Assignment 10

course Mth 174

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ì¤ÃÙqŽ˜°£Ô–„wïýºHïò¢áXŒassignment #010

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Physics II

02-25-2007

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10:15:06

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

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10:59:27

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 -->

$1000: 8.11 years

$2000: 4.46 years

$1000: P(t) = 1000, r = .05

F.V. Formula: Integral from 0 to M: P(t)e^[r(M-t]dt

= Integral from 0 to M: 1000e^[0.05(M-t)]dt

= 1000* Integral from 0 to M: e^[0.05(M-t)]dt

= (1000/-0.05)*e^[0.05(M-t)], evaluated from 0 to M

Set equal to $10,000 and M=t, and simplifying,

e^(.05M) = 1.5

.05M = ln 1.5

M = 8.11 years (approx.)

$2000: Following same process, except P(t) = 2000

Integral produces:

10,000 = (2000)/-0.05)*e^[0.05(M-t)], evaluated from 0 to M

e^(.05M) = 1.25

.05M = ln 1.25

M = 4.46 years (approx.)

The initial principle is the amount you start with in the account; you've treated it here as if it was the cash flow (and if that was the case your solution would be correct).

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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11:01:08

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

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RESPONSE -->

First in previous response.

Second: Integral (from 0 to M): 2000*e^[0.05*(M-t)]dt

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11:01:24

What did you get when you integrated?

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RESPONSE -->

See previous response

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11:15:21

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 -->

The amount deposited in interval dt is the amount deposited per year multiplied by dt [$/year (P(t) * years or fractions of years (dt) = [P(t)dt]$. This sum of money [P(t)dt]$ is then invested for the time T. The amount after T years is then P(t)dt*e^(rT) $, where r is the interest rate and e^(rt) is the multiplier (assuming continuous compounding).

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11:23:30

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 interpreted T to be the number years that this particular amount was invested, not the total time from year 0. Using T as the years from 0, and investing $1000 per year, then the value of the amount invested during dt near t is:

F.V. = 1000dt*e^[r*(T-t)]

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11:30:45

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

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The future value of each packet of money invested is approximated by the area of a rectangle under the function P(t) (from 0 to T) at x=t with width dt and height P(t). Summing the areas from each interval dt results in the Riemann sum for the entire time from 0 to T:

F.V. = Sum P(t)dt*e^[r*(T-t)]

For P(t) = 1000, the sum becomes:

F.V. = Sum $1000*dt*e^[r*(T-t)]

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11:33:45

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

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As dt approaches zero, the integral from 0 to T approximates the Riemann sum:

F.V. = Integral (from 0 to T) $1000*dt*e^[r*(T-t)],

where T = M and r = 0.05

Thus,

F.V. = Integral from 0 to M: $1000*e^[0.05*M-t)] dt

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11:35:38

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

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11:51:00

what is c in terms of k?

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I first found C(t), the cumulative death distribution function:

C(t) = Integral cte^(-kt) dt

C(t) - c* Integral te^(-kt) dt

Using formula #14 from the table with p(x)=t and a=-k

C(t) = c*[(-te^(-kt))/k - (e^(-kt))/(k^2)] + constant

Since C(0) = 0 (no one has died at time = 0)

C(0) = 0 = c*[(-(0)e^(-k(0)))/k - (e^(-k(0)))/(k^2)] + constant

0 = c*[0 - e^0/(k^2) + constant

Solving for constant:

Constant = c/(k^2), thus

C(t) = c*[(-te^(-kt))/k - (e^(-kt))/(k^2)] + c/(k^2)

Since C(t) approaches 1 as t approaches infinity,

lim (as t>>infinity) C(t) = 1 = c*(0 + 0) + c/(k^2)

(the two zeroes come from the two terms inside the brackets which, as t approaches infinity, approach zero because the denominators--moving the e^(-kt) to the denominator--grow much faster than the numerators--for the second term the numerator is a constant 1.)

Thus, 1 = c/(k^2)

c = k^2

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11:58:24

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

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RESPONSE -->

Using the function C(t) derived previously:

C(5) = 0.4 = c*[(-5e^(-5k))/k - (e^(-5k))/(k^2)] + c/(k^2)

Substituting k^2 for c

0.4 = (k^2)*[(-5e^(-5k))/k - (e^(-5k))/(k^2)] + (k^2)/(k^2)

0.4 = -5ke^(-5k) - e^(-5k) + 1

Sovling for k

k = -0.14 (approx.)

c = k^2 = 0.0198 (approx.)

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12:00:43

What is the cumulative death distribution function?

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RESPONSE -->

C(t) was found in previous reponse.

Substitutin sqrt k for c gives C(t) in terms of k:

C(t) = sqrt k * [(-te^(-kt))/k - (e^-(-kt))/(k^2)] + 1

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12:01:04

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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See previous response

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12:05:27

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

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RESPONSE -->

Integral given previously.

This integral is the integral of the death density curve because the area under the death density curve between any points is the total number of deaths betwen those two points. Thus the cumulative death distribution function is the integral of the death density function

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12:12:42

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

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12:14:00

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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I cannot find this problem anywhere in the chapter.

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12:15:07

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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RESPONSE -->

See previous response. This question appears to relate to section 8.8, which is not part of this assignment (and I could not find it in 8.8 anyway).

This problem was eliminated from the most recent edition of the text. That's a shame.

Here is a discussion FYI:

** 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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12:17:24

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

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I was familiar with the P.V. and F.V. formulas for single quantities, but the integrals for streams were new to me.

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