Query 4

#$&*

course MTH 173

5/27/2013 at 2:41AM

004. `query 4

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Question: `q query modeling project #2 #5. $200 init investment at 10%. What are the growth rate and growth factor for this function? How long does it take the principle to double? At what time does the principle first reach $300?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Growth rate = .10

Growth factor = 1.10

It takes around 7 years for it to double and it takes around 4 years for it to reach $300.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

** The growth rate is .10 and the growth factor is 1.10 so the amount function is $200 * 1.10^t.

This graph passes through the y axis at (0, 200), increases at an increasing rate and is asymptotic to the negative x axis.

For t=0, 1, 2 you get p(t) values $200, $220 and $242--you can already see that the rate is increasing since the increases are $20 and $22.

Note that you should know from precalculus the characteristics of the graphs of exponential, polynomial an power functions (give that a quick review if you don't--it will definitely pay off in this course).

$400 is double the initial $200. We need to find how long it takes to achieve this.

Using trial and error we find that $200 * 1.10^tDoub = $400 if tDoub is a little more than 7. So doubling takes a little more than 7 years. The actual time, accurate to 5 significant figures, is 7.2725 years. If you are using trial and error it will take several tries to attain this accuracy; 7.3 years is a reasonable approximation for trail and error.

To reach $300 from the original $200, using amount function $200 * 1.10^t, takes a little over 4.25 years (4.2541 years to 5 significant figures); again this can be found by trial and error.

The amount function reaches $600 in a little over 11.5 years (11.527 years), so to get from $300 to $600 takes about 11.5 years - 4.25 years = 7.25 years (actually 7.2725 years to 5 significant figures). **

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Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

OK

*********************************************

Question: `q At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

t = 20

$200 * 1.1^20 = $1340

$1340/2 = $670

t = 12.7

The principle will double between 12.7 and 20 years, or 7.3 years.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

** The t = 20 value is $200 * 1.1^20 = $1340, approx.

Half the t = 20 value is therefore $1340/2 = $670 approx..

By trial and error or, if you know them, other means we find that the value $670 is reached at t = 12.7, approx..

For example one student found that half of the 20 value is 1345.5/2=672.75, which occurs. between t = 12 and t = 13 (at t = 12 you get $627.69 and at t = 13 you get 690.45).˙ At 12.75=674.20 so it would probably be about12.72.˙

This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr.

This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. **

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Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

OK

*********************************************

Question: `q query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40%

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

?

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

2

.............................................

Given Solution:

** We find that for the first interest rate, 10%, we have amount = 1(1.10)^t. For the first 4 years the values at t=1,2,3,4 years are 1.10, 1.21, 1.33 and 1.46. By trial and error we find that it will take 7.27 years for the amount to double.

for the second interest rate, 20%, we have amount = 1(1.20)^t. For the first 4 years the values at t=1,2,3,4 years are 1.20, 1.44, 1.73 and 2.07. By trial and error we find that it will take 3.80 years for the amount to double.

Similar calculations tell us that for interest rate 30% we have $286 after 4 years and require 2.64 years to double, and for interest rate 40% we have $384 after 4 years and require 2.06 years to double.

The final 4-year amount increases by more and more with each 10% increase in interest rate.

The doubling time decreases, but by less and less with each 10% increase in interest rate. **

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Self-critique (if necessary):

I don’t understand. You wanted me to sketch the graph right?

------------------------------------------------

Self-critique Rating:

@&

The problem is as stated in your text. The very abbreviated summary of the problem given here is only for purpose of identification of the problem.

*@

3

*********************************************

Question: `q query #11. equation for doubling time

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(1 + r)^ t = 2

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

** the basic equation says that the amount at clock time t, which is P0 * (1+r)^t, is double the original amount P0. The resulting equation is therefore

P0 * (1+r)^t = 2 P0.

Note that this simplifies to

(1 + r)^ t = 2,

and that this result depends only on the interest rate, not on the initial amount P0. **

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Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

OK

*********************************************

Question: `q Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

$5000 * 1.08 ^ ( 2 + doubling time)

2 * ($5000 * 1.08 ^2) / $5000

1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2

1.08^2 * 1.08^doublingtime / 1.08^2 = 2 * 1.08^2 / 1.08^2

1.08^doublingtime = 2

Roughly 9 years

1.08^(9) = 1.999 or 2

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

**dividing the equation $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] by $5000 we get

1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2].

This can be written as

1.08^2 * 1.08^doublingtime = 2 * 1.08^2.

Dividing both sides by 1.08^2 we obtain

1.08^doublingtime = 2.

We can then use trial and error to find the doubling time that works. We get something like 9 years. **

STUDENT COMMENT

I have growth factor ^ time = 2 instead of g.f.^doublingtime =2, I don't understand how the calculation in the solution above was done.

INSTRUCTOR RESPONSE

If P(t) is the principle at clock time t, then the principle at clock time t = 2 is P(2).

To double, starting at t = 2, the principle would have to become 2 * P(2).

The clock time at which the doubling occurs, starting at t = 2, can be expressed as 2 + doublingTime.

Thus the statement that the principle doubles, starting at t = 2, is interpreted as

• P(2 + doublingTime) = 2 * P(2).

This functional equation will apply for any function, whether exponential or not.

In terms of the exponential function of this problem the equation for the specified conditions becomes

• $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2]

For an exponential function the doubling time is constant, so if P(t) is an exponential function we have

• P(t + doublingTime) = 2 * P(t)

for any starting time t. Your equation corresponds to starting time t = 0. Since the function is exponential your equation gives you the correct answer; had the function not been exponential your approach almost certainly wouldn't have worked.

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Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

OK

*********************************************

Question: `q Desribe how on your graph how you obtained an estimate of the doubling time.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I noticed a pattern that the initial amount would increase over time in an exponential manner. By this evidence, I concluded that from t = 0 to the clock time on the x-axis shows the doubling time.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

**In this case you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis.

The interval from t = 0 to the clock time found on the horizontal axis is the doubling time. **

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Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

OK

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

OK

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

&#Good work. Let me know if you have questions. &#

Query 4

#$&*

course MTH 173

5/27/2013 at 2:41AM

004. `query 4

*********************************************

Question: `q query modeling project #2 #5. $200 init investment at 10%. What are the growth rate and growth factor for this function? How long does it take the principle to double? At what time does the principle first reach $300?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Growth rate = .10

Growth factor = 1.10

It takes around 7 years for it to double and it takes around 4 years for it to reach $300.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

** The growth rate is .10 and the growth factor is 1.10 so the amount function is $200 * 1.10^t.

This graph passes through the y axis at (0, 200), increases at an increasing rate and is asymptotic to the negative x axis.

For t=0, 1, 2 you get p(t) values $200, $220 and $242--you can already see that the rate is increasing since the increases are $20 and $22.

Note that you should know from precalculus the characteristics of the graphs of exponential, polynomial an power functions (give that a quick review if you don't--it will definitely pay off in this course).

$400 is double the initial $200. We need to find how long it takes to achieve this.

Using trial and error we find that $200 * 1.10^tDoub = $400 if tDoub is a little more than 7. So doubling takes a little more than 7 years. The actual time, accurate to 5 significant figures, is 7.2725 years. If you are using trial and error it will take several tries to attain this accuracy; 7.3 years is a reasonable approximation for trail and error.

To reach $300 from the original $200, using amount function $200 * 1.10^t, takes a little over 4.25 years (4.2541 years to 5 significant figures); again this can be found by trial and error.

The amount function reaches $600 in a little over 11.5 years (11.527 years), so to get from $300 to $600 takes about 11.5 years - 4.25 years = 7.25 years (actually 7.2725 years to 5 significant figures). **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

OK

*********************************************

Question: `q At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

t = 20

$200 * 1.1^20 = $1340

$1340/2 = $670

t = 12.7

The principle will double between 12.7 and 20 years, or 7.3 years.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

** The t = 20 value is $200 * 1.1^20 = $1340, approx.

Half the t = 20 value is therefore $1340/2 = $670 approx..

By trial and error or, if you know them, other means we find that the value $670 is reached at t = 12.7, approx..

For example one student found that half of the 20 value is 1345.5/2=672.75, which occurs. between t = 12 and t = 13 (at t = 12 you get $627.69 and at t = 13 you get 690.45).˙ At 12.75=674.20 so it would probably be about12.72.˙

This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr.

This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

OK

*********************************************

Question: `q query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40%

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

?

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

2

.............................................

Given Solution:

** We find that for the first interest rate, 10%, we have amount = 1(1.10)^t. For the first 4 years the values at t=1,2,3,4 years are 1.10, 1.21, 1.33 and 1.46. By trial and error we find that it will take 7.27 years for the amount to double.

for the second interest rate, 20%, we have amount = 1(1.20)^t. For the first 4 years the values at t=1,2,3,4 years are 1.20, 1.44, 1.73 and 2.07. By trial and error we find that it will take 3.80 years for the amount to double.

Similar calculations tell us that for interest rate 30% we have $286 after 4 years and require 2.64 years to double, and for interest rate 40% we have $384 after 4 years and require 2.06 years to double.

The final 4-year amount increases by more and more with each 10% increase in interest rate.

The doubling time decreases, but by less and less with each 10% increase in interest rate. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

I don’t understand. You wanted me to sketch the graph right?

------------------------------------------------

Self-critique Rating:

@&

The problem is as stated in your text. The very abbreviated summary of the problem given here is only for purpose of identification of the problem.

*@

3

*********************************************

Question: `q query #11. equation for doubling time

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

(1 + r)^ t = 2

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

** the basic equation says that the amount at clock time t, which is P0 * (1+r)^t, is double the original amount P0. The resulting equation is therefore

P0 * (1+r)^t = 2 P0.

Note that this simplifies to

(1 + r)^ t = 2,

and that this result depends only on the interest rate, not on the initial amount P0. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

OK

*********************************************

Question: `q Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

$5000 * 1.08 ^ ( 2 + doubling time)

2 * ($5000 * 1.08 ^2) / $5000

1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2

1.08^2 * 1.08^doublingtime / 1.08^2 = 2 * 1.08^2 / 1.08^2

1.08^doublingtime = 2

Roughly 9 years

1.08^(9) = 1.999 or 2

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

**dividing the equation $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] by $5000 we get

1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2].

This can be written as

1.08^2 * 1.08^doublingtime = 2 * 1.08^2.

Dividing both sides by 1.08^2 we obtain

1.08^doublingtime = 2.

We can then use trial and error to find the doubling time that works. We get something like 9 years. **

STUDENT COMMENT

I have growth factor ^ time = 2 instead of g.f.^doublingtime =2, I don't understand how the calculation in the solution above was done.

INSTRUCTOR RESPONSE

If P(t) is the principle at clock time t, then the principle at clock time t = 2 is P(2).

To double, starting at t = 2, the principle would have to become 2 * P(2).

The clock time at which the doubling occurs, starting at t = 2, can be expressed as 2 + doublingTime.

Thus the statement that the principle doubles, starting at t = 2, is interpreted as

• P(2 + doublingTime) = 2 * P(2).

This functional equation will apply for any function, whether exponential or not.

In terms of the exponential function of this problem the equation for the specified conditions becomes

• $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2]

For an exponential function the doubling time is constant, so if P(t) is an exponential function we have

• P(t + doublingTime) = 2 * P(t)

for any starting time t. Your equation corresponds to starting time t = 0. Since the function is exponential your equation gives you the correct answer; had the function not been exponential your approach almost certainly wouldn't have worked.

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

OK

*********************************************

Question: `q Desribe how on your graph how you obtained an estimate of the doubling time.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

I noticed a pattern that the initial amount would increase over time in an exponential manner. By this evidence, I concluded that from t = 0 to the clock time on the x-axis shows the doubling time.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

3

.............................................

Given Solution:

**In this case you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis.

The interval from t = 0 to the clock time found on the horizontal axis is the doubling time. **

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary):

OK

------------------------------------------------

Self-critique Rating:

OK

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

OK

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

&#Good work. Let me know if you have questions. &#