Assignment 10

course Mth 174

I am looking forward to using your notes to help with my review for Test #2. I should be taking that on Thursday or Friday.

~??????j???yF??assignment #010

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Physics II

07-08-2008

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14:00:43

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

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I used the formula Present Value = [0, M] int( P(t)e^(-rt)

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

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

Divide by 1000 - > 10 = -e^(-.05t) / .05

Multiply by .05 - > .5 = -e^(-.05t)

To solve for t I used ln(e^x) = x -> ln(.5) = ln(-e^(-.05t))

-> ln(.5) = -(-.05t) -> ln(.5) = .05t

Divide by .05 -> t = ln(.5) / .05 = t = -13.86 years

I do realize that time cannot be negative, so my answer is obviously wrong. I just wanted to show my answer so you could critique it.

** 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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14:04:22

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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I used the formula Present Value = [0, M] int( P(t)e^(-rt)

10000 = int(2000*e^(-.05t))

The integral -> 10000 = 2000 * (-e^(-.05t) / .05)

Divide by 2000 - > 5 = -e^(-.05t) / .05

Multiply by .05 - > .25 = -e^(-.05t)

To solve for t I used ln(e^x) = x -> ln(.25) = ln(-e^(-.05t))

-> ln(.25) = -(-.05t) -> ln(.25) = .05t

Divide by .05 -> t = ln(.25) / .05 = t = -27.72 years

Once again, I know this is also wrong, but I am not sure where I messed up.

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14:05:50

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

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[0, M] ( 1000 * (-e^(-.05t) / .05))

[0, M] ( 2000 * (-e^(-.05t) / .05))

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14:06:05

What did you get when you integrated?

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14:14:20

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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-> Amount of money deposited is approx. the rate of deposits * Time

-> Which can be shown with P(t) dollars/year * 'dt years

-> equals P(t) * 'dt dollars

-> From the present, the deposit changes due to interest, in the interval of t to t + 'dt approx. P(t)*'dt*e^(-rt)

-> int(P(t)*e^(-rt))

The amount deposited at clock time t years will have T - t years still to grow.

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14:15:06

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 really doubt my answer is consistent with anything since it seems so wrong.

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14:15:11

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

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14:15:30

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

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It might be a good idea for you to review your work on q_a_10 from Calculus I, which addresses this situation, or even to repeat the exercise. This has the potential to give you a much better perspective on present value, future value, and the general process of setting up integrals.

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14:27:37

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

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14:29:23

what is c in terms of k?

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I am going to need quite a bit of feedback on this one.

f(t) = c*t*e^(-kt)

c in terms of k -> I just solved for c. I am not sure if that's correct however.

-> c = f(t) / (t * e ^ (-kt))

You need to consider the integral of f(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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14:34:46

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

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This also got me confused because I do not understand how we could figure out what c or k could be.

c = f(t) / (t * e ^ (-kt)) = f(5) / (5 * e ^ (-5k))

I realize that f(t) has something to do with 40% of the population having died, but my question is: 40% of what exactly. 40% of 1 million or 1 billion.

I didn't add in my attempt for solving for k because I ended it with the same useless answer that I got when i solved for c.

** See also the 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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15:05:57

What is the cumulative death distribution function?

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Once again, I am not positive in what I am doing, but my cumulative death distribution was the integral of the death density function.

int( c*t*e ^ (-kt))

-> since c is a constant i rewrote it as c * int(t * e ^ (-kt))

-> c * (t^2 / 2) * (-e ^ (-kt) / k) in the interval of t, t + 'dt, this is my cumulative death distribution function

** Right idea. Having evaluated c and k you find the cumulative function by integrating the density function. However you need to use the appropriate limits on your integral.

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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15:13:55

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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The total area under a probability distribution curve is one because it is the sum of all of the areas under the curve which means that it is the sum of each probability fraction. Not sure if I explained that correctly, but it equals one because it is all of the probabilities added together.

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15:14:01

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

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15:15:38

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

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My page 415 did not have any questions on it. It only had Example 1 for 8.7. I looked in the question section but it wasn't there either.

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15:15:42

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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15:15:45

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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15:16:15

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

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I know my answers were not correct, but I feel like with a little more practice I will grasp this. It didn't confuse me as much as other assignments.

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See my notes. You are getting a good bit of this, just falling short on a few details. I agree that you should be able to get it.