If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
004. `query 4
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
Your solution:
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.
Your solution:
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.
Your solution:
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.
Your solution:
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?
Your solution:
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?
Your solution:
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%
Your solution:
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
Your solution:
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%.
Your solution:
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.
Your solution:
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)?
Your solution:
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.?
Your solution:
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.
Your solution:
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?
Your solution:
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?
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):
Self-critique Rating:
Question:
`qexplain why the negative t axis is a horizontal asymptote
for this function.
Your solution:
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) )?
Your solution:
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) )?
Your solution:
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) )?
Your solution:
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.
Your solution:
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:
`q0.4.44 (was 0.4.40
find all real zeros of x^2+5x+6
Your solution:
Confidence Rating:
Given Solution:
`a We can factor this equation to get (x+3)(x+2)=0.
(x+3)(x+2) is zero if x+3 = 0 or if x + 2 = 0.
We solve these equations to get x = -3 and x = -2, which are
our two solutions to the equation.
COMMON ERROR AND COMMENT:
Solutions are x = 2 and x = 3.
INSTRUCTOR COMMENT:
This error generally comes after factoring the equation into
(x+2)(x+3) = 0, which is satisfied for x = -2 and for x = -3. The error could easily be spotted by
substituting x = 2 or x = 3 into this equation; we can see quickly that neither
gives us the correct solution.
It is very important to get into the habit of checking
solutions. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`qExplain how these zeros would appear on the graph of this
function.
Your solution:
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:
`q 0.4.50 (was 0.4.46
x^4-625=0
Your solution:
Confidence Rating:
Given Solution:
`a Common solution:
x^4 - 625 = 0. Add
625 to both sides:
x^4 = 625. Take
fourth root of both sides, recalling that the fourth power is `blind' to the
sign of the number:
x = +-625^(1/4) = +-[ (625)^(1/2) ] ^(1/2) = +- 25^(1/2) =
+-5.
This is a good and appropriate solution.
It's also important to understand how to use factoring,
which applies to a broader range of equations, so be sure you understand the
following:
We factor the equation to get
(x^2 + 25) ( x^2 - 25) = 0, then factor the second factor to
get
(x^2 + 25)(x - 5)(x + 5) = 0.
Our solutions are therefore x^2 + 25 = 0, x - 5 = 0 and x +
5 = 0.
The first has no solution and the solution to the second two
are x = 5 and x = -5.
The solutions to the equation are x = 5 and x = -5. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`q0.4.70 (was 0.4.66 P = -200x^2 + 2000x -3800. Find the x interval for which profit is
>1000
Your solution:
Confidence Rating:
Given Solution:
`a You have to solve the inequality
1000<-200x^2+2000x-3800, which rearranges to
0<-200x^2+2000x-4800.
This could be solved using the quadratic formula, but it's easier if we
simplify it first, and it turns out that it's pretty easy to factor when we do:
-200x^2+2000x-4800 =
0 divided on both sides by -200 gives
x^2 - 10 x + 24 = 0,
which factors into
(x-6)(x-4)=0 and has
solutions
x=4 and x=6.
So -200x^2+2000x-4800 changes signs at x = 4 and x = 6, and
only at these values (gotta go thru 0 to change signs).
At x = 0 the expression is negative. Therefore from x = -infinity to 4 the
expression is negative, from x=4 to x=6 it is positive and from x=6 to infinity
it's negative. Thus your answer would be
the interval (4,6).
COMMON ERROR: x = 4
and x = 6.
INSTRUCTOR COMMENT:
You need to use these values, which are the zeros of the function, to
split the domain of the function into the intervals 4 < x < 6,
(-infinity, 4) and (6, infinity). Each
interval is tested to see whether the inequality holds over that interval. COMMON ERROR:
Solution 4 > x > 6.
INSTRUCTOR COMMENT:
Look carefully at your inequalities. 4 > x > 6 means that 4 > x AND x
> 6. There is no value of x that is
both less than 4 and greater than 6.
There are of course a lot of x values that are greater than 4 and less
than 6. The inequality should have been
written 4 < x < 6, which is equivalent to the interval (4, 6). **
Self-critique (if necessary):
Self-critique Rating: