Query 26

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course Mth 163

12/18

026. `query 26

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Question: `qGive the rabbit populations for the first 12 months. Explain how each new population is obtained,

and what your method for obtaining the new population has to do with the assumed nature of rabbit reproduction.

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Your solution:

1 pair of new baby bunnies then 1 month later there is a pair of rabbits, so after one month there is 2 pair of rabbits. From there do simple addition

One month 1 + 1 = 2

Two months 2 + 1 =3

Three months 3 + 2 = 5

Four months 5 + 3 = 8

Five months 8 + 5 = 13

Six months 13 + 8 = 21

Seven months 21 + 13 = 34

Eight months 34 + 21 = 55

Nine months 55 + 34 = 89

Ten months 89 + 55 = 144

Eleven months 144 + 89 = 233

Twelve months 233 + 144 = 377

Needless to say, rabbits multiply quickly. haha

confidence rating #$&*: 3

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Given Solution:

`a** Our assumptions are that rabbits require a month after birth to mature, mature rabbbits produce a pair of newborns

every month starting the first month after they reach maturity, and rabbits never die. Every month the number of newborn

pairs is equal to the number of mature pairs in the preceding month, which is equal to the total number of pairs from the

month before that. Since all the rabbits from the preceding month are still present, the total number in the new month will

be equal to the total number from the preceding month plus the number of newborns, which is equal to the total number

from the preceding month plus the month before that.

If we start with 1 pair of newborns then 1 month later we have a pair of mature rabbits, so after another month we have 2

pairs of rabbits. Our first three numbers are therefore 1, 1 and 2.

In the next month we will have the 2 pairs we had in the preceding month plus 1 pair of newborns from the pair we had

the month before that for a total of 3 pairs.

In the next month we will have the 3 pairs we had in the preceding month plus 2 pairs of newborns from the pair we had

the month before that for a total of 5 pairs.

In the next month we will have the 5 pairs we had in the preceding month plus 3 pairs of newborns from the pair we had

the month before that for a total of 8 pairs.

The pattern continues

8 + 5 = 13

13 + 8 = 21

21 + 13 = 34

etc. . **

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Question: `qExplain how the rule a(n) = a(n-1) + a(n-2) is related to the enumeration of the rabbit population.

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Self-critique (if necessary):

A(n) is for number of rabbits total for months a(n-1) is for the previous month, and a(n-2) is for 2 previous months.

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Question: `qSTUDENT ANSWER: a(n) is total number of rabbits for month(n) a(n-1) is total for previous

month a(n-2) is total for 2 months previous which is total mature for 1 month previous

INSTRUCTOR COMMENT: Good. In terms of the solution given for the preceding exercise:

In the nth we will have the a(n-1) pairs we had in the preceding month plus a(n-2) pairs of newborns from the a(n-2) pairs

we had the month before that for a total of a(n) = a(n-1) + a(n-2) pairs.

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Question: `qWhat are your ratios a(n) / a(n-1) for n = 1, 2, 3, 4, ..., 10, and what does your graph of ratio vs. n

look like?

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Your solution:

Ratios are as follows: 1, 2, 1.5, 1.7, 1.6, 1.6, 1.6

They keep repeating after the first 1.6

The graph would be up and down at first and then level out.

confidence rating #$&*: 2

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Given Solution:

`a** The ratios are 1, 2, 1.5, 1.66, 1.6, 1.625, 1.615, 1.619, 1.6176 1.6181 1.61797 etc.

The graph is jagged, up one time, down the next, but jumping less and less each time.

If you were to make a horizontal line through two successive graph points, all subsequent points would be 'squeezed'

between these lines.

COMMON ERROR: The ratios are 1, 2, 1.5, 1.7, 1.6, 1.6, 1.6, ... .

You rounded off before you could see the difference in the last 3 results. If you don't see a difference in the ratios you

need to use more significant figures--carry your calculation out to enough decimal places that the differences show. **

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Self-critique (if necessary):

Needed to extend my decimals to get a more accurate answer.

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Question: `q problem 8 Use the Fibonacci sequence data points for n = 5 and n = 10 to obtain an

exponential model in the form y = a b^x.

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Your solution:

(5, 8) (10, 89)

8 = A b^5

89 = A b^10

11.125 = b^5

b = 1.619

8 = A * (1.619)^5

A = 8 /(1.619)^5 = .711.

y = .719 * 1.619^x

confidence rating #$&*: 3

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Given Solution:

`a** You would use the points (5, 8) and (10,89) for your y = A b^x model. You would get the equations

8 = A b^5

89 = A b^10

Dividing the second equation by the first you would get

11.125 = b^5, so

b = (11.125)^1/5 = 1.619.

Substituting into the first equation we get

8 = A * (1.619)^5

A = 8 /(1.619)^5 = .711.

So the model is y = .719 * 1.619^x. **

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Question: `qWhat are the values predicted by your model for the first 10 members of the Fibonacci sequence?

What is the average error of your approximation?

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Your solution:

x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

= 0.719, 1.163342, 1.882287356, 3.045540942, 4.927685244,

7.972994725, 12.90030547, 20.87269424, 33.77201928, 54.6431272.

Not sure what to do next.

confidence rating #$&*: 1

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Given Solution:

`a** For x = 0, 1, 2, ..., 9 the formula gives us values 0.719, 1.163342, 1.882287356, 3.045540942, 4.927685244,

7.972994725, 12.90030547, 20.87269424, 33.77201928, 54.6431272.

These numbers differ from the Fibonnaci numbers 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 by

-0.281, 0.163342. -0.117712644, 0.045540942, -0.072314756, -0.027005275, -0.099694535, -.127305757,

-0.22798071, and -0.356872798.

These are the errors in the function. The average of these errors is easily found by adding the errors and dividing by 10,

obtaining an average of about -.016. **

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Your solution:

Need to get the average of the fibonnaci numbers and divide by 10 to get the average.

confidence rating #$&*: ok

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Given Solution:

`aIf mature rabbits were to produce two baby pairs per month, what then would be the definition of the Fibonacci-type

sequence that models the situation where we start with 1 pair of baby rabbits, and what would be population be at the

end of each of the first 6 months?{}{}** Remember that the total number of pairs in month n-2 will be the number of

mature pairs in month n - 1, which will produce the newborns in month n.

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Your solution:

1 + 2 * 1 = 3

3 + 2 * 1 = 5

5 + 2 * 3 = 11

11 + 2 * 5 = 21

21 + 2 * 11 = 43

43 + 2 * 21 = 85

85 + 2 * 43 = 171

…….

confidence rating #$&*: 3

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Given Solution:

`aThe model would give 2 baby pairs for each mature pair; in month n the number of mature pairs is a(n-2), so there

would be 2 a(n-2) baby pairs in month n. The a(n-1) pairs from the previous month would also survive, giving a

population of a(n) = a(n-1) + 2 a(n-2).

The numbers would be 1, 1, 3, 5, 11, 21, 43, ...... . e.g, 1 + 2 * 1 = 3; 3 + 2 * 1 = 5; 5 + 2 * 3 = 11; 11 + 2 * 5 = 21;

21 + 2 * 11 = 43 etc.. **

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ok

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Question: `qHow would the Fibonacci model change if rabbits required two months to mature, with each

mature pair still producing 1 pair of baby rabbits per month?ith each mature pair still producing 1 pair of baby rabbits per

month?

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Your solution:

A(n-1)+ (n-3)

A(n-3) would be the previous 3 months.

confidence rating #$&*: 3

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Given Solution:

`a** The number of baby pairs in month n would be the number of mature pairs from the month before, which would be

the total number of pairs from 2 months previous to that, so we would be looking back 3 months to the population a(n-3).

There would be a(n-3) mature pairs in month n-1 which would give a(n-3) baby pairs in month n. The rabbits from

month n-1 would still survive so we would have a(n) = a(n-1) + a(n-3).

Note that to get started with this model we would need the populations for the first three months. **

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Question: `qproblems 12-16

How did you describe how to reason out the dosage D required for a desired level L of a drug if proportion r of the drug

present immediately after a dose has been removed at the time of the next dose.

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Your solution:

(L + D) * ( 1 - r) = L is the equation

confidence rating #$&*: 2

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Given Solution:

`a** If proportion r has been removed to give concentration L, then 1-r of the amount immediately after the dose will

remain.

The amount immediately after the dose is L + D so we have (L + D) * ( 1 - r) = L.

You then solve this equation for L:

Expanding the left-hand side we have L(1-r) + D(1-r) = L, or L - r L + D - r D = L.

Rearranging the equation to isolate L we get

- r L = - ( D - r D), or

L = ( D - r D) / r

= ( 1-r) / r * D. **

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Your solution:

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Given Solution:

`aStarting with a dose of 429 mg at the instant the concentration of a drug reaches 1000 mg, and knowing that 30% of

the drug is removed in 6 hours, what exponential function models the amount of drug vs. time for the 6-hour interval

between doses?

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Your solution:

***DID NOT HAVE TIME TO FINISH THIS QUERY. THANK YOU FOR EVERYTHING THIS SEMESTER!! Heather ****

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Given Solution:

`a** We need an exponential function y = A * b^t (note that when t = 0 y = A) such that when t = 6, y is 30% less than

A, or y = .7 A.

Thus .7 A = A b ^ 6 and b = .7 ^(1/6) = .942.

Our model is thus y = A * .942 ^ t.

Since when t = 0 we have y = 1429 mg, the model is y = 1429 mg * .942^t. **

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Question: `qWhat dosage is required to maintain a minimum level of 800 mg if 70% of the drug is removed

between doses?

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Your solution:

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Given Solution:

`a** The minimum level occurs just before the next dose, and must be 70% of the maximum level which occurs just after

the dose.

If D is the dose and 800 mg is the level just before the does, then just after the dose the level is 800 mg + D.

This level falls to 70%, i.e., to .70 ( 800 mg + D), just before the next dose. This level should be 800 mg.

We therefore want

( D + 800 mg) ( 1 - .7) = 800 mg.

Subtracting 1 - .7 to get .3, and multiplying this through the grouped expression on the left we get

.3 D + 800 mg * .3 = 800 mg so

.3 D = 800 mg - .3 * 800 mg = 560 mg and

D = 560 mg / .3 = 1867 mg (check my arithmetic). **

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Question: `qWhat will be the maintenance dose of a drug, given a dose of 500 mg with 35% of the drug

removed between doses?

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Your solution:

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Given Solution:

`a** The way this question is stated, with the word 'dose' used for two different quantities, is incorrect. With this

miswording you have the choice of lett 500 mg stand for the dose, or for the maintenance level.

If 500 mg is read as the dose, then the maintenance level is

L = ( 1-r) / r * D = (1 - .35) / .35 * D = .65 / .35 * D = 930 mg, approx..

If you read this as a 500 mg maintenence level then we L = 500 mg and r = .35 so we have

L = ( 1-r) / r * D so that

D = r / (1-r) * L = .35 / (1-.35) * 500 = .35 / .65 * 500 = 269.23.

This is the dose required to maintain a 500 mg level.

If you read this as saying that

I believe you might have added the .3 * 800 mg to both sides, which would have given you something a lot like the 3467

mg you got. **

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Question: `qIf a patient starts with no drug in her body and takes a 500 mg dose every six hours, losing 40% of

the after-dose amount during that time, then how much drug will remain six hours after the initial dose?

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Your solution:

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Given Solution:

`a** In the time between doses concentration will fall to .60 of the after-dose concentration.

Just before the second dose the level will be .60 * 500 mg = 300 mg. Just after the second dose the level will be 300 mg

+ 500 mg = 800 mg.

Just before the third dose the level will be .60 * 800 mg = 480 mg. Just after the third dose the level will be 480 mg +

500 mg = 980 mg.

Just before the fourth dose the level will be .60 * 980 mg = 588 mg. Just after the fourth dose the level will be 588 mg +

500 mg = 1088 mg.

Just before the fifth dose the level will be .60 * 1088 mg = 653 mg approx.. Just after the fifth dose the level will be 653

mg + 500 mg = 1153 mg. approx.

Just before the sixth dose the level will be .60 * 1153 mg = 692 mg approx.. **

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Question: `qWhat sort of function do you think would model the just-before-dose drug concentration?

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Your solution:

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Given Solution:

`a** The increases in mg are 180, 108, 65, 39, 24.

108/180 = .6;

65/108=/6;

39/65=.6;

24/39=.6.

The ratio sequence for the increases is constant. This implies an exponential function. **

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Question: `qproblem 19

Determine the minimum concentration vs. dosage cycle of a drug for six dosage cycles, starting with the first dose, if the

dosage is 750 mg and if 55% of the maximum concentration of the drug is removed during each cycle.

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Your solution:

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Given Solution:

`a** After 55% of the first 750 mg dosage is removed there will be about 338 mg left;

after the next 750 mg dose there will be 1088 mg, which after removal of 55% will leave about 490 mg. etc.

This gives us minimum concentrations of about 338 mg, 490 mg, 558 mg, 588 mg, 602 mg, 609 mg, and 611 mg. **

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Question: `qWhat is the function that models the minimum concentration during a cycle vs. the number of the

cycle?

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Your solution:

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Given Solution:

`a** As before you can find that the ratios of the sequence of differences are constant, so the function exponentially

approaches the limiting concentration.

The limiting after-dose concentration L is that for which the amount removed during a cycle is equal to the dosage. At

that level the amount removed will be .55 L, which will equal the 750 mg dose. So we have

.55 L = 750 mg, which gives us

L = 750 mg / .55 =1363.6 mg approx..

The minimum concentration will be 1363.6 mg - 750 mg = 613.6 mg, approx..

Our concentrations of 0, 338, 490, 558, 588, 602, 609, 611 mg lie 276, 124, 56, 25, 11, 5 and 2 mg below the limiting

minimum concentration. We find the exponential function that models these differences.

The differences are modeled by an exponential function y = A b^n, with y = 276 when n = 1 and y = 124 when n = 2.

Solving the equations

276 = A * b^1

124 = A * b^2

for A and b we obtain

A = 613.6 and b = .45.

The function for the differences is therefore

difference = 613.6 * .45^n.

The concentrations lie below the limiting minimum concentration of 613.6 by this amount, so our exponential function is

concentration = 613.6 - 613.6 * .45^n, or if we factor out the 613.6

concentration = 613.6 ( 1 - .45^n),

where the concentration is in mg and n stands for the number of the last dose taken. **

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Question: `q problem 21

If the dosage is 500 mg and the proportion removed during a dosage cycle is .6, then in terms of a(n-1), write an

expression for the drug concentration just after dose number n-1

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Your solution:

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Given Solution:

`a** If we have a(n-1) just after dose n - 1 then just before dose n we have .40 * a(n-1).

So just after dose n we will have a(n) = .40 * a(n-1) + 500 mg.

This recurrence relation together with the initial condition a(0) = 0 defines the situation.

}If a(0) = 0 then we have

a(1) = .40 * a(0) + 500 mg = 500 mg

a(2) = .40 * a(1) + 500 mg = .40 * 500 mg + 500 mg = 700 mg

a(3) = .40 * a(2) + 500 mg = .40 * 700 mg + 500 mg = 780 mg

etc. **

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Question: `qproblem 22

What is the dosage and what proportion of a drug is removed if the recurrence relation for the concentration a(n) just

before the nth dose is a(n) = .3(a(n-1) + 800), a(1) = 0 ?

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Your solution:

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Given Solution:

`aThe .3 indicates that .3 of the drug remains between doses so that .7 or 70% is removed.

The 800 indicates that a does of 800 units is added to the existing concentration with each new does.

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Question: `q problem 23

Write the general recurrence relation for initial concentration 0, dosage D and proportion removed = r.

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Your solution:

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Given Solution:

`a** If a(n) stands for the amount just after dose number n then

(amt just after dose n) = dose + (amt after proportion r is removed from dose n - 1)

we have

a(n) = D + (1 - r) * a(n-1). **

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

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If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

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Question: The recurrence relation for the rabbit population model, which assumes that each pair newborn rabbits require 2 months to reach sexual maturity and afterwards produce a pair of offspring each month, is a(n) = a(n-1) + a(n-2).

What would be the recurrence relation for dadwattedwabbits, which require 2 months to reach maturity and afterwards produce two pairs of offspring every month?

Starting with a single pair of dadwattedwabbits, what would the population be after each of the first ten months?

What are the ratios a(n+1) / a(n) for the first ten months? What do you think the approximate limiting value of the ratio would be?

Based on this ratio, how much longer do you think it will be before the population of dadwattedwabbits reaches a million?

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Your solution:

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Question: Determine the minimum concentration vs. dosage cycle of a drug for six dosage cycles, starting with the first dose, if the dosage is 500 mg and if 65% of the maximum concentration of the drug is removed during each cycle.

What function models the minimum concentration vs. the number of the cycle?

According to this function, what is the limiting value of the minimum concentration?

After how many cycles is the minimum concentration within 4 milligrams of the limiting value?

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Your solution:

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