the one right below it and so on and you will determine the first term.
** 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). **
STUDENT QUESTION:
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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21:23:32
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 -->
to verify that the ball dropped from height of h feet reaches the ground in 1/4sqrt(h) you would have to use the average velocity that it drops at. which is easy because it would just be that it drops at a rate of h feet per (1/4)sqrt(h).
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21:23:57
What geometric series gives the time and how does this geometric series yield the above result?
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do not totally understand this question at all, so i have no response to it i am very confused here with this.
** 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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21:24:20
How far does the ball travel on the nth bounce?
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i know that it doesnt travel very far because it has already being boucing for a while before this.
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21:24:24
How long does it takes a ball to complete the nth bounce?
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21:26:03
with what integral need you compare the sequence and did it converged or diverge?
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i do not see this problem in the book anywhere so i dont know where it comes from cause it is different problem than 9.2 problem 21 in the book that i have.
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21:26:12
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 -->
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21:26:12
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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See my notes and let me know if you have questions.
Note that you are still operating the program in such a way that it doesn't include all the information in the SEND file. The only explanation I know of is that you might be manually clearing the contents of the Question box. If this is the case, please be sure you don't do so.
I've emailed you regarding your question about your progress.