Assignment 10

course MTH 174

4:15 p.m. As you will see the questions in this query proved elusive to me and I pretty much bombed this Query. I just hope that it doesn't impact my grade too negatively and that I will be able to work any problems like this that might appear on the test. You can see from the time stamps and how late this assignment is that I really put the effort into these problems but just couldn't figure them out. Most of the other assigned problems were no trouble it was just the ones that found their way into this query.

|̂~ʭzw~assignment #010

010. `query 10

Physics II

06-30-2010

yg糑̡w芭b

assignment #010

|̂~ʭzw~

Physics II

06-30-2010

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12:30:10

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12:30:10

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12:30:14

Query problem 8.6.8 (8.4.8 in 3d edition) $1000/yr continuous deposit at 5%

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12:30:16

how long does it take the balance to reach $10000, and how long would take if the account initially had $2000?

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12:30:19

What integral did you use to solve the first problem, and what integral did use to solve the second?

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12:30:22

What did you get when you integrated?

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12:30:23

06-30-2010 12:30:23

What did you get when you integrated?

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NOTES ------->

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12:30:59

describe your density function in detail -- give its domain, the x coordinates of its maxima and minima, increasing and decreasing behavior and concavity.

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12:30:59

describe your density function in detail -- give its domain, the x coordinates of its maxima and minima, increasing and decreasing behavior and concavity.

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12:30:59

describe your density function in detail -- give its domain, the x coordinates of its maxima and minima, increasing and decreasing behavior and concavity.

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12:31:00

describe your density function in detail -- give its domain, the x coordinates of its maxima and minima, increasing and decreasing behavior and concavity.

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12:31:00

describe your density function in detail -- give its domain, the x coordinates of its maxima and minima, increasing and decreasing behavior and concavity.

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saږꍛBvXǐY~

assignment #010

|̂~ʭzw~

Physics II

06-30-2010

z|nЕŅxN

assignment #010

|̂~ʭzw~

Physics II

06-30-2010

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15:36:18

Query problem 8.6.8 (8.4.8 in 3d edition) $1000/yr continuous deposit at 5%

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I have worked for aq very long time on this problem and it continues to elude me. I try to use the Future value integral but keep getting nonsense answers. I don't know what I'm doing wrong. From trial and error I get a value of about 8.11 years. And for the $2000 starting value I'm not very sure on what to do. I read the notes and the section several times but neither go into very much detail in the sections in this assignment.

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15:36:26

how long does it take the balance to reach $10000, and how long would take if the account initially had $2000?

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Read my last entry

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15:39:46

What integral did you use to solve the first problem, and what integral did use to solve the second?

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I tried using the future value integral but couldn't get anywhere with it except by trial and error. I just can't get the M-t part to come out and I end up with e^(.05M)*e^(-.05M) which is one and the whole thing breaks down right there. I tried doing it with M=T and t but that gives a crazy answer like T=277t. I haven't had this much trouble with anything we've done so far.

For a full explanation of this type of integral see http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa10. You can probably understand it by just reading, but if you want to work it out and submit it you're welcome.

To develope the future-value integral, you partition the entire time interval into subintervals. The typical subinterval will have a duration `dt. Let t_i be any clock time within the subinterval. All clock times and time intervals are in years.

We will let T be the duration of the entire interval, and let t = 0 at the beginning of the interval. The rate is symbolized by r, and the final value of the money that flows in during the interval by `dy.

The amount of money, in dollars, which flows into the stream during this interval is 1000 * `dt.

The money that flows in at instant t_i has time T - t_i to grow. Assuming `dt to be small, all points within the subinterval have very nearly the same time to grow. SO the money flowing into the subinterval will grow to about

`dy = 1000 * `dt * e^(r(T - t_i)).

Now if we sum up the `dy contributions of all subinterval, we get a Riemann sum

sum(`dy) = sum ( 1000 * e^(r(T - T_i) `dt).

This sum approaches the integral

int(1000 * e^r(T - t) dt, t from 0 to T).

The integral is easily evaluated. We get

1000 / r (e^(r T) - 1).

If r = .05, we the result is

20 000 (e^(.05 T) - 1)

Setting this equal to 10 000 we get

20 000 (e^(.05 T) - 1) = 10 000 so that

e^(.05 T) - 1 = 1/2 and

e^(.05 t) = 1.5. Taking the log of both sides we get

.05 t = ln(1.5) so that

t = 20 ln(1.5) = 8.109.

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15:41:59

What did you get when you integrated?

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Like I said I used trial and error and got about 8.11 for the first integral and couldn't figure out the second one. I just don't see a formula in this section to work for this. I know that the initial 2000 will figure into all subsequent interest accrua but don't see how to combine this with the future value integral.

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15:44:23

Explain how you would obtain the expression for the amount after T years that results from the money deposited during the time interval `dt near clock time t.

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This too eludes me. I'm having a hard time applying the equations from this section for some reason.

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15:46:42

The amount deposited in the time interval `dt of the previous question is $1000 * `dt and it grows for T - t years. Use your answer consistent with this information?

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This still doesn't help that much. I tried working with T-t and like I said I didn't end up with anything remotely correct.

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15:47:12

Explain how the previous expression is built into a Riemann sum.

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I'm still drawing a blank on this.

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15:50:42

Explain how the Riemann sum give you the integral you used in solving this problem.

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It seems the Riemann sum should be something like Sum ( 1000e^(-r*t_1) 'dt. Which leads to the integral in the book: Int( 1000e^(-rt) dt, 0 M). Which I fumbled around with for a long time and have just run out of time to work on anymmore.

You're pretty close. I think you'll understand the information in the solution I appended.

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15:55:46

query 8.7.20 (8.6.20 ed editin) death density function f(t) = c t e^-(kt)

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Most of this section wasn't bad, especially when working with the graphs and the other questions, except this one. I just don't know exaclty where to begin. I tried using the integral of this and setting equal to 1 but that very quickly became quite complicated. And I could never quite get the t out of the expression to just have c and k.

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15:56:03

what is c in terms of k?

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See previous response.

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15:56:25

If 40% die within 5 years what are c and k?

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Without having found c in terms of k I couldn't work this out.

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15:57:06

What is the cumulative death distribution function?

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F(x)= Int(cte^(-kt), 0, 5)

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16:01:03

If you have not already done so, explain why the fact that the total area under a probability distribution curve is 1 allows you to determine c in terms of k.

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I thought this might help my understading but when I tried to evaluate the integral for the area and set it equal to 1 I got a very complicated answer and still couldn't see how to get the t out of the equation.

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16:02:00

What integral did you use to obtain the cumulative death distribution function and why?

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Without getting c and k values here I couldn't find the cumulative distribution function.

** The integral of any probability density function should be 1, which is equivalent for the present problem to saying that every sick individual will eventually die. Thus the integral of c t e^(-kt), from t=0 to infinity, is 1. This is the relationship you solve for c.

An antiderivative of c t e^(-kt) is F(t) = - c (t e^(-kt) / k + e^(-kt) / k^2) = -c e^(-kt) ( k t + 1) / k^2.

lim{t -> infinity)(F(t)) = 0.

F(0) = c(- 1/k^2) = -c/k^2.

So the integral from 0 to infinity is c / k^2.

This integral must be 1.

So c / k^2 = 1 and

c = k^2 . **

** See also the previous note, where we see that in order for this function to be a probability distribution, the constant c must take the value k^2.

We now have the information that 40% die within 5 years, so that the integral of f(t) from 0 to 5 is .4. This integral is in terms of c and k and yields an equation relating c and k. Combining this information with our previously found relationship between c and k you can find both c and k.

We have for the proportion dying in the first 5 years:

integral ( k^2 t e^-(kt) dt, t from 0 to 5) = .4.

Using the antiderivative

F(t) = -c e^(-kt) ( k t + 1) / k^2 = - k^2 e^(-kt) ( k t + 1) / k^2

= -e^(-kt) ( kt + 1)

we get

F(5) - F(0) = .4

1 - e^(-5 k) ( 5 k + 1) = .4

e^(-5 k) ( 5 k + 1) = .6.

This equation presents a problem because it can't be solved exactly.

If we graph the left-hand side as a function of k we see that there are positive and negative solutions. We are interested only in positive solutions because otherwise the limit of the original antiderivative at infinity won't be 0 and the integral will be divergent.

Solving approximately using Derive, with trial interval starting at 0, we get k = .4045.

** the cumulative function is just the integral of the density function--the integral from 0 to t of f(x), where x is our 'dummy' integration variable. We have

P(t) = cumulative distribution function = integral ( k^2 x e^-(kx) dx, x from 0 to t).

Using the same antiderivative function as before this integral is

P(t) = F(t) - F(0) = -e^(-kt) ( kt + 1) - (- e^(-k*0) ( k*0 + 1) ) = 1 - e^(-kt) ( kt + 1).

Note that for k = .4045 this function is

P(t) = 1 - e^(-.4045 t) ( -.4045 t + 1).

You can check that for this function, P(5) = .4 (40% die within 5 years) and lim{t -> infinity}(P(t)) = 1.

**

**

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16:05:45

query problem page 415 #18 probability distribution function for the position of a pendulum bob

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This question and some of the earlier ones didn't give me as much trouble as the last two questions did. Here the graph of position at a certain time on the x-axis and percent chance of finding the bob in a specific position on the y-axis you would get a steadily decreasing concave down graph with the highest distribution the closer to the initial position (when the pendulum is straight up and down) since the pendulum passes through here more often than the outer edges with the passage of time.

Actually it passes through the central positions more quickly, and is more likely to be observed further away.

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16:08:19

describe your density function in detail -- give its domain, the x coordinates of its maxima and minima, increasing and decreasing behavior and concavity.

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As stated above the graph would be decreasing more and more as you move along the x-axis concave down. The maxima would be at position 'zero' and the minima at the outer extemity postions. The domain would be maximum position change possible for the specific pendulum.

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16:09:21

Where is the bob most likely to be found and where is at least likely to be found, and are your answers consistent with your description of the density function?

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It is most likely found near the zero postion, straigh up and down, and least likely to be in it's outer reaches, especially as time goes on. This is consistent with my descriptions above.

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16:13:48

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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This wasn't an extremely difficult assignment except for the more rudimentary questions which weren't in this query, maybe my good work with the previous and, hopefully, future assignments will minimize the negative effects of having performed so poorly here. I really worked a long time on these sections, especially the questions here, and I don't know what it is that I'm not getting. This is the hardest part of distance classes beacaus I have to take the test tomorrow and probably won't have time to get any solution explanations from you.

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I wish I could have gotten this to you a day earlier. However you have the right ideas and have done very well on previous assignments, so you might well come out OK on the test.

See my inserted notes.

I've debated with myself about the nature of the Query for the second half of this course. In the past I've chosen to provide the solutions, from this point on, only after having receive student attempts. This generally gives me a better picture of what students really understand about the assignments, since student answers are often influenced by the given solutions. My premise is that anyone who has been successful to this point will profit more from this approach, and from the insights I can offer when I see work that hasn't been shaped by the given solutions. However I'm about 60-40 on this decision, and would be glad to have your input. It would be easy to make the change.