Assignment - 16

course Mth163

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.

016. `query 16

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Question: `qbehavior and source of exponential functions problem 1, perversions of laws of exponents

Why is the following erroneous: a^n * b^m = (ab) ^ (n*m)

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Your solution:

here we see with distribution and exponents that if you have 2^1 * 4^3 it would be the same as (2*4) ^ (1 +3)

Confidence rating: positive

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Given Solution:

`aSTUDENT RESPONSE: my example was (4^2)(5^3) did not equal 20^6

INSTRUCTOR COMMENT

** more generally a^n * b^n = a^(n+m), which usually does not equal (ab) ^ (n * m) **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qWhy is the follow erroneous: a^(-n) = - a^n

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Your solution:

this is wrong if you have say 4 ^-5, to get rid of the neg. exponent you would have to make it a fraction like so: 1 / 4^5

Confidence rating: positive

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Given Solution:

`aSTUDENT RESPONSE: 2^-3 is not equal to -2^3

INSTRUCTOR COMMENT:

** A more general counterexample: a^(-n) = 1 / a^n is positive when a is positive whereas -a^n is negative when a is

positive **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qWhy is the following erroneous: a^n + a^m = a^(n+m)

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Your solution:

If we say 3^4 + 3^6 = 3^(4=6) = 3 ^10, we would get 59,049: but if we check it to the way the original problem was written 3^4 + 3^6 = 81 + 729 = 810 – by far not the same answer, because you are adding the products you must work you problem and exponential functions out thy way it, is written.

Confidence rating: positive

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Given Solution:

`aSTUDENT RESPONSE: (5^3)+(5^4) is not equal to 5^7

INSTRUCTOR COMMENT:

(5^3) * (5^4) = (5^7) since (5*5*5) * (5*5*5*5) = 5*5*5*5*5*5*5 = 5^7.

However (5^3) + (5^7) = 5*5*5 + 5*5*5*5*5 = 5*5*5( 1 + 5*5) = 5^3( 1 + 5^2), not 5^7.**

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Your solution:

???? Why do you write it out as 5*5*5(1+5*5) = 5^3(1+5^2) and how did you get that from the original problem????

If you factor 5*5*5 out of 5*5*5 + 5*5*5*5*5 you get 5*5*5 ( 1 + 5*5).

If you aren't sure of why, multiply that product out:

5*5*5 ( 1 + 5*5) = 5*5*5*1 + 5*5*5 * 5*5 = 5*5*5 + 5*5*5*5*5.

The factorization used here can be generalized to prove that a^3 + a^4 is not equal to a^(3 + 4), and more generally that a^b + a^c is cannot be identified with a^(b + c).

*&$*&$

Confidence rating:

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Given Solution:

`aSTUDENT RESPONSE: Why is the following erroneous: a^0 = 0

4^0 is not equal to 0

INSTRUCTOR COMMENT:

** a^(-n) * a^n = 0 but neither a^(-n) nor a^n need equal zero **

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Self-critique (if necessary):

ok, I didn’t have the problem but I do understand the rule that any factor to the power of 0 is equal to one.

Self-critique rating:

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Question: `q Why is the following erroneous: a^n * a^m = a^(n*m).

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Your solution:

This is wrong because any same factor multiplied together with exponents is the same as that factor to the exponents added together for example:

if we have 5^2 * 5 ^4 = 5^(2*4) = 5^8 = 390,625 this would not be the same as (5*5) * (5*5*5*5) = 25 * 625 = 15,625; which is the same as 5^2 * 5^4 = 5^(2+4) =5^6 = 15625

Confidence rating: positive

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Given Solution:

`aSTUDENT RESPONSE: (4^7)(4^2) is not equal to 4^14

INSTRUCTOR COMMENT:

Right. Generally a^n * a^m = a^(n+m), not a^(n*m).

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qproblem 2. Graph and describe

Give basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = 1200 (2^(.12 t)

)

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Your solution:

using t = 0, 1, 2, 3 to find y and make a graph;

I have coordinates(0,1200),(1,1304),(2,1417),(3,1540)

the two basic points are: (0,1200) and (1,1304)

the basic ratio of these y values is 1304 / 1200 = 1.09 approx

Confidence rating: unsure

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Given Solution:

`aSTUDENT RESPONSE

(0,1200),(1,1304)

negative x-axis

ratio=163/150

INSTRUCTOR COMMENT:

the precise ratio is 2^.12, which is probably pretty close to 163/150

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Self-critique (if necessary):

ok, I understand after some calculation that you reduced 1304/1200 by 8 to get 163/150 which is still 1.09 approx. Thank-you for your help on my question form.

???? Does it matter which form you write this ratio in?

If you're looking for an exact result then the fraction would be the most useful form.

If, as in most applications, you're dealing with approximate numbers in the first place, then the approximation is the more desirable form.

Self-critique rating:

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Question: `qgive basic points and asymptote; y values corresponding to the basic points, ratio of these y values:

y = 400 ( 1.07 ) ^ t

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Your solution:

using t = 0, 1, 2, 3: I get (0,400), (1, 428), (2, 458), (3,490)

Basic point are ( 0,400) and (1,428)

The y ratio is 428/400 = 1.07

Confidence rating:

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Given Solution:

`aSTUDENT RESPONSE

(0,400),(1,428)

Neg. x-axis

1.07 or 107/100 is ratio

INSTRUCTOR COMMENT: that ratio is correct and is of course equal to 1.07

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qgive basic points and asymptote; y values corresponding to the basic points, ratio of these y values:

y = 250 ( 1 - .12 ) ^ t

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Your solution:

using t = 0, 1, 2, 3: I get (0,250), (1,220), (2,194), (3,170)

The basic points being, (0,250) and (1,220)

the ratio of the y values is: 220 / 250 = .88

Confidence rating:

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Given Solution:

`aSTUDENT RESPONSE

The basic points are (0,250),(1,220)

The positive x-axis is the horizontal asymptote

The ratio of y values at the basic points is 220 / 250 = .88.

INSTRUCTOR COMMENT: Note also that the ratio .88 is equal to 1-.12; .88 is the growth factor and .12 is the

growth rate.

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Self-critique (if necessary):

ok note that growth factor is 1 - .12 and the growth rate is .12

Self-critique rating:

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Question: `qgive basic points and asymptote; y values corresponding to the basic points, ratio of these y values:

y = .04 ( .8 ) ^ t

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Your solution:

using t = 0, 1, 2, 3: giving you (0, .04), (1,,032), (2,.0256), (3,.02048)

The basic points being: (0,.04) and (1,.032)

The ratio being: .032 / .04 = .8

Confidence rating:

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Given Solution:

`aSTUDENT RESPONSE

(0,.04),(1,.032) are the basic points and the asymptote is the positive x-axis.

The ratio is .32 / .4 = .8.

The pattern is that the ratio is equal to b, where b is the base for the form y = A b ^ x.

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Self-critique (if necessary):

ok, the graph is a positive asymptote towards the x axis because it gets closer and closer to the x axis on the positive side.

*remember ratio is equal to b, and b is the base form for y = A b^x*

Self-critique rating:

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Question: `q problem 3. y = f(x) = 5 (1.27^x).

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Self-critique (if necessary):

Self-critique rating:

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Question: `qWhat is the ratio between the y values at x = 0 and at x = 1?

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Your solution:

in y = f(x) = 5(1.27^x)…if x = 0 and 1, solve for y

y = 5(1.2^0)…..5, giving you (0,5)

y = 5(1.2^1)….6, giving you (1,6.35)

the ratios between y at x = 0 and 1 is: 6.35 / 5 = 1.27

Confidence rating: positive

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Given Solution:

`a** f(1) / f(0) = 5 * 1.27^1 / ( 5 * 1.27^0) = 1.27 **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qWhat is the ratio between the y values at to x = 3.4 and x = 4.4?

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Your solution:

in y = f(x) = 5(1.27^x)….x = 3.4 and 4.4

The y ratio is [5(1.27^4.4)] / [5(1.27^3.4)]…..1.27

Confidence rating: positive

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Given Solution:

`a** f(4.4) / f(3.4) = 5 * 1.27^4.4 / ( 5 * 1.27 ^3.4) = 1.27. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qVerify that the ratio of y values is again the same for your own points where x differs by 1 unit.

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Your solution:

For my points of x = 2 and 3 with y = 8.0645 and 10.24195:

I have a y ratio 10.24195 / 8.0645 = 1.27

Confidence rating: positive

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Given Solution:

`a** STUDENT RESPONSE: My points were 4.5 and 5.5, and the y ratio was again 1.27 **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qWhat is the ratio of y values when x values are separated by two units?

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Your solution:

To find the ratio of the y values, you need to find the difference between the two y’s. the x values would be: y = f(x) and y1 = f(x1) + 2

The ratio would be y1/ y = (x1 +2) / x

Now plugging that into your ratio factor of 1.27^x you would have:

1.27^(x+2) / 1.27^x….The laws of exponents says you would subtract the exponents:

1.27^(x + 2 – x) = 1.27^2 = 1.61 approx.

Confidence rating: mostly

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Given Solution:

`a** If x values are separated by 2 units then the ratio is 1.27^(x+2) / 1.27^x = 1.27^(x+2 - x) = 1.27^2 = 1.61 approx.

**

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `q problem 4. Ratio of y values at x = x1 and x = x1+1

What does your result tell you about how the ratio depends on the x value x1?

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Your solution:

if y = 1.27 ^x then y1 = 1.27 ^x1: to find the ratio between x = x1 and x = x1+1

you would have to take 1.27^(x1 + 1) / 1.27^x1 ….1.27^(x1 +1- x1)….1.27 ^ 1 = 1.27

This tells us like y = A b^x that the ratio where your answer is b; that your x values has nothing to do with the ratios.

Confidence rating: positive

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Given Solution:

`a** If y = A b^x then the value at x1 is A b^x1 and the value at x1 + 1 is A b ^(x1 + 1). The ratio of these values is

A b^(x1+1) / A b^x1 = b^(x1 + 1 - x1) = b^1 = b.

The ratio should have been b, where b is the base in the form y = A b^x. This is the same as in all previous examples,

which shows that there is no dependence on x1. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qproblem 5. y = 3 (2 ^ (.3 x) ).

What is the ratio of the two basic-point y values?

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Your solution:

x = 0 and 1

giving you the basic points of (0,3) and (1,3.69)

the y ratio would be: 3(2^(.3*1)) / 3(2^(.3*0))……1.23 approx.

Confidence rating: positive

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Given Solution:

`a** The basic points are the x = 0 and x = 1 points. The corresponding y values are 3(2^.(3*0) ) = 3 and 3(2^(.3 *1) )

= 3 * 2^.3 = 3.69 approx.

The ratio of these values is 3.69 / 3 = 1.23. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qWhat is the y = A b^x form of this function?

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Your solution:

If I take 3(2^(.3x) and put it in the form of y = A b^x:

I would get y = 3(1.23^x)

A = 3 and b = 1.23

Confidence rating:

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Given Solution:

`a** 3 (2 ^ (.3 x) ) = 3 (2^.3)x = 3 * 1.23^x, approx.

This is in the form y = A b^x for A = 3 and b = 1.23. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qWhat does the value of 2 ^ .3 have to do with this situation?

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Your solution:

in the form y = A b^x….. 2^.3 is the value of b

Confidence rating: positive

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Given Solution:

`a** CORRECT STUDENT RESPONSE: this is the b value in the form y = A b^x. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qproblem 6 P(n+1) = (1+r) P(n), with r = .1 and P(0) = $1000.

What are P(1), P(2), ..., P(5)?

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Your solution:

In P(n + 1) = (1 + r)P(n)……r = .1 and P(0) = $1000 solve P(1), P(2), P(3)

We have to solve for 0 first P(0 + 1) = (1 + .1)* P(0)….P(1) = 1.1 * 1000….P(1) = $1100

P(1 + 1) = (1 + .1) * P(1)…..P(2) = 1.1 * 1100……P(2) = $1210

P(2 +1) = (1+.1) * P(2)……P(3) = 1.1 * 1210….P(3) = $1331

P(3+1) = (1+.1) * P(3)……P(4) = 1.1 * 1331…..P(4) = $1464.1

P(4+1) = (1+.1) * P(4)……P(5) = 1.1 * 1464.1…..P(5) = 1610.51

Confidence rating: positive

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Given Solution:

`a** If n = 0 we get

P(0 + 1) = (1 + .1) * P(0), or P(1) = 1.1 * P(0). Since P(0) = $1000 we have P(1) = 1.1 * $1000 = $1100.

If n = 1 we get

P(1 + 1) = (1 + .1) * P(1), or P(2) = 1.1 * P(1). Since P(1) = $1000 we have P(2) = 1.1 * $1100 = $1210.

If n = 2 we get

P(2 + 1) = (1 + .1) * P(2), or P(3) = 1.1 * P(2). Since P(2) = $1000 we have P(3) = 1.1 * $1210 = $1331.

If n = 3 we get

P(3 + 1) = (1 + .1) * P(3), or P(4) = 1.1 * P(3). Since P(3) = $1000 we have P(4) = 1.1 * $1331 = $1464/1.

If n = 4 we get

P(4 + 1) = (1 + .1) * P(4), or P(5) = 1.1 * P(4). Since P(4) = $1000 we have P(5) = 1.1 * $1464.1 = $1610.51. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qproblem 8. Q(n+1) = .85 Q(n), Q(0) = 400.

What are Q(n) for n = 1, 2, 3 and 4 /

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Your solution:

In Q(n +1) = .85 Q(n)….Q(0) = 400

Q(0 +1) = .85 * Q(0)…..Q(1) = .85 * 400….Q(1) = 340

Q(1+1) = .85 * Q(1)….Q(2) = .85 * 340….Q(2) = 289

Q(2+1) = .85 * Q(2)…..Q(3) = .85 * 289…..Q(3) = 245.65

Q(3+1) = .85 * Q(3)….Q(4) = .85 * 245.65….Q(4) = 208.8025

Confidence rating: positive

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Given Solution:

`a** For n = 0 we have Q(0 + 1) = .85 Q(0) so Q(1) = .85 * 400 = 340.

For n = 1 we have Q(1 + 1) = .85 Q(1) so Q(2) = .85 * 340 = 289.

For n = 2 we have Q(2 + 1) = .85 Q(2) so Q(3) = .85 * 400 = 245.65.

For n = 3 we have Q(3 + 1) = .85 Q(3) so Q(4) = .85 * 400 = 208.803. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qWhat is the growth rate for this equation?

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Your solution:

the growth rate is ..85

Confidence rating: positive

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Given Solution:

`a** The growth factor is .85, which is 1 + growth rate. It follows that the growth rate is .85 - 1 = .15 **

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Self-critique (if necessary):

ok so the growth factor is .85, which is 1 + growth rate, so to find this growth rate we would have to take .85 – 1 = -.15 because if you added the .15 to 1 to get your growth factor you would actually have 1.15 that is why it has to be -.15.

Self-critique rating:

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Question: `qproblem 9. interest rate 12%, initial principle $2000.

What is your difference equation?

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Your solution:

Your difference equation would be:

P(n + 1) = (1+r) P(n)……r = .12 and P(0) = $2000

Confidence rating: mostly

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Given Solution:

`a** The growth rate is 12% = .12

The growth factor is therefore 1 + .12 and the difference equation is

P(n+1)=(1+.12)P(n), P(0)=2000. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qHow did you use your difference equation to find the principle after 1, 2, 3 and 4 years and what

did you get?

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Your solution:

Taking P(n+1) = (1+.12) P(n)…n = 1, 2, 3, 4 …..P(0) = 2000

P(0+1) = (1+.12)* P(0)…P(1) = 1.12 * 2000….P(1) = $2240

P(1+1) = (1+.12)P(1)….P(2) = 1.12 * 2240…P(2) = $2508.8

P(2+1) = (1+.12)P(2)…P(3) = 1.12 * 2508.8….P(3) = 2809.856

P(3+1) = (1+.12)P(3) ….P(4) = 1.12 * 2809.856….P(4) = 3147.13872

Confidence rating: positive

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Given Solution:

`a** STUDENT RESPONSE

P(0+1)=(1+.12)2000 and so on up to P(4) was found.

P1=2240

P2=2508.8

P3=2809.856

P4=3147.03872 **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qproblem 11. Texcess(t) = 50 (.97 ^ t).

What is your estimate of the time required to fall to 1/8 of the original value?

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Your solution:

You have to find the original value first by saying t = 0:

50(.97^0) = 50

In order to find the fall of 1/8 of that value, you must take 1/8 * 50 = 6.25

to find the time required to fall 6.25 you would have an equation like this:

50(.97^t) = 6.25 to solve for t you would have to divide both sides by 50:

.97^t = 6.25 / 50…. .97^t = .125; now we have to find the time that works:

.97^40 = .296 approx.

.97^ 60 = .161 approx.

.97 ^ 65 = .138 approx.

.97^ 68 =- .126 approx

.97^68.2 = .125 approx or .97^68.3 = .125 approx

t = between 68.2 and 68.3

Confidence rating: positive

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Given Solution:

`a** The original value takes place at t = 0 and is Texcess(0) = 50 * .97^0 = 50.

1/8 of the original value is therefore 1/8 * 50 = 6.25.

You have to find t such that 50 * .97^t = 6.25. Dividing both sides by 50 you get

.97^t = 6.25 / 50 or

.97^t = .125.

Use trial and error to find t:

Try t = 10: .97^10 = .74 approx. That's too high.

Try t = 100: .97^100 = .04 approx. That's too low.

So try a number between 10 and 100, probably closer to 100.

Try 70: .97^70 = .118. Lucky guess. That's close to .125 but a little low.

{Try 65: .97^65 = .138. Too high.

Try a number between 65 and 70, closer to 70 but not too much closer.

Try 68: .97^68 = .126. That's good to the nearest whole number.

The process could be continued and refined to get more accurate values of t. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qWhat are your ratios of temperature excess to average rate, and are they nearly constant?

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Your solution:

With the function of 50 * .97^t…t = 0, 4, 7, 9 with the resulting values of:

50, 44.2646405, 40.39914224, 38.01155293:

Giving you the 1st average rate change of:

-5.7353595, -3.86549826, -2.38758931

Giving us or second rate change of:

1.86986124, 1.47790895

???? I know that temperature difference2 / T D1 = ave. rate 2 / ave. rate 1, I don’t quite know if this is what you are asking and exactly how to find the TD ????

Confidence rating: unsure

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Given Solution:

`a** If the function is 50 (.97^t) then at t = 0, 17, 34, 51, 68 we have function values 50, 29.79130219, 17.75043372,

10.57617070, 6.301557949 and average rates of change -1.18756929, -0.7082863803, -0.4220154719,

-0.2514478090, -0.1498191533.

Trapezoids have 'average altitudes' 39.89565109, 23.77086796, 14.16330221, 8.438864327.

Ratio of temp excess to ave rate, using 'average altitudes' as temp excess, are therefore 39.9 / (-1.19), 23.8 / (-.708),

14.2 / (-.422), 8.44 / (-.251). These quantities vary slightly but all are close to the same value around 33. **

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Self-critique (if necessary):

???? I don’t understand why you used a trapezoid and the altitude for the ave rate. I would like to know how to actually use the data given to figure out the average rate. Could you please explain ????

I've edited the 'given solution' in response to your question. Here's a copy:

If the function is 50 (.97^t) then at t = 0, 17, 34, 51, 68 we have function values

temp excesses: 50, 29.79130219, 17.75043372, 10.57617070, 6.301557949

and average rates of change

ave rates: -1.18756929, -0.7082863803, -0.4220154719, -0.2514478090, -0.1498191533.

On the corresponding trapezoidal graph of temperature excess vs. clock time we have four trapezoids, and their slopes correspond to the average rates of change.

The 'altitudes' of the trapezoids correspond to the temperature excesses.

Each trapezoid has a single slope, but two 'altitudes', corresponding to the fact that each interval has two temperature excesses but only a single average rate.

To calculate the desired ratio for an interval, we need a single value of the temperature excess to compare with the single rate.

Rather than using either of the two temp excesses or 'altitudes', it's more appropriate to use their average.

The four trapezoids have 'average altitudes' 39.89565109, 23.77086796, 14.16330221, 8.438864327, each corresponding to the average of the initial and final 'tempearture excesses' on the associated interval.

 

Ratio of temp excess to ave rate, using 'average altitudes' as temp excess, are therefore 39.9 / (-1.19), 23.8 / (-.708), 14.2 / (-.422), 8.44 / (-.251).

These quantities vary slightly but all are close to the same value around 33. **

Self-critique rating:

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Question: `qWhat are your estimates of the times required to fall to half of the three values?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

to fall half of the original value of 50:

½ * 50 = 25….50(.97^t) = 25…..97^t = .5……t = approx.22.75

half of that half time is ½ * 25 = 12.5:

50(.97^t) = 12.5…..97^t = .25….t = approx. 45.51

and the half time of that half time is ½ * 12.5 = 6.25:

50(.97^t) = 6.25……97^t = .125…..t = approx. 68.26

the two time differences intervals is 21.76 and 23.75, which are very close showing some consistency.

Confidence rating:

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Given Solution:

`a** STUDENT RESPONSE: The temperature falls to 50/2 = 25 at t = 22.75.

The temperature falls to 25/2 = 12.5 at t = 45.51

The temperature falls to 12.5/2 = 6.25 at t = 68.26.

The time interval required for each subsequent fall is very close to 22.75, demonstrating that the half-life is constant. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qGive the original and the simplified equation to determine the time required for Texcess to fall to half

its original value.

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Your solution:

The original equation is: Texcess(t) = 50(.97^t) = 50(.97^0) = 50 and half of that is 25.

Then you would simplify your new equation which is: 50(.97^t) = 25

.97^t = ½ you time after trial and error is approx 23.

Confidence rating:

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Given Solution:

`a** Texcess has an original value at t = 0, which gives us Texcess(0) = 50 * .97^0 = 50. Half its original value is

therefore 25.

So our equation is

25 = 50 * .97^t.

This equation is simplified by dividing both sides by 50 to get

.97^t = 1/2. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qproblem 12. Texcess(t) = 50(.97^t), room temperature {{ 25 if you used Celsius and 75 if you

used Farenheit in your observations

What function Temp(t) gives temperature as a function of time?

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Your solution:

Considering my observations were in Fahrenheit I would simply add that to my original equation giving you: 50(.97^t) + 75

Confidence rating:

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Given Solution:

`a** Using 75 for room temperature and realizing that temperature is room temperature + temperature excess we obtain

the function

Temp(t)=50(.97^t)+75.**

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Self-critique (if necessary):

ok, note temp. is room temp. + excess.

Self-critique rating:

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Question: `qIdentify the values of A, b and c in the generalized form y = A b^x + c.

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Your solution:

A = 50

b= .97

c = 75

Confidence rating: positive

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Given Solution:

`a** Since the function is Temp(t) = 50(.97^t)+75 , the y = A b^x + c has y = Temp(t), A = 50, b = .97 and c = 75. **

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Self-critique (if necessary):

ok

Self-critique rating:

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Question: `qproblem 14. Antiobiotic removal, 40 mg/hour when there are 200 milligrams present

At what rate would antibiotic be removed when there are 70 milligrams present?

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Your solution:

Since this is a proportionality question you would use y = k x: y is your amount of removal and x is amount, then we solve for k:

40 = k * 200…….k = 40 / 200 …..k = .2

Function y = .2 x: then we substitute the amount of 70 in the x place:

y = .2 * 70…. y = 14 mg / hr

Confidence rating: positive.

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Given Solution:

`a** If the rate of removal is directly proportional the quantity present then we have

y = k x

where y is the rate of removal and x the amount present.

Since y = 40 when x = 200 we have

40 = k * 200 so that

k = 40/200 = .2.

Thus y = .2 x.

If x = 70 then we have

y = .2 * 70 = 14.

When there are 70 mg present the rate of removal is 14 mg/hr. **

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

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Self-critique (if necessary):

ok

Self-critique rating:

"

&#Your work looks good. See my notes. Let me know if you have any questions. &#