Assignment 16 Query-again

course mth 163

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016. `query 16

Precalculus I

03-12-2007

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17:00:42

behavior 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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RESPONSE -->

In order to be able to multiply the powers, you must have the same variable. For example (4^3) * (5^2) is not the same as (4*5)^(3*2). To see this we cna multiply both out:

(4*4*4) * (5*5) = 64 * 25 = 1600

the other is:

(4*5)^(3*2) = (20)^(6) = 64000000

confidence assessment: 3

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17:01:07

STUDENT 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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RESPONSE -->

self critique assessment: 3

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17:05:02

Why is the follow erroneous: a^(-n) = - a^n

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RESPONSE -->

Again a substitution example works well here, but a negative exponent makes a fraction, not the variable negative:

2^(-3) = 1/(2)^3 = 1/8 or .125

- 2 ^ 3 = -8

confidence assessment: 3

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17:05:13

STUDENT 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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RESPONSE -->

self critique assessment: 3

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17:06:53

Why is the following erroneous: a^n + a^m = a^(n+m)

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RESPONSE -->

Again, a substitute:

2^2 + 2^4 = 4 + 16 = 20

2^(2+4) = 2^(6) = 64

confidence assessment: 3

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17:07:04

STUDENT 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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RESPONSE -->

self critique assessment: 3

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17:09:55

STUDENT RESPONSE: Why is the following erroneous: a^0 = 0

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RESPONSE -->

I just know that a^0 = 1 I am not sure how to explain this.

self critique assessment: 1

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17:10:51

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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RESPONSE -->

self critique assessment: 2

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17:13:54

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

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RESPONSE -->

4^2 * 4^4 = 16 * 256 = 4096

4^(2*4) = 4^8 = 65536

They are not equal.

confidence assessment: 3

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17:14:04

STUDENT 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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RESPONSE -->

self critique assessment: 3

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17:16:57

problem 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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RESPONSE -->

I was slightly confused....Do we use the -3, -2, -1 , 0, 1, 2, & 3 or am I missing something?

confidence assessment: 1

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17:55:13

STUDENT 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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RESPONSE -->

I understand where the numbers came from.

self critique assessment: 1

The basic points of an exponential are the x = 0 and x = 1 points, and the horizontal asymptote.

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18:04:11

give 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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RESPONSE -->

The points are (0, 400) & (1, 428)

negative x-axis

ratio: 107/100

confidence assessment: 2

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18:04:23

STUDENT 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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RESPONSE -->

self critique assessment: 3

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18:09:55

give 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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RESPONSE -->

(0, 250) & (1, 220)

Positive x axis

ratio:22/25

confidence assessment: 3

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18:11:27

STUDENT 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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RESPONSE -->

self critique assessment: 3

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18:16:15

give basic points and asymptote; y values corresponding to the basic points, ratio of these y values: y = .04 ( .8 ) ^ t

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RESPONSE -->

(0, .04) (1, .032)

Positive x-axis

Ratio: .8

confidence assessment: 3

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18:16:24

STUDENT 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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RESPONSE -->

self critique assessment: 3

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18:25:48

problem 3. y = f(x) = 5 (1.27^x).

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RESPONSE -->

(0, 5) (1, 6.35)

Positive x axis asymtote

ratio: 127/100

confidence assessment: 3

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18:29:34

What is the ratio between the y values at x = 0 and at x = 1?

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RESPONSE -->

The ratio as I showed in the last answer:

127/100 or 1.27

confidence assessment: 3

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18:29:38

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

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RESPONSE -->

self critique assessment: 3

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18:35:57

What is the ratio between the y values at to x = 3.4 and x = 4.4?

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RESPONSE -->

First you find the y values:

y = 5(1.27^3.4)

y = 5(2.25389)

y = 11.27

(3.4, 11.27)

y = 5(1.27^4.4)

y = 5(2.86244)

y = 14.31

(4.4, 14.31)

The ratio is the same as before: 14.31/11.27 = 1.27

confidence assessment: 3

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18:36:38

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

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RESPONSE -->

self critique assessment: 3

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18:42:21

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

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RESPONSE -->

In the first example with the points (0, 5) (1, 6.35) the ratio was 1.27. In the previous example with points (3.4, 11.27) & (4.4, 14.31) the ratio is 1.27

confidence assessment: 2

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18:42:52

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

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RESPONSE -->

I didn't realize I was supposed to pick two more points.

self critique assessment: 3

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18:45:01

What is the ratio of y values when x values are separated by two units?

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RESPONSE -->

I chose to use (0, 5) & (2, 8.0645)

the ratio was 1.6

confidence assessment: 2

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18:45:12

** 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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RESPONSE -->

self critique assessment: 3

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19:10:41

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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RESPONSE -->

the ratio will be the same as long as there is one other unit is higher or lower than the first.

confidence assessment: 1

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19:10:51

** 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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RESPONSE -->

self critique assessment: 3

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19:13:44

problem 5. y = 3 (2 ^ (.3 x) ).

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

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RESPONSE -->

you get as the basic points:

(0, 3) and (1, 3.69)

The ratio is 1.23

confidence assessment: 2

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19:13:48

** 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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RESPONSE -->

self critique assessment: 3

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19:15:11

What is the y = A b^x form of this function?

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RESPONSE -->

I am not sure.

confidence assessment: 0

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19:17:33

** 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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RESPONSE -->

I think I understand the explaination.

self critique assessment: 1

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19:20:19

What does the value of 2 ^ .3 have to do with this situation?

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RESPONSE -->

If you need to find other pointsyou can use it to find b

confidence assessment: 1

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19:20:35

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

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RESPONSE -->

My answer was close.

self critique assessment: 3

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19:29:07

problem 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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RESPONSE -->

After substituting:

P(1) = 1100

P(2) = 1210

P(3) = 1331

P(4) = 1464.10

P(5) = 1610.51

confidence assessment: 2

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19:30:17

** 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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RESPONSE -->

I didn't show everything multiplied out (some of the other problems didn't do this)

self critique assessment: 3

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19:42:33

problem 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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RESPONSE -->

Substitute n = 0 we get

Q(0 + 1) = (.85) * Q(0), or Q(1) = .85 * Q(0). Since Q(0) = 400 we have Q(1) =. 85 * 400 = 340

n = 1:

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

n = 2:

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

n = 3:

Q(3 + 1) = (.85) * Q(3), or Q(4) = .85 * 245.65 = 208.80

confidence assessment: 3

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19:42:45

** 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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RESPONSE -->

self critique assessment: 3

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19:44:35

What is the growth rate for this equation?

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RESPONSE -->

.85

confidence assessment: 1

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19:45:40

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

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RESPONSE -->

I got growth rate and growth factor backward.

self critique assessment: 2

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19:48:55

problem 9. interest rate 12%, initial principle $2000.

What is your difference equation?

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RESPONSE -->

y(t) = P(1+r)^t or in this case:

y(t) = 2000(1.12)^t

confidence assessment: 2

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19:49:26

** 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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RESPONSE -->

The equation I gave is equal to this one.

self critique assessment: 1

Be sure you're aware of the important distinction:

The difference equation is P(n+1)=(1+.12)P(n), P(0)=2000; this would be the correct answer to the question.

y(n) = 2000(1.12)^n is the solution to the equation, not the equation from which the solution is obtained.

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19:54:27

How 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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RESPONSE -->

Substituted in :

n = 0:

P(0 + 1) = (1.12) * Q(0), or Q(1) = 1.12 * 2000 =2240

n = 1:

P(1 + 1) = (1.12) * Q(1), or Q(2) = 1.12 * 2240 =2508.80

n = 2:

P(2 + 1) = (1.12) * Q(2), or Q(3) = 1.12 * 22508.80 = 2809.86

n = 0:

P(3 + 1) = (1.12) * Q(33), or Q(4) = 1.12 * 2809.86 =3147.04

confidence assessment: 3

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19:54:39

** 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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RESPONSE -->

self critique assessment: 3

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20:04:17

problem 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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RESPONSE -->

I think I know how to do this, but I want to check out the explaination.

confidence assessment: 0

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20:07:00

** 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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RESPONSE -->

This explaination makes sense to me. It was not what I had, but I understand how to work the problem.

self critique assessment: 2

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21:05:21

What are your ratios of temperature excess to average rate, and are they nearly constant?

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RESPONSE -->

not sure where to get the numbers.

confidence assessment: 0

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21:06:26

** 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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RESPONSE -->

I guess just pick some numbers? Then look at what info they give.

self critique assessment: 1

Part of the problem statement includes

'Sketch a graph of this function and estimate the time required for the temperature excess to fall to 1/8 of its original value.

Divide this time range into 4 equal segments. For each segment determine the ratio of the average rate at which temperature changes to the average temperature excess for the segment. How nearly constant are your ratios?'

If the time to fall to 1/4 the original value is about 68, then the numbers result from dividing this into four equal intervals of length 68/4 = 17.

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21:09:10

Give the original and the simplified equation to determine the time required for Texcess to fall to half its original value.

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RESPONSE -->

Texcess(t) = (1/2)50(.97^t)

Texcess(t) = 25(.97^t)

confidence assessment: 1

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21:10:00

** 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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RESPONSE -->

I didn't quite have it right. I should have had 25 on the other side and then divided.

self critique assessment: 1

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21:18:00

problem 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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RESPONSE -->

The temperature problems have given me trouble tonight. I will have to look back over it.

confidence assessment: 0

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21:24:00

** 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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RESPONSE -->

The formula makes sense, I just didn't have a clue.

self critique assessment: 1

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21:25:23

Identify the values of A, b and c in the generalized form y = A b^x + c.

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RESPONSE -->

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

A=50

b = .97

c = 75

confidence assessment: 3

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21:25:29

** 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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RESPONSE -->

self critique assessment: 3

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21:48:26

problem 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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RESPONSE -->

Since this is a proportionality problem:

40 = k* 200^3

40 = 8000000k

.000005 = k

r = (.000005) * (70^3)

r = (.000005) * 343000

r = 1.75

confidence assessment: 1

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21:50:05

** 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. **

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RESPONSE -->

I cubed 200 and I shouldn't have.

self critique assessment: 2

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"

Most of your work is good; look back over the problems and be sure you understand. Let me know if you have specific questions.

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