Assignment 10

course Mth 174

Once again told you why this is late for in the e-mail sent to you.

???????????assignment #010??n??????g???~w???Physics II

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

10-17-2007

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00:02:06

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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We first look at the present value of 1000 a year and from one year from now with the interest right the amount would be

1000e^(-0.05(1)) = 951.23

1000e^(-0.05(2)) = 904.84

1000e^(-0.05(3)) = 860.71

1000e^(-0.05(4)) = 818.73

1000e^(-0.05(5)) = 778.80

1000e^(-0.05(6)) = 740.82

1000e^(-0.05(7)) = 704.69

1000e^(-0.05(8)) = 670.32

1000e^(-0.05(9)) = 637.63

1000e^(-0.05(10)) = 606.53

1000e^(-0.05(11)) = 576.95

1000e^(-0.05(12)) = 548.81

1000e^(-0.05(13)) = 522.05

now add them all up and add 1000 to them and come up with 10322.11 so it takes a little under 13 years to come up with the money. Brought if the account had initially 2000 the years would be cut in half.

** In a short time interval `dt at t years after the start the amount deposited will be 1000 * `dt.

Suppose that T stands for the time required to grow to $10,000. Then the $1000 * `dt will be able to grow for T – t years at a 5% continuous rate. It will therefore grow to $1000 * `dt * e^(.05 ( T – t) ).

Adding up all the contributions and taking the limit as `dt -> 0 we get the integral of 1000 e^(.05 ( T – t) ) with respect to t, integrated from t = 0 to t = T.

An antiderivative is -1000 / .05 e^(.05 T – t)); evaluating at the limits and subtracting we get 20,000 ( e^(.05 T) – e^0) = 20,000 (e^(.05 T) – 1)

Setting this equal to 10,000 we get e^(.05 T) – 1 = .5, or e^(.05 T) = 1.5.

Taking ln of both sides gives .05 T = ln(1.5). Thus T = ln(1.5) / .05 = 8.1093.

It takes 8.1093 years for the principle to reach $10,000 with initial principle $1000.

If initial principle is $2000 then the equation becomes

2000 e^(.05 T) + 20,000 ( e^(.05 T) - 1 ) = 10,000, or
22,000 e^(.05 T) = 30,000 so
e^(.05 T) = 30/22,
.05 T = ln(30/22)
T = ln(30/22) / .05 = 6.2. **

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00:02:19

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

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shown in previous answer

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00:02:27

What did you get when you integrated?

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previous screen

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00:02:35

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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previous screen

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00:03:17

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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dont understand the question

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00:03:28

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

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dont know because didnt understand the question

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00:03:31

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

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00:04:24

what is c in terms of k?

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i didnt really understand these problems really well at all..

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 .

01:21:07

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01:21:07

01:21:10

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

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01:21:10

** see previous note. 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 will give you an equation relating c and k. Combining this information with your 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 lilmit 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.

01:25:22

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What is the cumulative death distribution function?

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01:25:22

** 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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00:04:32

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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00:04:33

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

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00:05:12

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

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there is no problem #18 on page 415 in my book

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00:05:14

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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00:05:15

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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00:05:17

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

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00:05:33

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

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i had problems explained in the answers so if you would please help me out a little bit.

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Assignment 10

See my notes and let me know if you have questions.