Assignment 10

course Mth 174

I am resending this assignment.

_ꦬz뼸|f唡assignment #010

د`SB

Physics II

06-15-2009

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14:53:06

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

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I wasn't exactly sure how to find the total number of years but this is how I attempted the problem:

Present value = integral from t to 0 is 1000 e^-.05t dt =

1000 (- e^-.05t)/.05

I then plugged this in to the future value equation:

Future value - 10000

10000 = (1000 (- e^-.05t)/.05) * e^-.05t

I'm not sure what to do from here.

** 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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14:53:12

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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14:53:28

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

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Present value = integral from t to 0 is 1000 e^-.05t dt =

1000 (- e^-.05t)/.05

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14:53:46

What did you get when you integrated?

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Present value = integral from t to 0 is 1000 e^-.05t dt =

1000 (- e^-.05t)/.05

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14:54:39

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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I tried using the present value equation to form an integral but I couldn't figure out how to get t by itself to figure out the value.

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14:55:17

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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so far it was

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14:56:23

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

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This creates a Riemann sum because you take the continuous stream and break it into parts of change in t and then add the parts together.

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14:57:26

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

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The Riemann sum gives you te total present value that allows you to put it into an integral, which allows you to find the future value.

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15:44:38

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

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15:46:01

what is c in terms of k?

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f(t) = cte^-kt

c = 1/ (cte^-kt)

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15:51:26

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

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c = 1/ (cte^-kt) so I plugged this back in for c in the antiderivitive.

integral from t + change in t to t of cte^-kt change in t

= c * t^2/2 * (-e^-kt)/ k

= 1/ (cte^-kt) * t^2/2 * (-e^-kt)/ k

I factored out a t from the numerator and an e^-kt and ended up with -t/2k = -5/2k so k = -2/5

Then I plugged k back in the antiderivative:

c * 5^2/2 * (-e^(2/5)(5)) / (-2/5)

c25 = 18.47

c= .7388

** 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.

**

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15:55:27

What is the cumulative death distribution function?

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integral from t + change in t to t of P(x)dx

P(x) is the antiderivitive of p(x) so I think it would be:

integral from t + change in t to to of c * t^2/2 * (-e^-kt)/k

** 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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15:56:12

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'm not sure how to explain this.

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15:57:15

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

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I used integral from t + change in t to t from c * t^2/2 * (-e^-kt)/k, because this is the antiderivitive and P(x) is the antiderivitive of p(x).

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16:02:15

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

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I couldn't find this one in my book but I think it would be the integral from 1 + change in t to t of P(x) dx where P(x) represent the position of the pendulum bob.

** 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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16:02:23

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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16:02:35

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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16:03:24

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

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