Assignment 10

course MTH 174

߾a\|ϚӑFassignment #010

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Physics II

12-02-2007

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16:40:46

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

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the integral from 0 to M of 1000e^(.05*(M - t)) dt = 10,000

M = 8.1093 years

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16:45:57

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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If the account initially had $2000,

the integral from 0 to M of 2000 + 1000e^(.05*(M - t)) dt = 10,000

M = 3.24088 years

** 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 this amount grows for the entire time T, while the income stream grows as before; 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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16:46:02

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

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16:49:18

What did you get when you integrated?

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For the first, I got:

-20,000 + 20,000 e^(0.05M) = 10,000

For the second, I got:

2000M - 20,000 + 20,000 e^(0.05M) = 10,000

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17:14:57

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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M = T

P(t) = dollars / year

(dollars / year) * (some fraction of years) = dollars deposited during that time interval.

P(t)`dt = money deposited at `dt near clock time t

For continuous compounding the amount is multiplied by e^(r * t). The amount after M years is the future value which equals P(t)`dt multiplied by e^(r * t). After M years (final clock time), the number of years that has passed is M - t.

P(t)`dt * e^(r * (M - t))

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17:16:35

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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Use your answer consistent with this information?

Use = Was ?

Right. That was originally dictated using dictation software and the error slipped by.

Yes, my answer was consistent with that information.

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17:20:37

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

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the sum of all the intervals as `dt approaches zero of P(t)`dt * e^(r * (M - t)) = the future value after M years.

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17:42:02

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

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The limit of the sum of P(t)`dt * e^(r * (M - t)) as `dt approaches 0 =

the integral from 0 to M of P(t) * e^(r * (M - t)) dt

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01:08:58

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

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(a) the integral from 0 to infinity of cte^(-kt) dt = 1

antiderivative of cte^(-kt) with respect to t is cte^(-kt) * t

c = 1 / te^(-kt) * t

Right idea but the derivative of your antiderivative function has two terms (product rule) and can't equal the original 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 . **

(b) .4 = the integral from 0 to 5 of cte^(-kt) dt

c(5)e^(-k(5))(5) = .4

c = .016e^(5k)

k = -(ln (1 / c) - 4.13517) / 5 for c > 0

(c) c * te^(-kt) * t

c = .016e^(5k)

(.016e^(5k)) * te^(-kt) * t

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01:09:13

what is c in terms of k?

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RESPONSE -->

c = 1 / te^(-kt) * t

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01:09:34

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

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.4 = the integral from 0 to 5 of cte^(-kt) dt

c(5)e^(-k(5))(5) = .4

c = .016e^(5k)

k = -(ln (1 / c) - 4.13517) / 5 for c > 0

** 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.

**

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01:09:52

What is the cumulative death distribution function?

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RESPONSE -->

c * te^(-kt) * t

** 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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01:17:02

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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Knowing that the total area is 1, from 0 to infinity, gives us something to set the integral equal to. Whereas, without any other data, we would only have the indentity cte^(-kt) = f(t).

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01:33:06

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

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The antiderivative of the density function is the distribution fuction:

(a) the integral from 0 to infinity of cte^(-kt) dt = 1

antiderivative of cte^(-kt) with respect to t is cte^(-kt) * t

c = 1 / te^(-kt) * t

Why?

The value of the cumulative fuction at any given point is the area up to that point of the density function.

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01:35:14

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

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Does not apply to 4th edition text. Unknown problem.

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01:35:16

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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01:35:18

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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01:35:23

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

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