#$&*

course MTH 271

Question: `qQuery class notes #04 explain how we can prove that the rate-of-depth-change function for depth function y = a t^2 + b t + c is y' = 2 a t + b

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

To start:

Average rate of change is equal to [ a (t+`dt)^2 + b (t+`dt) + c - ( a t^2 + b t + c ) ] / `dt

=[ a t^2 + 2 a t `dt + a `dt^2 + b t + b `dt + c - ( a t^2 + b t + c ) ] / `dt

=[ 2 a t `dt + a `dt^2 + b `dt ] / `dt

= 2 a t + b t + a `dt.

confidence rating #$&*:

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

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

Given Solution:

`a You have to find the average rate of change between clock times t and t + `dt:

ave rate of change = [ a (t+`dt)^2 + b (t+`dt) + c - ( a t^2 + b t + c ) ] / `dt

= [ a t^2 + 2 a t `dt + a `dt^2 + b t + b `dt + c - ( a t^2 + b t + c ) ] / `dt

= [ 2 a t `dt + a `dt^2 + b `dt ] / `dt

= 2 a t + b t + a `dt.

Now if `dt shrinks to a very small value the ave rate of change approaches y ' = 2 a t + b. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `q explain how we know that the depth function for rate-of-depth-change function y' = m t + b must be y = 1/2 m t^2 + b t + c, for some constant c, and explain the significance of the constant c.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y = a t^2 + b t + c= y ' (t) = 2 a t + b= y ' (t)=mt+b

confidence rating #$&*:

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

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

Given Solution:

`a Student Solution: If y = a t^2 + b t + c we have y ' (t) = 2 a t + b, which is equivalent to the given function y ' (t)=mt+b .

Since 2at+b=mt+b for all possible values of t the parameter b is the same in both equations, which means that the coefficients 2a and m must be equal also.

So if 2a=m then a=m/2. The depth function must therefore be y(t)=(1/2)mt^2+bt+c.

c is not specified by this analysis, so at this point c is regarded as an arbitrary constant. c depends only on when we start our clock and the position from which the depth is being measured. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `q Explain why, given only the rate-of-depth-change function y' and a time interval, we can determine only the change in depth and not the actual depth at any time, whereas if we know the depth function y we can determine the rate-of-depth-change function y' for any time.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

As long as we do not know the initial depth, we are unable to calculate the change in depth.

confidence rating #$&*:

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

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

Given Solution:

`a Given the rate function y' we can find an approximate average rate over a given time interval by averaging initial and final rates. Unless the rate function is linear this estimate will not be equal to the average rate (there are rare exceptions for some functions over specific intervals, but even for these functions the statement holds over almost all intervals).

Multiplying the average rate by the time interval we obtain the change in depth, but unless we know the original depth we have nothing to which to add the change in depth. So if all we know is the rate function, have no way to find the actual depth at any clock time. **

STUDENT COMMENT:

Not sure I understand this I read you solution and it makes more sense.

INSTRUCTOR RESPONSE:

Here's an analogy:

If I tell you that I drove down the interstate at 50 mph for 1/2 hour, then at 70 mph for 1 hour, could you tell me at what milepost I ended?

You couldn't because I didn't tell you where I started.

But you could tell me how far I traveled.

In other words, if I give you the rate information and time intervals, you can tell me how far I went. But if that's the only information I give you, you can't tell me where I started or ended.

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `q In terms of the depth model explain the processes of differentiation and integration.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Differentiation is the change in the rate of depth that is found from using depth data

Integration is being able to find possible depth but not actual depth from a given rate of change

confidence rating #$&*:

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

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

Given Solution:

`a Rate of depth change can be found from depth data. This is equivalent to differentiation.

Given rate-of-change information it is possible to find depth changes but not actual depth. This is equivalent to integration.

To find actual depths from rate of depth change would require knowledge of at least one actual depth at a known clock time. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Growth rate: .10

Growth factor: 1.10

It takes a little more than 7 years to double the principle

confidence rating #$&*:

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

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

Given Solution:

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

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

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

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

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

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

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

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

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The principle is equal to half its t=20 value is 12.7

confidence rating #$&*:

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

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

Given Solution:

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

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

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

For example one student found that half of the 20 value is 1345.5/2=672.75, which occurs. between t = 12 and t = 13 (at t = 12 you get $627.69 and at t = 13 you get 690.45). At 12.75 we get 674.20 so it would probably be about 12.72

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

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

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

10%

1= 1.10

2=1.21

3=1.33

4=1.46

20%

1=1.20

2=1.44

3=1.73

4=2.07

30%

confidence rating #$&*:

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

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

Given Solution:

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

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

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

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

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

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qquery #11. equation for doubling time

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

=(1 + r)^ t = 2

confidence rating #$&*:

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

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

Given Solution:

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

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

Note that this simplifies to

(1 + r)^ t = 2,

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

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

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

1.08^doublingtime = 2

confidence rating #$&*:

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

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

Given Solution:

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

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

This can be written as

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

Dividing both sides by 1.08^2 we obtain

1.08^doublingtime = 2.

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

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The graph would move vertically toward the horizontal axis then moving down the axis

confidence rating #$&*:

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

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

Given Solution:

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

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

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `q#17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour. What is the function Q(t)?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Growth rate =-.11

Growth factor= 1 + r = 1 + (-.11) = .89\

= Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t

confidence rating #$&*:

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

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

Given Solution:

`a Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have

Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or

Q(t)=550(.89)^t **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qHow much antibiotic is present at 3:00 p.m.?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Q(5) = 550 mg * .89^5 = 307.123mg

confidence rating #$&*:

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

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

Given Solution:

`a 3:00 p.m. is 5 hours after the initial time so at that time there will be

Q(5) = 550 mg * .89^5 = 307.123mg

in the blood **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qDescribe your graph and explain how it was used to estimate half-life.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The graph we see

confidence rating #$&*:

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

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

Given Solution:

`a Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point.

The vertical coordinate represents the quantity Q(t), so we find the point on the vertical axis which is half as high as the vertical coordinate of our starting point. We then draw a horizontal line directly over to the graph, and project this point down.

The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qWhat is the equation to find the half-life? What is the most simplified form of this equation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Formula for half life= `a Q(doublingTime) = 1/2 Q(0)

confidence rating #$&*:

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

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

Given Solution:

`a Q(doublingTime) = 1/2 Q(0)or

550 mg * .89^doublingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have

.89^doublingTime = .5.

We can use trial and error to find an approximate value for doublingTIme (later we use logarithms to get the accurate solution). **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `q#19. For the function Q(t) = Q0 (1.1^ t), a value of t such that Q(t) lies between .05 Q0 and .1 Q0.

For what values of t did Q(t) lie between .005 Q0 and .01 Q0?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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

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

Given Solution:

`a Any value between about t = -24.2 and t = -31.4 will result in Q(t) between .05 Q0 and .1 Q0.

Note that these values must be negative, since positive powers of 1.1 are all greater than 1, resulting in values of Q which are greater than Q0.

Solving Q(t) = .05 Q0 we rewrite this as

Q0 * 1.1^t = .05 Q0. Dividing both sides by Q0 we get

1.1^t = .05. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get

t = -31.4 approx.

Solving Q(t) = .1 Q0 we rewrite this as

Q0 * 1.1^t = .1 Q0. Dividing both sides by Q0 we get

1.1^t = .1. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get

t = -24.2 approx.

(The solution for .005 Q0 is about -55.6, for .01 is about -48.3

For this solution any value between about t = -48.3 and t = -55.6 will work). **

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

Self-critique (if necessary):

I had little bit of an issue with this one� Im still actually struggling with it. I have a tutoring session tomorrow night so I plan on brining this question up then.

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

Self-critique Rating:

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

Question: `qexplain why the negative t axis is a horizontal asymptote for this function.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

as t approaches infinity 1.1^t also approaches infinity

Because the numbers are negatively growing, they are approaching zero

confidence rating #$&*:

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

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

Given Solution:

`a The value of 1.1^t increases for increasing t; as t approaches infinity 1.1^t also approaches infinity. Since 1.1^-t = 1 / 1.1^t, we see that for increasingly large negative values of t the value of 1.1^t will be smaller and smaller, and would in fact approach zero. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `q#22. What value of b would we use to express various functions in the form y = A b^x? What is b for the function y = 12 ( e^(-.5x) )?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

`a 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x

y = A b^x for b = .61

confidence rating #$&*:

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

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

Given Solution:

`a 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx.

So this function is of the form y = A b^x for b = .61 approx.. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qwhat is b for the function y = .007 ( e^(.71 x) )?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

`a 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x

y = A b^x for b = 2.041

confidence rating #$&*:

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

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

Given Solution:

`a 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx.

So this function is of the form y = A b^x for b = 2.041 approx.. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qwhat is b for the function y = -13 ( e^(3.9 x) )?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

`a 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x

y = A b^x for b = 49.4

confidence rating #$&*:

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

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

Given Solution:

`a 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x, approx.

So this function is of the form y = A b^x for b = 49.4 approx.. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qList these functions, each in the form y = A b^x.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y=12(.6065^x)

y=.007(2.03399^x]

y=-13(49.40244^x)

confidence rating #$&*:

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

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

Given Solution:

`a The functions are

y=12(.6065^x)

y=.007(2.03399^x) and

y=-13(49.40244^x) **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qExplain how these zeros would appear on the graph of this function.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

On the X-axis the zeros will show up at -2 and -3

confidence rating #$&*:

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

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

Given Solution:

`a We've found the x values where y = 0. The graph will therefore go through the x axis at x = -2 and x = -3. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qQuery class notes #04 explain how we can prove that the rate-of-depth-change function for depth function y = a t^2 + b t + c is y' = 2 a t + b

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

To start:

Average rate of change is equal to [ a (t+`dt)^2 + b (t+`dt) + c - ( a t^2 + b t + c ) ] / `dt

=[ a t^2 + 2 a t `dt + a `dt^2 + b t + b `dt + c - ( a t^2 + b t + c ) ] / `dt

=[ 2 a t `dt + a `dt^2 + b `dt ] / `dt

= 2 a t + b t + a `dt.

confidence rating #$&*:

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

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

Given Solution:

`a You have to find the average rate of change between clock times t and t + `dt:

ave rate of change = [ a (t+`dt)^2 + b (t+`dt) + c - ( a t^2 + b t + c ) ] / `dt

= [ a t^2 + 2 a t `dt + a `dt^2 + b t + b `dt + c - ( a t^2 + b t + c ) ] / `dt

= [ 2 a t `dt + a `dt^2 + b `dt ] / `dt

= 2 a t + b t + a `dt.

Now if `dt shrinks to a very small value the ave rate of change approaches y ' = 2 a t + b. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y = a t^2 + b t + c= y ' (t) = 2 a t + b= y ' (t)=mt+b

confidence rating #$&*:

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

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

Given Solution:

`a Student Solution: If y = a t^2 + b t + c we have y ' (t) = 2 a t + b, which is equivalent to the given function y ' (t)=mt+b .

Since 2at+b=mt+b for all possible values of t the parameter b is the same in both equations, which means that the coefficients 2a and m must be equal also.

So if 2a=m then a=m/2. The depth function must therefore be y(t)=(1/2)mt^2+bt+c.

c is not specified by this analysis, so at this point c is regarded as an arbitrary constant. c depends only on when we start our clock and the position from which the depth is being measured. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

As long as we do not know the initial depth, we are unable to calculate the change in depth.

confidence rating #$&*:

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

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

Given Solution:

STUDENT COMMENT:

Not sure I understand this I read you solution and it makes more sense.

INSTRUCTOR RESPONSE:

Here's an analogy:

If I tell you that I drove down the interstate at 50 mph for 1/2 hour, then at 70 mph for 1 hour, could you tell me at what milepost I ended?

You couldn't because I didn't tell you where I started.

But you could tell me how far I traveled.

In other words, if I give you the rate information and time intervals, you can tell me how far I went. But if that's the only information I give you, you can't tell me where I started or ended.

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `q In terms of the depth model explain the processes of differentiation and integration.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Differentiation is the change in the rate of depth that is found from using depth data

Integration is being able to find possible depth but not actual depth from a given rate of change

confidence rating #$&*:

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

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

Given Solution:

`a Rate of depth change can be found from depth data. This is equivalent to differentiation.

Given rate-of-change information it is possible to find depth changes but not actual depth. This is equivalent to integration.

To find actual depths from rate of depth change would require knowledge of at least one actual depth at a known clock time. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Growth rate: .10

Growth factor: 1.10

It takes a little more than 7 years to double the principle

confidence rating #$&*:

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

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

Given Solution:

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

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

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

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

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

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The principle is equal to half its t=20 value is 12.7

confidence rating #$&*:

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

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

Given Solution:

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

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

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

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

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

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

10%

1= 1.10

2=1.21

3=1.33

4=1.46

20%

1=1.20

2=1.44

3=1.73

4=2.07

30%

confidence rating #$&*:

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

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

Given Solution:

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

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

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qquery #11. equation for doubling time

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

=(1 + r)^ t = 2

confidence rating #$&*:

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

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

Given Solution:

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

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

Note that this simplifies to

(1 + r)^ t = 2,

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

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

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

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

1.08^doublingtime = 2

confidence rating #$&*:

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

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

Given Solution:

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

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

This can be written as

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

Dividing both sides by 1.08^2 we obtain

1.08^doublingtime = 2.

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

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

Self-critique (if necessary):

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

Self-critique Rating:

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

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The graph would move vertically toward the horizontal axis then moving down the axis

confidence rating #$&*:

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

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

Given Solution:

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

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

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

Self-critique (if necessary):

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

Self-critique Rating:

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

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Growth rate =-.11

Growth factor= 1 + r = 1 + (-.11) = .89\

= Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t

confidence rating #$&*:

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

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

Given Solution:

`a Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have

Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or

Q(t)=550(.89)^t **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qHow much antibiotic is present at 3:00 p.m.?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Q(5) = 550 mg * .89^5 = 307.123mg

confidence rating #$&*:

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

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

Given Solution:

`a 3:00 p.m. is 5 hours after the initial time so at that time there will be

Q(5) = 550 mg * .89^5 = 307.123mg

in the blood **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qDescribe your graph and explain how it was used to estimate half-life.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The graph we see

confidence rating #$&*:

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

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

Given Solution:

`a Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point.

The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qWhat is the equation to find the half-life? What is the most simplified form of this equation?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Formula for half life= `a Q(doublingTime) = 1/2 Q(0)

confidence rating #$&*:

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

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

Given Solution:

`a Q(doublingTime) = 1/2 Q(0)or

550 mg * .89^doublingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have

.89^doublingTime = .5.

We can use trial and error to find an approximate value for doublingTIme (later we use logarithms to get the accurate solution). **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `q#19. For the function Q(t) = Q0 (1.1^ t), a value of t such that Q(t) lies between .05 Q0 and .1 Q0.

For what values of t did Q(t) lie between .005 Q0 and .01 Q0?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:

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

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

Given Solution:

`a Any value between about t = -24.2 and t = -31.4 will result in Q(t) between .05 Q0 and .1 Q0.

Note that these values must be negative, since positive powers of 1.1 are all greater than 1, resulting in values of Q which are greater than Q0.

Solving Q(t) = .05 Q0 we rewrite this as

Q0 * 1.1^t = .05 Q0. Dividing both sides by Q0 we get

1.1^t = .05. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get

t = -31.4 approx.

Solving Q(t) = .1 Q0 we rewrite this as

Q0 * 1.1^t = .1 Q0. Dividing both sides by Q0 we get

1.1^t = .1. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get

t = -24.2 approx.

(The solution for .005 Q0 is about -55.6, for .01 is about -48.3

For this solution any value between about t = -48.3 and t = -55.6 will work). **

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

Self-critique (if necessary):

I had little bit of an issue with this one� Im still actually struggling with it. I have a tutoring session tomorrow night so I plan on brining this question up then.

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

Self-critique Rating:

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

Question: `qexplain why the negative t axis is a horizontal asymptote for this function.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

as t approaches infinity 1.1^t also approaches infinity

Because the numbers are negatively growing, they are approaching zero

confidence rating #$&*:

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

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

Given Solution:

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `q#22. What value of b would we use to express various functions in the form y = A b^x? What is b for the function y = 12 ( e^(-.5x) )?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

`a 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x

y = A b^x for b = .61

confidence rating #$&*:

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

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

Given Solution:

`a 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx.

So this function is of the form y = A b^x for b = .61 approx.. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qwhat is b for the function y = .007 ( e^(.71 x) )?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

`a 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x

y = A b^x for b = 2.041

confidence rating #$&*:

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

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

Given Solution:

`a 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx.

So this function is of the form y = A b^x for b = 2.041 approx.. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qwhat is b for the function y = -13 ( e^(3.9 x) )?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

`a 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x

y = A b^x for b = 49.4

confidence rating #$&*:

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

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

Given Solution:

`a 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x, approx.

So this function is of the form y = A b^x for b = 49.4 approx.. **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qList these functions, each in the form y = A b^x.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

y=12(.6065^x)

y=.007(2.03399^x]

y=-13(49.40244^x)

confidence rating #$&*:

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

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

Given Solution:

`a The functions are

y=12(.6065^x)

y=.007(2.03399^x) and

y=-13(49.40244^x) **

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

Self-critique (if necessary):

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

Self-critique Rating:

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

Question: `qExplain how these zeros would appear on the graph of this function.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

On the X-axis the zeros will show up at -2 and -3

confidence rating #$&*:

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

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

Given Solution:

`a We've found the x values where y = 0. The graph will therefore go through the x axis at x = -2 and x = -3. **

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

Self-critique (if necessary):

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

Self-critique Rating:

Self-critique (if necessary):

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

Self-critique rating:

Self-critique (if necessary):

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

Self-critique rating:

#*&!

&#This looks good. Let me know if you have any questions. &#