query16

#$&*

course Mth163

July 4,2013 8:06PM

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

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

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)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

a^n * b^m = (ab) ^ (n+m)

the above is the correct version you are supposed to add not multiply

&&&&

confidence rating #$&*: 3

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

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

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) **

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

Self-critique (if necessary):

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

Self-critique rating:OK

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

a^(-n) = - a^n

this is incorrect because a^-n=1/(a^n)

where a neg power makes the answer a fraction

&&&&

confidence rating #$&*: 3

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

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

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 **

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

Self-critique (if necessary):

&&&

The whole reason we make it a fraction so that a stays positive

&&&

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

Self-critique rating:3

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

a^n *a^m = a^(n+m)

the only way you would add two powers would be when you are multiplying the main numbers (in the form after you have already used the powers on them which would be the same as having a combined power )

&&&&

confidence rating #$&*: 3

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

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

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.**

STUDENT QUESTION:

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?

INSTRUCTOR RESPONSE:

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).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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

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

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 **

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

Self-critique (if necessary):

&&&&

I did not see a place to put my solution in so I will put it here

a^0 = 0 does not equal zero, anything raised to the zero power equals one

&&&&

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

Self-critique rating:OK

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

: a^n * a^m = a^(n*m). erroneous beauce you do not multiply these you add them

a^n * a^m = a^(n+m)

you would only mutlptiy these two powers when ((a^b)^c) = a^(b*c)

&&&

confidence rating #$&*:

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

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

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).

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

Self-critique (if necessary):

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

Self-critique rating:OK

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

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) )

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

Asymptote:1200

Basic points (0,1200) (1,1304.8)(-1,1104.23)

1304.8/1200=1.09

1104.23/1304.8=.8

&&&&

confidence rating #$&*: 3

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

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

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

STUDENT QUESTION:

Does it matter which form you write this ratio in?

INSTRUCTOR RESPONSE:

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 (if necessary):

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

Self-critique rating:OK

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

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

y = 400 ( 1.07 ) ^ t

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&&

Asymptote: (400,0)

@&

(400, 0) is a point, not an asymptote.

An asymptote is a line a graph approaches, more and more closely and with no limit on how close, but never reaches.

*@

Basic point (0,400) vertex

Other two basic points (1,428)(-1,373.8)

-337.8/428=-.873

&&&&

@&

This is a ratio, but the ratio implied here is a ratio between successive points. This ratio skips a point.

Also the 373.8 is not negative. Ideally you would have sketched a rough graph showing these points, which would avoid the error in reading that value as a negative.

Otherwise, since the question didn't specify just which ratio, what you did makes sense.

However the ratios as illustrated in the worksheets are the ratios as you move to the right. So you divide the later value by the earlier.

When you do so for two consecutive basic points, as you see, the ratio is the same as the base of your exponential function (which in this case is 1.07).

The result would be the same whichever of the two pairs of consecutive basic points you used (i.e., it would be the same whether you used the t = -1 and t = 0 points or the t = 0 and t = 1 points). In fact it will be the same for any two points where the t value changes by +1.

This is a fundamental property of the exponential function. When the t value changes by a given amount, it doesn't matter which t value you started at, you get the same ratio.

*@

confidence rating #$&*: 2

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

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

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

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

Self-critique (if necessary):

&&&

-400 is approaching neg x axis

428/400=107/100=1.07

&&&&

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

Self-critique rating:3

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

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

y = 250 ( 1 - .12 ) ^ t

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

(0,0)

The asymptote approaches zero on the vertical (positive ) y axis

(1,220)

(-1,-1833.33)

@&

Only one of your y values is correct.

Can you show how you use the order of operations to obtain these results?

*@

Ratio of first and second y values

220/0

Ratio=220

@&

220 / 0 is not equal to 220.

What is the correct result for 220 / 0?

*@

&&&&

confidence rating #$&*:

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

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

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.

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

Self-critique (if necessary):

&&&&

I did not use the AOS to get my basic points I was just plugging in zero and one and neg one

I will now plug my numbers into AOS

I know what 250 is probably the a but is the b the 1+.12??? and then I guess c would be zero ???

Never mind the c value wouldn’t even matter because AOS doesn’t even deal with c, that I know of anyway

&&&&

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

Self-critique rating:1

@&

Can you tell me what AOS stands for? I've seen that abbreviation, but only within the last month or two.

*@

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

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

y = .04 ( .8 ) ^ t

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

Using AOS

.a=.04

.b=.8

.c=0

-.8/(2*.04)=x=-100

@&

-.8 / (2 * .04) is not equal to -100.

There is no mention of x, which would need to be defined before being mentioned.

There is no mention of 2 in the question, and 2 does not seem relevant to the calculation of the basic points.

The purpose of this calculation is unclear.

*@

(-100,1.96*10^8)

(-101,2.45*10^8)

(-99,1.57*10^8)

@&

These would be points on the graph of the function, and their t values are separated by 1.

They do have a common ratio of .8, as will any set of points for which t values differ by 1.

However these are not the basic points of the graph.

What are the basic points of the graph of an exponential function?

*@

&&&&

confidence rating #$&*: 1

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

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

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.

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

Self-critique (if necessary):

&&&&

Can you go step by what you did here????

&&&&

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

Self-critique rating:2

@&

If necessary, on your revision, having seen your answers to these questions, I'll do so.

*@

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

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

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

Self-critique (if necessary):

&&&

There is no place to put a solution so I will put it here

Approaches zero at pos y axis

Vertex (-.127,4.9)

1unit right (.873,5.9)

5.9/4.9=1.2

@&

The graph goes through the positive y axis at one of the basic points.

However the value of the function at that point is over 6. The value of the function does not approach zero anywhere near that point.

The graph of this function has a horizontal asymptote. It cannot have a vertex.

When x is -.127, the value of y is not 4.9.

When x is .873, the value of y is not 5.9.

Can you show in detail how you did these calculations?

*@

@&

The basic points are at x = -1, x = 0 and x = 1.

Can you show in detail how you calculate the y values for these x values?

Can you explain how you know, in terms of the behavior of the numbers you get when the function is evaluated for an appropriate sequence of x values, why the graph has an asymptote at the x axis, and on which side of the x axis the asymptote occurs?

What sequence of x values would you use?

How do you know without actually evaluating these x values why the y values must approach zero?

*@

&&&&

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

Self-critique rating:N/A

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

5.9/4.9=1.2

&&&&&

confidence rating #$&*: 3

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

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

Given Solution:

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

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

Self-critique (if necessary):

&&&&

I did that and my values were 5.9/4.9=1.2, I left off the 7

&&&

@&

OK, you're doing the right thing here.

I see how you could be 5.9.

I don't see how you could have gotten 4.9.

*@

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

Self-critique rating:OK

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

12.3/11.3=1.09

&&&

confidence rating #$&*: 3

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

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

Given Solution:

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

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

Self-critique (if necessary):

&&&&

f(4.4) / f(3.4) = 5 * 1.27^4.4 / ( 5 * 1.27 ^3.4) =12.3/11.3=1.09

those are my values what values are you getting after 5 * 1.27^4.4 / ( 5 * 1.27 ^3.4) ,I did the same thing and I did not get 1.27 I got 1.0885

never mind I redid it and got 1.27, I must have misplaced a number before …

&&&&&

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

Self-critique rating:3

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&&

This is really vague are you referring to the pervious problem???

@&

These questions always refer back to the problems you did on the worksheets. The statements are sufficient to identify the question, which will appear in sequence with the previous questions on the problems you did from the worksheet.

*@

If you are then I got 1.27 when I re-plugged the equation back in

f(4.4) / f(3.4) = 5 * 1.27^4.4 / ( 5 * 1.27 ^3.4) = 14.3/11.26=1.27 approx. (technically 1.26998 rounded gives us 1.27)

@&

Good.

Note also that

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

1.27^4.4 / (1.27^3.4) =

1.27^(4.4 - 3.4) =

1.27^1 = 1.27.

*@

&&&&

confidence rating #$&*:

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

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

Given Solution:

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

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

Self-critique (if necessary):

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

Self-critique rating:OK

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&&

Ohh on the pervious one I used your number but my x,y’s are (8,33.8)(10,54.6) (at least I think so its hard to tell I did not label my answer/work paper any way the ratio of these y values are 54.6/33.8=1.6

&&&&

confidence rating #$&*:

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

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

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.

**

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

Self-critique (if necessary):

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

Self-critique rating:OK

@&

Good.

Be sure you know how to prove that this is the case for any value of x, as is done in the solution below.

*@

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

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

I do not have drive but my equation looks like this 5(1.27^(x1+1))

&&&&&

confidence rating #$&*: 3

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

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

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. **

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

Self-critique (if necessary):

&&&&

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

b^(x1 + 1 - x1) once you get to these points the x’s cancel out

&&&&&

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

Self-critique rating:3

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

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&&

3.9/2.9=ratio of y values =1.34

&&&&

confidence rating #$&*:

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

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

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. **

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

Self-critique (if necessary):

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

Self-critique rating:OK

@&

It isn't clear what calculation you are using to get the numbers 3.9 and 2.9.

You do need to indicate how you are calculating your numbers. This is especially important since you have apparently made some calculation and other number-related errors and I can't tell what it is you are missing.

*@

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&&

3(1.23^x)

&&&&

confidence rating #$&*:

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

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

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. **

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

Self-critique (if necessary):

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

Self-critique rating:OK

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

2 ^ .3=b

&&&&

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

Given Solution:

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

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

Self-critique (if necessary):

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

Self-critique rating:OK

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

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)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

P(1)=(1.1^1)(PO=$1000)

P(1)=$1100

P(2)=(1.1^2)(PO=$1000)

P(2)=$1210

P(3)=(1.1^3)(PO=$1000)

P(3)=$1331

P(4)=(1.1^4)(PO=$1000)

P(4)=$1464.10

P(5)=(1.1^5)(PO=$1000)

P(5)=$1610.51

&&&&

confidence rating #$&*: 3

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

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

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. **

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

Self-critique (if necessary):

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

Self-critique rating:OK

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

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

Q(1)=.85^1(400)=340

Q(2)=.85^2(400)=289

Q(3)=.85^3(400)=245.65

Q(4)=.85^4(400)=208.8

&&&&

confidence rating #$&*: 3

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

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

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. **

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

Self-critique (if necessary):

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

Self-critique rating:OK

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

Growth rate is -.15 if Q(n+1)=.85

Then you would have had to add by 1 to get .85(growth factor) so just do the problem in reverse .85-1=-.15

&&&&

confidence rating #$&*: 3

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

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

Given Solution:

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

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

Self-critique (if necessary):

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

Self-critique rating:OK

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

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

What is your difference equation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

(1.12^t)$2000

&&&&

confidence rating #$&*: 3

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

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

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. **

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

Self-critique (if necessary):

&&&

I just fast frwd some of those steps but I used the same process

&&&

@&

This is OK when you are clearly doing the right things and consistently getting answers by a process that agrees with the given solution.

*@

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

Self-critique rating:OK

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

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

did you get?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

.t=years

.t=1,2,3,4

(1.12^1)$2000=$2240

(1.12^2)$2000=$2508.80

(1.12^3)$2000=$2809.86

(1.12^4)$2000=$3147.04

&&&&

confidence rating #$&*: 3

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

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

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 **

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

Self-critique (if necessary):

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

Self-critique rating:OK

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

Question: `qproblem 11. Texcess(t) = 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

(50 (.97 ^ t))^1/8

&&&&

confidence rating #$&*:3

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

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

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. **

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

Self-critique (if necessary):

&&&&

I did not have the materials needed for the experiment

&&&&

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

Self-critique rating:N/A

@&

That's, OK.

However you can still estimate the time required for the given function to fall to 1/8 of its initial value.

*@

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

Im going to make an estimate and say they were because a potato should have an overall constant temp itself

&&&

confidence rating #$&*: N/A

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

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

Given Solution:

`a** 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 (if necessary):

&&&&

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

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

You minus actual intial temp by room temp to get temp excess

&&&&

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

you would just divide each set product by 2

&&&&

confidence rating #$&*: 3

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

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

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. **

STUDENT QUESTION

I don’t understand how to get the t values.

INSTRUCTOR RESPONSE

You evaluate the function

Texcess(t) = 50 (.97 ^ t)

at various values of t until you find the result you're looking for.

For example the temperature falls to 25/2 = 12.5 at some value of t. You can plug in t = 10 but the result will be greater than 12.5. You could plug in higher values of t until you find a result that's lower than 12.5. Then you can keep trying numbers in between until you find a t value that gets you reasonably close to T = 12.5. That occurs around t = 45.51.

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

Self-critique (if necessary):

&&&

[50 (.97 ^ t)]^(1/2)

&&&

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

Self-critique rating:3

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

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

its original value.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

[50 (.97 ^ t)]^(1/2)

&&&&

confidence rating #$&*: 3

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

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

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. **

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

Self-critique (if necessary):

@&

&#Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the parts of the given solution on which your solution didn't agree, and if necessary asking specific questions (to which I will respond).

&#

*@

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

Self-critique rating:OK

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

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

50(.97^t)+75

&&&&

confidence rating #$&*:

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

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

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.**

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

Self-critique (if necessary):

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

Self-critique rating:OK

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

a= 50

b= .97

c= 75

&&&&

confidence rating #$&*: 2

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

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

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. **

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

Self-critique (if necessary):

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

Self-critique rating:OK

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

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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

40/200=k200/200

k= .2

.2*70=14mg/hour

.2*90=18mg/hour

Its like saying 7*2 and 9*2

&&&

confidence rating #$&*: 3

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

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

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.

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

Self-critique (if necessary):

&&&

Why is it when you multiply .2by 70 its like multiplying 7*2???

@&

.2 means 2/10.

2/10 * 70 = 2 * 70 / 10 = 2 * (70 / 10) = 2 * 7.

*@

&&&

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

Self-critique rating:OK

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

Question: Explain why valid or not valid. if invalid give valid form for both sides:

(n^p)^q = n^(p+q)

2^(.3 * x) = (2^.3) * x

2^(.3 + x) = 2^.3 * 2^x

a^(b^c) = (a^b)^c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

(n^p)^q = n^(p+q) when raising a power to a number you do not just switch the numbers you raising with a power it should look like this q^(n^p)=q^(n*p) also when you raise a power to a power you can just multiply these two powers instead

2^(.3 * x) = (2^.3) * x wrong

2^(.3 * x) = (2^.3) ^ x right

2^(.3 + x) = 2^.3 * 2^x right

a^(b^c) = (a^b)^c right

&&&&&

confidence rating #$&*: 3

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

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.

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

Question: What equation would you solve to get the half-life of the function y = 3500 * 0.88^t?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

y = 3500 * 0.88^t

(3500 * 0.88^t)/2 or (3500 * 0.88^t)^1/2

@&

This isn't correct.

This is important, so please review the assignment and revise this.

*@

&&&&

confidence rating #$&*: 3

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

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

Question: What is the ratio f(x + .5) / f(x) for the function f(x) = 25 * (1/4)^x?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

[25(1/4)^(x+.5)]/(25*1/2^(x)

If you plug 1 into x your final answer would be

.25

@&

Good, as far as this answer goes, but you should not have to plug in 1. You have no guarantee that the ratio doesn't depend on the value of x.

How do you simplify your expression

[25(1/4)^(x+.5)]/(25*1/2^(x))?

*@

&&&

confidence rating #$&*: 3

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

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

Question: What are the growth rate and growth factor of the function y = 140 * (1 - .08)^x?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

Growth rate is .08 or 8% and growth factor is .92

&&&&

confidence rating #$&*: 3

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

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

Question: What is the y = A b^x form of the function y = 3 * 2^(2 x)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&&

3(4^x)

&&&&

confidence rating #$&*: 3

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

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

Question: The function y = A b^x has value y = 5 when x = 4 and y = 6 when x = 5. What is the value of b?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

&&&

3(4^x)

.b still stays 4 (if we are using the pervious problem ) or if we are not suing the previous problem b stays b for original value

The values of b could as be (not original value(s))

b^4=b

b^5=b

(ignore everything up there)

[5= A b^4]-[ 6 = A b^5]

I am going to get rid of the exponents on both sides

(1.25=ab)-(1.2=ab)

b=.05

@&

If you subtract those two equations you get

5 - 6 = A b^4 - A b^5.

This leaves you with a single equation having two unknowns, A and b.

This solution doesn't work.

This is very important. Please review and revise this.

*@

&&&&

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

Self-critique rating:N/A

""

"

Self-critique (if necessary):

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

Self-critique rating:

""

"

Self-critique (if necessary):

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

Self-critique rating:

#*&!

@&

You're doing well with much of this assignment, but need a little more work on parts of it.

Please revise by answering the questions I've posed, and making the revisions I've specified in my notes.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

&#

*@