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course MTH 174

9/11/10

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

010. Income Streams

`qNote that there are 11 questions in this assignment.

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Question: `q001. Suppose that you expect for the next 6 years to receive a steady stream of extra income, at the rate of $20,000 per year. This income is expected to 'flow in' at a constant rate, day by day. Suppose furthermore that you expect that your money will grow at a constant annual rate of 8%.

Assuming you do not use any of the money or interest, the question we want to answer is how much you would therefore expect to have at the end of 6 years. This problem is complicated by the fact that the money that goes in, say, today will earn interest over a longer period than the money you earn tomorrow. In fact, the money you receive in one minute will earn interest for a different length of time than the money you receive in a different minute.

To begin to answer the problem you could start out by saying that while some of the money will earn interest for 6 years, some will go in at the very end of the 6-year period and will therefore not earn any interest at all, so the average time period for earning interest will be 3 years. You could then calculate the interest on the full amount for 3 years, and get a fairly good idea of the final amount. As we will see, you can't get a precise estimate this way, but it gives you a reasonable starting estimate.

What would be the simple interest on the total 6-year amount of money flow for 3 years, and what would be the final amount?

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Your solution:

You earn 20000 a year for 6 years

=120000 at the end of the 6 year period

take the average interest earned over the 6 year period will be 6/2 = 3 years

.08*3= 24%

.24* 120000= 28,800

120000+28800= 148,800 over the 6 year period

confidence rating #$&* 3

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Given Solution:

`aThe total money flow would be $20,000 per year for 6 years, or $120,000. 8% simple interest for 3 years would be 3 * 8% = 24%, and 24% of $120,000 is $28,800, so your first estimate would be $28,800. You therefore expect to end up with something not too far from $148,800.

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Self-critique (if necessary):

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Question: `q002. Money usually doesn't earn simple interest. Interest is almost always compounded. If the interest on $120,000 was compounded annually at 8% for 3 years, what would be the final amount?

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Your solution:

120000(1.08)^3=151,165 dollars at the end of the 6 year period.

confidence rating #$&* 3

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Given Solution:

`aThere is a more efficient way to calculate this, but we'll see that shortly.

8% of $120,000 is $9600, so after 1 year we will have $120,000 + $9600 = $129,600.

Note that this result could have been obtained by multiplying $120,000 by 1.08. We will use this strategy for the next two years.

After another year we will have $129,600 * 1.08 = $139,968. If you wish you can take 8% of $129,600 and add it to $129,600; you will still get $139,968.

After a third year you will have $139,968 * 1.08 = $151,165. Note that this beats the $148,800 you calculated with simple interest. This is because each year the interest is applied to a greater amount than before; previously the interest was just applied to the starting amount.

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Question: `q003. Money can earn interest that is compounded annually, quarterly, weekly, daily, hourly, or whatever. For given interest rate the maximum interest will be obtained if the money is compounded continuously.

When initial principle P0 is compounded continuously at rate r for t interest periods (e.g., at rate 8% = .08 for t years), the amount at the end of the time is P = P0 * e^(r t).

If the $120,000 was compounded continuously at 8% annual interest for 3 years, what would be the amount at the end of that period?

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Your solution:

P=120000(e)^(.08*3)=152550 at the end of the 6 year period

confidence rating #$&* 3

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Given Solution:

`aWe would have P0 = $120,000, r = .08 and t = 3. So the amount would be

P = $120,000 * e^(.08 * 3) = $120,000 * e^(.24) = $152,549.

This beats the annual compounding by over $1,000.

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Self-critique (if necessary):

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Question: `q004. This keeps getting better and better. We end up with more and more money. Now let's see how much money we would really end up with, if the money started compounding continuously as soon as it 'flowed' into the account.

As a first step, we note that 6 years is 72 months. About how much would the money flowing into the account during the 6th month be worth at the end of the 72 months?

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Your solution:

120000/72=1666.67 flowing in per month

after 6 months we would have 1666.67*6=10000

P=1666.67(e)^(.08*5.5)

P=2587.85

confidence rating #$&* 3

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Given Solution:

`aFlowing into the account at a constant rate of $20,000 per year, we see that the monthly flow is $20,000 / 12 = $1,666.67.

From the beginning of the 6th month to the end of the 72nd month is 67 months.

From the end of the 6th month to the end of the 72nd month is 66 months.

The $1666.67 flow into the account between the beginning and the end of the 6th month.

Some of this money has as much as 67 months to grow, and some as little as 66 months. The growth isn't linear, but we won't be far off if we 'split the difference' and assume that the $1666.67 grows for 66.5 months.

66.5 months is 66.5/12 = 5.54 years, approximately. So the $1,666,67 would grow at the continuous rate of 8 percent for 5.54 years. This would result in a principle of

$1,666,67 * e^(.08 * 5.54) = $2596.14.

STUDENT COMMENT:

I don’t really know why we would use the midpoint of the month. Is that exactly when the money “flows in?”

INSTRUCTOR RESPONSE: That's pretty much it, though the word 'exactly' isn't quite accurate.

We want to use a single time to indicate 'when' the money can be regarded as being 'in' the account. Since money 'flows in' from the beginning of the month to the end, the midpoint of the month is a better choice than either the beginning of the end.

STUDENT QUESTION

I am confused on why 5.54 was used for the time to me it seems like if we are starting with 6 years and subtract the first 6

months we should be looking at a time interval of 5.5 years???????????

INSTRUCTOR RESPONSE

The interval runs from t = 5 months to t = 6 months before the end. The midpoint of the interval is at t = 5.5 months. On the average the money that flows in during this interval grows for 66.5 months.

Or as stated in the given solution:

'Some of this money has as much as 67 months to grow, and some as little as 66 months. The growth isn't linear, but we won't

be far off if we 'split the difference' and assume that the $1666.67 grows for 66.5 months.'

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Self-critique (if necessary):

I see where I made the mistake and didn't average between 66 and 67 months. I do see how you came up with the 5.54 months as an average

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Question: `q005. How much will the money flowing into the account during the 66th month be worth at the end of the 72 months?

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Your solution:

If the money comes in at the beginning of the 66th month it still has 6 months to grow

1666.67(.08*.5)=1734.69

confidence rating #$&* 3

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Given Solution:

`aAgain $1,666,67 would flow into the account during this month. The midpoint of this month is 65.5 months after the start of the 72-month period, so the money will have an average of about 6.5 months, or 6.5 / 12 = .54 years, approx., to grow. Thus the money that flows in during the 66th month will grow to $1,666,67 * e^(.08 * .54) = $1737, approx..

STUDENT QUESTION

I don’t understand why you can’t calculate monthly since it is broken down to 1666.67

monthly. Also why didn’t we convert back to yearly after finding 5.54 yearly.????

INSTRUCTOR RESPONSE

$1666.67 'flows in' during the month. We use the monthly rate to calculate this.

Then the money grows for about .54 years. For convenience and understanding we used the number of months to calculate the number of years. Since the interest rate is a yearly rate, we use it with the time in years.

We could have used the monthly interest rate with the number of months. 8% / 12 = .667%, or .00667. Our calculation would be $1666.67 * e^(.000667 * 6.5). Because of roundoff with the .667% this comes out around $1740, a little higher than our original result.

STUDENT QUESTION

I am confused on why 5.54 was used for the time to me it seems like if we are starting with 6 years and subtract the first 6

months we should be looking at a time interval of 5.5 years?

INSTRUCTOR RESPONSE

Or as stated in the given solution:

'Some of this money has as much as 67 months to grow, and some as little as 66 months. The growth isn't linear, but we won't

be far off if we 'split the difference' and assume that the $1666.67 grows for 66.5 months.'

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Self-critique (if necessary):

I used .5 because that would be a full 6months for the money to grow. I see I should have taken the average here again

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Question: `q006. We could do a month-by-month calculation and add up all the results to get a pretty accurate approximation of the amount at the end of the 72 months. We would end up with $152,511, not quite as much as our last estimate for the entire 72 months.

Of course this approximation still isn't completely accurate, because the money that comes in that the beginning of a month earns interest for longer than the money that comes in at the end of the month. We could chop up the 72 months into over 2000 days and calculate the value of the money that comes in each day. But even that wouldn't be completely accurate.

We now develop a model that will be completely accurate. We first imagine a short time span near some point in the 72 months, and calculate the value of the income flow during that time span. We are going to use symbols because our calculation asked to apply to any time span at anytime during the 72 months.

We will use t for the time since beginning of the 6-year period, to the short time span we are considering; in previous examples t was the midpoint of the specified month: t = 5.5 months in the first calculation and 66.5 months in the second.

We will use `dt (remember that this stands for the symbolic expression delta-t) for the duration of the time interval; in the previous examples `dt was 1 month.

For a time interval of length `dt, how much money flows? Assume `dt is in years. If this money flow occurs at t years from the beginning of the 6-year period, then how long as it have to grow?

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Your solution:

20000*dt

confidence rating #$&* 3

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Given Solution:

`aThe amount in 1 year is $20,000 so the amount in `dt years is $20,000 * `dt.

Money that flows in a little before or a little after the instant t years from the beginning of the 6-year period will have (6-t) years, a little more or a little less depending on 'before' or 'after', to grow.

STUDENT QUESTION

(6-t) , does this mean that it has anywhere between the beginning and the end of the 6 year span???

INSTRUCTOR RESPONSE

From the t year instant to the end of the 6-year period, the number of years that intervene, and during which the money in question can grow, is 6 - t.

For example, money that flows into the account during the few days prior to the t-year instant has a little more than 6 - t years to grow; money that flows in during the first few days after this instant has a little less than 6 - t years to grow.

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Question: `q007. How much will the $20,000 `dt amount received at t years from the start grow to in the remaining (6-t) years?

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Your solution:

20,000 dt e^(.08 (6-t))

confidence rating #$&* 3

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Given Solution:

`aAmount P0 will grow to P0 e^( r t) in t years, so in 6-t years amount $20,000 `dt will grow to $20,000 `dt e^(.08 (6-t) ).

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Question: `q008. So for a contribution at t years from the beginning of the 6-year period, at what rate would we say that money is being contributed to the final t = 6 year value? Let y stand for the final value of the money, and `dy for the contribution from the interval `dt.

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Your solution:

20,000 e^(.08 ( 6-t))=dy/dt as dt approaches 0

confidence rating #$&* 3

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Given Solution:

`aThe money that flows in during time interval `dt will grow to

`dy = $20,000 `dt * e^(.08 (6-t) ),

so the rate is about

`dy / `dt = [ $20,000 `dt * e^(.08 (6-t) ) ] / `dt = $20,000 e^(.08 ( 6-t) ).

As `dt shrinks to 0, this expression approaches the exact rate y ' = dy / dt at which money is being contributed to the final value.

STUDENT COMMENT:

I do not really understand the last statement in the solution.

INSTRUCTOR RESPONSE:

This explanation leaves out some technicalities, but the last statement comes down to something very much like the following:

The shorter the time interval `dt, the closer everything in the interval is to the midpoint and the less the rate varies over the interval. For both of these reasons, the approximation improves greatly as `dt shrinks.

If `dt keeps shrinking, it approaches zero and the approximation approaches an exact value.

STUDENT QUESTION

why could we not use 20000 * e ^.08 (6-t) where t = 6 because it’s the last year and the interest wont increase because it goes in last?

INSTRUCTOR RESPONSE

We are only talking about the amount that flows in during some unspecified time interval `dt, presumed to be a short interval. $20 000 is the amount that flows in during a year, not during the interval `dt.

The amount of money that 'flows into' the account during time interval `dt is $20 000 / year * `dt. Assuming `dt is in years, and taking for granted that the amount is in dollars, we just write this as 20 000 * `dt.

STUDENT QUESTION

Why is t not just always 0 if we are looking for a change as t approaches 0??????

INSTRUCTOR RESPONSE

This is a very good question, and a universal point of confusion. It confused Western Civilization for thousands of years (see also 'Zeno's Paradox').

It's `dt that approaches zero.

If all the `dt's are zero then the total amount is zero. We don't care about the obvious (and uniformative) thing that happens when all the `dt's are zero; what we care about is what happens as `dt approaches zero.

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Self-critique (if necessary):

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Question: `q009. As we just saw the rate at which money accumulates is y' = dy / dt = 20,000 e^(.08(6-t)).

How do we calculate the total quantity accumulated given the rate function and the time interval over which it accumulates?

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Your solution:

If you subsitute a time in for t you will get an instantaneous rate for dy/dt this is the derivative of the function

if you take the derivative of this function you will have the antiderivative which will give you a quantity(accumulation of money)

confidence rating #$&* 2

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Given Solution:

`aThe rate function is the derivative of the quantity function. An antiderivative of a rate-of-change function is a change-in-quantity function. We therefore calculate the total change in a quantity from the rate function by finding the change in this antiderivative.

STUDENT COMMENT:

I am not sure what this question is asking. I do know that y’ refers to the derivative, but I know that as the rate of

change. Is it that same as the rate at which the money accumulates? And to find the total accumulation, wouldn’t we just use the rate function, which I guess would be the antiderivative of the derivative we just found?

INSTRUCTOR COMMENT:

Very good. Just a little clarification:

y ' is the rate of change of y with respect to t. Since y is the future value of the money flow, y ' is the rate of change of future value with respect to clock time. That would indeed be what we mean in this context by 'the rate at which the money accumulates'. For the 6-year cycle, the rate at which future value is accumulating is y = dy/dt = $20,000 e^(.08 ( 6-t) ).

y ' is the derivative of y, so y is an antiderivative of y '. So you are right about the antiderivative.

STUDENT QUESTION

why would we need the ant derivative?

INSTRUCTOR RESPONSE

What we know is the derivative function y ' of the accumulating amount function y. We know y ' and we want to find y. That is, we know the derivative and we want to find the function of which this is the derivative.

STUDENT QUESTION

Is there a difference in the antiderivative and the Integral???? I don’t think there is but I am concerened that I have

mixed up the meaning of the integral….

INSTRUCTOR RESPONSE

This isn't something you're expected to know just yet (soon but not right now), but along with many students in this course you have apparently already had a calculus course.

To clarify:

An antiderivative is a function; the term is pretty much equivalent to 'definite integral'.

The definite integral is the change in value of an antiderivative function.

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Self-critique (if necessary):

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Question: `q010. If the money that flows into the account t years along the 6-year cycle adds to the final value of the money at the previously mentioned rate y' = dy / dt = 20,000 e^(.08(6-t)), then what function describes the final value of the money accumulated through time t?

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Your solution:

the antiderivative of the y' function(this would be the accumulation of money over t years)

confidence rating #$&* 2

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Given Solution:

`aThe function must be an antiderivative of the rate function y ' = $20,000 e^(.08 ( 6-t) ).

The y ' function can be expressed as y ' = $20,000 e^(.08 * 6) e^(-.08 t)

= $20,000 * 1.616 e^(-.08 t)

= $32,321 e^(-.08 t).

At this point the qa gets a little ahead of itself, since the course hasn't yet introduced you to the chain rule (coming up very soon in the qa's), much less to antidifferentiation (coming up later in the course). However it will be worth your time to make a note of this solution and come back to it after your introduction to integration, about 2/3 of the way through the course. And if you've had a calculus course you should be in a position to understand this now (just remember to think in terms of the chain rule):

The general antiderivative of e^(-.08 t) is -1/.08 e^(-.08 t) + c = -12.5 e^(-.08 t) + c.

The general antiderivative of y ' = $32,321 e^(-.08 t) is y = $32,321 * -12.5 e^(-.08 t) + c = -404,012.5 e^(-.08 t) + c.

The value of the constant c could be determined if we knew, for example, the amount of money present at t = 0. However as we will see in the next step, the value of c doesn't affect the change in the quantity we are considering in this problem.

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Question: `q011. How much money accumulates during the 6 years?

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Your solution:

-404160 e^(-.08 * 6 ) - (-404,160 e^(-.08 * 0 )) = 154072

confidence rating #$&* 2

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Given Solution:

y represents the money accumulated through t years. We subtract the money accumulated at 0 years from the money accumulated at 6 years to get the amount accumulated during the time interval from t = 0 to t = 6 years.

The result is -404,160 e^(-.08 * 6 ) + c - [ -404,160 e^(-.08 * 0 ) + c ] = $154,072.

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Self-critique (if necessary):

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&#Very good responses. Let me know if you have questions. &#

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