Assignment 10 Query

course Mth 174

dDzɶ۽ۇw{]ވassignment #010

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Physics II

10-24-2007

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21:50:28

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

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RESPONSE -->

If I understood this problem correctly, what I need to do here is find a future value of $10,000 with a continuous yearly deposit of $1000 into an account bearing 5% interest. Okay, here goes:

a) I can find future value by taking the present value of a continuous income account and multiplying it by e^(rM):

Future = e^(rM) * int(P(t)e^(-rt)dt, t, 0, M) = 10,000

Since $1000 is the yearly continuous income and r= 0.05, the function is P(t) = 1000 and the integral becomes:

int( 1000e^(-0.05t)dt, t, 0, M)

Now I can take that and try and solve it for M by finding the antiderivative:

int = -20,000e^(-0.05t) from 0 to M

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

Now we can multiply this by e^(0.05M) to find the future value:

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

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

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

30,000e^(-0.05M) = 20,000

e^(-0.05M) = 2/3

-0.05M = ln(2/3)

M = ln(2/3) / -0.05 = 8.1093 years

So it will take just a bit more than 8 years to have $10,000 in the account. This matches with the integral over that period:

int( 1000e^(0.05(8.1093 - t)) dt, t, 0, 8.1093)

b) Since it starts with a $2000 balance, I think all I need to do is add the future balance of the $2000 to the future balance of the continual $1000 deposit, and then go through the same process that I used on the prior answer:

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

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

30,000e^(-0.05M) = 22,000

e^(-0.05M) = 22/30

-0.05M = ln(22/30)

M = ln(22/30) / -0.05 = 6.203099 years.

Whew!

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21:51:17

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 -->

According to my calculations, it will take 8.1093 years for the balance to reach $10,000 with no initial deposit.

With an initial deposit of $2000, it will take 6.2031 years.

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21:52:47

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 problem: e^(rM) * int( 1000e^(-0.05t) dt, t, 0, M)

Second problem: e^(rM) * int( 1000e^(-0.05t) dt, t, 0, M) + 2000e^(rM)

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21:54:35

What did you get when you integrated?

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RESPONSE -->

For the first problem: 9999.996757

Second problem: 10000.00215

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21:56:28

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 -->

I think that the t variable in the Present balance equation would turn into (M - t), or the total time minus the point at which the interval is calculated.

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21:58:49

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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RESPONSE -->

Yes. I used a slightly different flavor of the future value equation, namely:

e^(rM) * Present value

which is equal to:

int( P(t)e^(r (M - t) ) dt, t, 0, M)

I think using the alternate form made it easier to calculate the value of M.

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22:03:22

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

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RESPONSE -->

The growth factor is the rate of growth at a point t, and since it is continuous it uses logarithmic growth. Therefore, the equation is actually:

1000 e^(0.05 (T - t) ) 'dt

Summing this over the entire period creates the Riemann sum:

sum( 1000 e^(0.05 (T - t) ) 'dt )

This is like finding the area of a strange shape where the wierd side is given to us by the formula 1000 e^(0.05 (T - t) ) 'dt.

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22:04:47

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

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RESPONSE -->

To get to the integral here, all we need to do is take the limit of the Riemann sum as t->0.

Excellent work throughout. The given solution doesn't add anything to yours, but I'm including it for your reference:

** 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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22:50:41

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

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RESPONSE -->

Ah, the weird one. (Yes, I did misspell ""weird"" previously.)

a) The first part is to find c in terms of k. I take this to mean I solve for c and leave k on the other side of the equation. However, I am at a loss as to how to figure this out. Since this is a density function, the only value that I know it will have is:

int( f(t) 'dt, t, -infinity, infinity) = 1

But do I need to find the antiderivative of this and evaluate it, or am I missing something simpler?

The domain of the function is t >= 0 so the integral would just go from 0 to infinity. The integral from -infinity to infinity diverges.

Basically you just set the integral equal to 1 and solve for c.

b) I assume that once I find c in terms of k, then I would set the antiderivative of the function equal to 0.4, insert 5 for t, and solve.

c) The cumulative death distribution function would be the antiderivative of the original function with values for c and k inserted.

** 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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22:50:46

what is c in terms of k?

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RESPONSE -->

I am not sure.

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22:50:52

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

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RESPONSE -->

Again, not sure.

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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22:50:57

What is the cumulative death distribution function?

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RESPONSE -->

Not sure.

** 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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23:19:21

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 -->

Hmm. I guess my previous guess was right. Now, to figure it out...

Well, I think I got most of the way there. I used this relationship:

1 = int( cte^(-kt) 'dt, t, -infinity, infinity)

1 = -(c / k) e^(-kt) between -infinity and infinity

1 = - (c / k) e^(-k * infinity) + (c/k)e^(-k * -infinity)

(-k * infinity turns the first function into 0)

1 = 0 + (c/k)e^(-k * -infinity)

1 = (c/k)e^(-k * -infinity)

c = k e^(-k * infinity) (Sign change on infinity)

c = k e^(-k * infinity)

But doesn't this mean that c = 0? Hmm. I think I am still lost.

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23:20:54

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

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RESPONSE -->

The integral of the function:

c t e^(-kt) 'dt

should be the cumulative death distribution function. The death density function is actually a rate and is therefore the derivative of the cumulative death distribution function.

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23:21:25

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

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RESPONSE -->

There is no problem #18 on p. 415.

This problem was unfortunately eliminated with the new edition. It shouldn't have been included in the Query.

However the solution is interesting:

** 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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23:21:36

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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RESPONSE -->

I don't think I have enough info to work on this.

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23:21:44

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 -->

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23:22:56

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

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RESPONSE -->

Again, interesting stuff, but I preferred the previous section dealing with volumes and such.

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There are actually some close connections between physics and probability, both making extensive use of moment integrals of the form int( x^n p(x) , with respect to x).

In any case, excellent work and see my notes.