assignment 10

course Mth 174

?e??w?w?????wassignment #010

?????S??B+???Physics II

04-01-2009

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20:56:32

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

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RESPONSE -->

P = B / (1 + r) ^t

This formula doesn't apply to a continuous income stream.

0 = 10,000/ (1 + .05) ^t

(1 + .05) ^ t = 10,000

1.05 ^t = 10,000

When you start with $2000=

2000= 10,000 / (1 + .05) ^ t

(1.05) ^t * 2000 = 10,000

1.05^t = 5

** 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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20:58:48

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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RESPONSE -->

P = B / (1 + r) ^t

0 = 10,000/ (1 + .05) ^t

(1 + .05) ^ t = 10,000

1.05 ^t = 10,000

When you start with $2000=

2000= 10,000 / (1 + .05) ^ t

(1.05) ^t * 2000 = 10,000

1.05^t = 5

I'm not sure what to do from here.

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21:01:47

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

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Future value = integral of P(t) e^r(M - t) dt dollars

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21:03:45

What did you get when you integrated?

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Interal of 1000 (1) e^((.05) (M-t)dt dollars

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21:04:22

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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RESPONSE -->

you could obtain the expression for the amount after T years using the Future and Present integrals.

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21:05:20

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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I understand what the integrals and expressions mean, but I do not understand how to isolate the time T.

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21:06:25

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

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The expression is built into a Riemann sum because to find how many years it takes, you must break the process of taking .05% of each years added amount and then add together to get the total amount.

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21:07:04

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

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The Riemann sum gives important information to formulate the integral by evaluating (M -t)

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21:23:24

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

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RESPONSE -->

a) f(t) = cte^ (-kt)

c = te^ (-kt)

b) f(t) = c5e^(-(.4) t)

c=5e^(-.4t)

f(t) = cte^(-kt)

f(t) = 5e^(-.4t) * 5e^(-.4t)

c) Integral of cte^(-kt) * change in t =

Integral of 5e^(-.4t) * 5e^(-.4t) * change in t

** 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 . **

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21:23:34

what is c in terms of k?

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RESPONSE -->

a) f(t) = cte^ (-kt)

c = te^ (-kt)

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21:23:45

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

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RESPONSE -->

a) f(t) = cte^ (-kt)

c = te^ (-kt)

b) f(t) = c5e^(-(.4) t)

c=5e^(-.4t)

f(t) = cte^(-kt)

f(t) = 5e^(-.4t) * 5e^(-.4t)

** see also my 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.

**

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21:23:53

What is the cumulative death distribution function?

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RESPONSE -->

c) Integral of cte^(-kt) * change in t =

Integral of 5e^(-.4t) * 5e^(-.4t) * change in t

** the cumulative function is an 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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21:25:48

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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RESPONSE -->

When the total area is 1, you can determine c in terms of k because there are no sudden changes in the nature of the curve.

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21:26:43

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

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RESPONSE -->

Integral of cte^(-kt) * change in t =

Integral of 5e^(-.4t) * 5e^(-.4t) * change in t

I used this integral because it evaluates the fraction of those who die in relation to the rate of death in a period of 5 years.

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21:27:23

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

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Is this for 3rd edition?

Yes it is. However take a look at the solution. This sort of thinking can be helpful on tests.

** Imagine a pendulum swinging freely back and forth.

The pendulum bob is most likely to be found where it is moving most slowly, and least likely where it is moving most quickly.

It's slowest near its ends and quickest near the center.

So the probability density would be low in the middle and high near the ends.

Velocity increases smoothly so the probability would decrease smoothly from x = -a to x = 0, then increase smoothly again from x = 0 to x = a. **

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21:27:28

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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21:27:30

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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21:27:35

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

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