If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
016. `query 16
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)
Your solution:
Confidence Assessment:
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:
Question: `qWhy is the follow erroneous: a^(-n) = - a^n
Your solution:
Confidence Assessment:
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):
Self-critique rating:
Question: `qWhy is the following erroneous: a^n + a^m = a^(n+m)
Your solution:
Confidence Assessment:
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).
Your solution:
Confidence Assessment:
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):
Self-critique rating:
Question: `q Why is the following erroneous: a^n * a^m = a^(n*m).
Your solution:
Confidence Assessment:
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:
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) )
Your solution:
Confidence Assessment:
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:
Question: `qgive basic points and asymptote; y values corresponding to the basic points, ratio of these y values:
y = 400 ( 1.07 ) ^ t
Your solution:
Confidence Assessment:
Given Solution:
You get two separate equations.
Two equations are necessary to find the values of the two parameters A and b.
You would not use two different values of y in the same equation.
ERRONEOUS STUDENT SOLUTION (including error), instructor notes in bold
Asymptote: (400,0)
<h3>@&
(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.
*@</h3>
Basic points at x = -1, 0 and 1.
y values are 400 * 1.07^-1 = 373.8, 400 * 1.07^0 = 400, 400 * 1.07^1 = 428.
Basic point (0,400) vertex
Other two basic points (1, 428)(-1, 373.8)
Ratio: -337.8/428=-.873
<h3>@&
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.
*@</h3>
`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
ERRONEOUS STUDENT RESPONSE AND INSTRUCTOR COMMENT:
Self-critique (if necessary):
Self-critique rating:
Question: `qgive basic points and asymptote; y values corresponding to the basic points, ratio of these y values:
y = 250 ( 1 - .12 ) ^ t
Your solution:
Confidence Assessment:
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):
Self-critique rating:
Question: `qgive basic points and asymptote; y values corresponding to the basic points, ratio of these y values:
y = .04 ( .8 ) ^ t
Your solution:
Confidence Assessment:
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):
Self-critique rating:
Question: `q problem 3. y = f(x) = 5 (1.27^x).
Self-critique (if necessary):
Self-critique rating:
Question: `qWhat is the ratio between the y values at x = 0 and at x = 1?
Your solution:
Confidence Assessment:
Given Solution:
`a** f(1) / f(0) = 5 * 1.27^1 / ( 5 * 1.27^0) = 1.27 **
Self-critique (if necessary):
Self-critique rating:
Question: `qWhat is the ratio between the y values at to x = 3.4 and x = 4.4?
Your solution:
Confidence Assessment:
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):
Self-critique rating:
Question: `qVerify that the ratio of y values is again the same for your own points where x differs by 1 unit.
Your solution:
Confidence Assessment:
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:
Question: `qWhat is the ratio of y values when x values are separated by two units?
Your solution:
Confidence Assessment:
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:
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?
Your solution:
Confidence Assessment:
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):
Self-critique rating:
Question: `qproblem 5. y = 3 (2 ^ (.3 x) ).
What is the ratio of the two basic-point y values?
Your solution:
Confidence Assessment:
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:
Question: `qWhat is the y = A b^x form of this function?
Your solution:
Confidence Assessment:
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:
Question: `qWhat does the value of 2 ^ .3 have to do with this situation?
Your solution:
Confidence Assessment:
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:
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)?
Your solution:
Confidence Assessment:
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:
Question: `qproblem 8. Q(n+1) = .85 Q(n), Q(0) = 400.
What are Q(n) for n = 1, 2, 3 and 4 ?
Your solution:
Confidence Assessment:
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:
Question: `qWhat is the growth rate for this equation?
Your solution:
Confidence Assessment:
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:
Question: `qproblem 9. interest rate 12%, initial principle $2000.
What is your difference equation?
Your solution:
Confidence Assessment:
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):
Self-critique rating:
Question: `qHow did you use your difference equation to find the principle after 1, 2, 3 and 4 years and what
did you get?
Your solution:
Confidence Assessment:
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:
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?
Your solution:
Confidence Assessment:
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):
Self-critique rating:
Question: `qWhat are your ratios of temperature excess to average rate, and are they nearly constant?
Your solution:
Confidence Assessment:
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):
Self-critique rating:
Question: `qWhat are your estimates of the times required to fall to half of the three values?
Your solution:
Confidence Assessment:
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):
Self-critique rating:
Question: `qGive the original and the simplified equation to determine the time required for Texcess to fall to half
its original value.
Your solution:
Confidence Assessment:
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):
Self-critique rating:
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?
Your solution:
Confidence Assessment:
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:
Question: `qIdentify the values of A, b and c in the generalized form y = A b^x + c.
Your solution:
Confidence Assessment:
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:
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?
Your solution:
Confidence Assessment:
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):
Self-critique rating:
(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.
Your solution:
Confidence Assessment:
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?
Your solution:
Confidence Assessment:
Question: What is the ratio f(x + .5) / f(x) for the function f(x) = 25 * (1/4)^x?
Your solution:
Confidence Assessment:
Question: What are the growth rate and growth factor of the function y = 140 * (1 - .08)^x?
Your solution:
Confidence Assessment:
Question: What is the y = A b^x form of the function y = 3 * 2^(2 x)?
Your solution:
Confidence Assessment:
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?
Your solution:
Self-critique rating: