## Microsoft word - math%20iii%20unit%203%20te.doc

CULMINATING ACTIVITY: TRAVELING TO EXPONENTIA
This culminating task addresses all standards in the unit and is designed to serve as an
assessment task. Some parts of the items below are very explicit in order to assess student
mastery of particular topics. However, in general, the questions do not give the same level of
guidance as the learning tasks. In the learning tasks, students are guided quite explicitly to
solve problems and consider the meanings of the procedures. Such guidance is not included in
the assessment task in order to measure whether students have learned how to compare
graphs, use rational exponents, and understand exponential and logarithmic expressions and
functions As with the learning tasks, the process standards are embedded throughout the task.

A new solar system was discovered far from the Milky Way in 1999. One of the planets in the system, Exponentia, has a number of unique characteristics. After much preparation, NASA sent a group of astronauts to explore the planet. Motion sickness: The time finally came for the astronauts to travel to Exponentia. Before
leaving, the astronauts decided to take medication for motion sickness. Suppose that 85%
of this medication remains in the bloodstream after 1 hour. Three different doses are
taken by the astronauts. Natalie took 25 mg; Jason took 50 mg; and Derrick took 75 mg.
a. Write exponential equations for the amount of dimenhydrinate, motion sickness medication, in each astronaut’s bloodstream after x hours. Solution: Natalie: y = 25(.85)x; Jason: y = 50(.85)x; Derrick: y = 75(.85)x. Note that x is
number of hours.

b. How will the graphs of the three equations be similar? Different? Specifically address domain, range, asymptotes, intercepts, and rate of increase/decrease. Explain what accounts for any differences. Solution: The graphs will have the same domain, range, asymptotes, and rate of decrease. The
domains are all real numbers. The ranges are y > 0. The asymptotes are y = 0. The rate of
decrease is 85%. Because the starting amounts are different, the three graphs will have
different y-intercepts. Natalie’s is (0, 25); Jason’s is (0, 50); and Derrick’s is (0, 75).

c. Express the equation representing Derrick’s remaining dimenhydrinate as a function of Natalie’s equation. Then express Derrick’s remaining medication as a function of Jason’s equation. What graphical transformation does this correspond to? Solution: Derrick’s equation is 3 times Natalie’s. If Natalie’s equation is called N(x), then
Derrick’s equation, D(x), can be expressed as follows: D(x) = 3N(x). Similarly, D(x) = 1.5J(x).
These are vertical stretches of N(x) and J(x).

d. Find the half-life for each concentration. That is, when does the amount in each person’s bloodstream reach half of the original concentration? Explain your solution. Kathy Cox, State Superintendent of Schools Unit 3: Page 59 of 66
Solution: The answer will be the same for all three astronauts because the equations simplify
to the same equivalent equation: .5 = .85x. Using the method of rewriting exponential
equations as equations with both sides having base of 10 through the use of common
logarithms, we get 10-.301 = 10-.0706x. So x = 4.265 hours.

e. At what level might you consider each person’s bloodstream cleared of the motion sickness medication? Why did you choose this level? Solution: Answers will vary. One example: When there is less than 1 mg of the medication
remaining, I would consider it cleared. Technically, it will never reach 0 and this was pretty
close to 0.

f. Using your established criteria, find how long it takes for the dimenhydrinate to clear each person’s bloodstream. Are the three answers the same? Why or why not? Solution: Answers will be based on criteria in (e). Answers based on above: Natalie: x = 19.81
hours; Jason: x = 24.07 hours; Derrick: x = 26.57 hours. Because each person started with a
different dosage, the time it takes for the remaining mg to be less than 1 mg will be greater for
those who took a higher dosage.

Loudness of sound: While on the mission, the astronauts were exposed to a variety of
loud and soft sounds. The loudness of sound, D, measured in decibels (dB) is given by
the formula D = 10 log (I / 10-16), where I is the intensity measured in watts per square
cm (w/cm2). The denominator, 10-16, is the approximate intensity of the least sound
audible to the human ear.6
a. Describe how the graph of the loudness of sound, D, differs from the graph of f(x) = log x. Specifically, name the transformations necessary for f(x) to be transformed into D and state how each transformation alters the characteristics of the logarithm graph (domain, range, asymptote, intercept, increasing/decreasing). Graph D. Solution: The 10 is a vertical stretch and will change all the y-values by a scale factor of 10. It
does not change the domain, range, asymptote, or intercept, but it does make the graph
increase faster. 1016 is a horizontal shrink by a factor of 1016, so the intercept will now be
(1016, 0).

(Note: The graphing program being used for this teachers’ edition can not handle the
numbers needed for this graph. Student graphs should have an x-intercept of (1016, 0).
Students should substitute in some reasonable values, e.g. 1013, 1011, etc., to help them obtain
an accurate graph.)

6 Problems 2 and 3 are adapted from Discovering Advanced Algebra: An Investigative Approach from Key Curriculum Press, 2004. Kathy Cox, State Superintendent of Schools Unit 3: Page 60 of 66
b. If a normal conversation is held at an intensity of 3.16 x 10-10 w/cm2, how many decibels is this? Show your work. Simplify as much as possible before using the calculator. (You should use some properties of exponents.) Solution: D = 10 log (I / 10-16) = 10 log (3.16 x 10-10 / 10-16) = 10 log (3.16 x 106) 65 dB
c. Suppose the whisper of the ventilation system in the space shuttle had an intensity of 10-15 w/cm2. How many decibels is this? Do not use a calculator. Explain how you determined the answer. Solution: D = 10 log (I / 10-16) = 10 log (10) = 10 dB. We know that log (10) means the
exponent, a, that the base 10 is raised to in order for 10a = 10. So a = 1. Therefore, the
equation is now 10 (1) = 10 dB.

d. The loudest a rock concert may be held is 120 dB. This is also how loud a space shuttle launch is from a viewing area for non-essential NASA personnel. What is the intensity of the launch from this site? Leave your answer in exponential notation. (You will need to use that logarithms and exponential functions are inverses.) 7 Solution: D = 10 log (I / 10-16)
12 = log (I/10-16)
1012 = I/10-16
10-4 w/cm2 = I
e. Closer to the launch pad, the sound reaches decibels levels as loud as 180 to 200 dB. What is the range of sound intensity at this site? Leave your answer in exponential notation. Solution: 180 < 10 log(I / 10-16) < 200
18 < log(I / 10-16) < 20
102 < I < 104 w/cm2
f. How many more times more intensity is the launch at the closer observation site than at the farther site in part (c)? Express your answer in exponential notation. Solution: Between 106 and 108 times the w/cm2.
3. Intensity of light: The intensity, I, of a beam of light after passing through t cm of liquid
is given by I = Ae-kt, where A is the intensity of the light when it enters the liquid and k is a constant for the particular liquid. During the planetary rover’s exploration of Exponentia, it encountered a number of different liquid like substances. The NASA scientists would like the astronauts to run some tests on these liquids. a. To make sure that they understood how to use the formula, the scientists asked the astronauts the following questions: If the light intensity in a lake is measured at a 7 Information obtained from the Space Shuttle Recording Project. Kathy Cox, State Superintendent of Schools Unit 3: Page 61 of 66
depth of 50 cm, and the light intensity was 80% of the light intensity at the surface, what is the constant k for the lake water? Solution: .8 = 1e-k(50)
loge.8 = -50k
ln .8 = -50k
k = .0045
b. If the light intensity is measured as 1% of the intensity at the surface of the lake, Solution: .01 = 1e-.0045(t)
loge.01 = -.0045t
ln .01 = -.0045t
t = 1023.27 cm (or
10.2327 m)
c. Suppose that the intensity of light measured 1 m under the surface of the liquid found on Exponentia is 35% of the intensity of light measured at the surface. What is the constant k? Solution: .35 = e-k(100)
ln .35 = -100k
k = .0105
d. At what depth would you need to measure the lake and the substance on Exponentia for the light intensity to be half of what it is at the surface? Use your answers to determine which allows light through more easily. Solution: Lake: .5 = e-.0045t
ln .5 = -.0045t
154.03 cm or 1.54 m
Exponentia: .5 = e-.0105t
ln .5 = -.0105t
66.01 cm
Because it takes less depth for the light to reach 50% of its intensity on Exponentia, this
substance is more resistant to allowing light through. So the lake water allows light
through more easily.

e. Describe the transformations of the graph of the light intensity in the lake from the graph of y = ex. How will they be similar or different? Solution: The equation for light intensity in the lake is I = e-.0045t. The negative sign
reflects the graph of y = ex over the y-axis, so it is now decreasing. The .0045 is a
horizontal stretch by a factor of 1/.0045 = 2000/9.This makes the decrease much slower
than the original increase. The domain, range, intercept, and asymptote all remain
unchanged.

f. How will the graphs of the light intensity for the lake water and light intensity for Solution: Because the horizontal stretch is only 1/.0105 = 2000/21 95.24 in the
Exponentia graph but it is 2000/9
222.2 in the lake graph, the Exponentia graph will not
take as long to decrease. All other characteristics will be the same.

g. Consider the equations of the light intensity in the lake water and in the substance discovered on Exponentia. Write one equation as a transformation of the other. Kathy Cox, State Superintendent of Schools Unit 3: Page 62 of 66
Solution: If the lake equation is L(x) = e-.0045x and the Exponentia equation is E(x) = e-
.0105x, we can write L(x) = E((7/3)x) or E(x) = L((3/7)x).

4. As part of an ongoing space project, the astronauts grew a number of plants while on board the space shuttle. They were charged with keeping up with records on the plants’ growth. Suppose that the plant was measured at noon each day, with the following results. Height (cm)
a. Write an exponential equation to model the growth pattern. (Hint: By how much, in relation to the previous day, is the height changing each day?) Specify what each variable means and the units of measurement for each variable. Solution: y = 2.5(2x), where x is the day at noon and y is the height in centimeters.
b. Suppose that the astronauts forgot to measure the plant on the 5th day. Assuming the rate of growth remained consistent, what was the height of the plant on day 5 at noon? Solution: y = 2.5(25) = 80 cm
c. On the 8th day of the mission, the astronauts, following orders, removed part of the plant from the shuttle and placed it on Exponentia so that they could determine how the plant grows in the new atmosphere. However, the astronauts forgot to measure the height before removing it from the shuttle. If they removed it at 6 pm on the 8th day, what should the height have been? Solution: y = 2.5(28.25) = 761.09 cm or 7.6109 m
d. After being placed in Exponentia’s atmosphere, the plant continued to grow. The following table provides the data for the plant after removal from the shuttle. Write an exponential equation to model the growth pattern. Exponentia, at
6pm)
Height (cm)

Solution: y = 761(1.5)x
Kathy Cox, State Superintendent of Schools Unit 3: Page 63 of 66
e. Explain how the graphs of the equations from (a) and (d) will compare. Specially address their domains, ranges, asymptotes, intercepts, and rate of change. Solution: Because the starting values are different, the y-intercepts will be different. At time 0
for the original plant, it was 2.5 cm tall. At time 0 for the removed plant, it was 761 cm tall. So
the first intercept is (0, 2.5) and the second is (0. 761). Also the rates of change will be
different. The plant on the shuttle grows faster and so will increase faster than the plant on
Exponentia. Otherwise, the domain, range, and asymptotes remain the same.

f. How long will it take the plant that remained on the shuttle to triple its height? How long, from the time it was removed from the shuttle, will it take the plant on Exponentia to triple in height? Solution: Shuttle: 7.5 = 2.5(2x)
3 = 2x
log (3) = log (2) * x
1.58 days
Exponentia: 2283 = 761(1.5x)
3 = 1.5x
log (3) = log (1.5)
2.71 days
Explain why there is / is not a difference in your two answers. If they are different, give a scenario for which they would have the same tripling time. Solution: There is a difference because the rates are different. If they had the same rate, say 2,
then their tripling times would be the same.

g. Assuming that the plants survive, how long will it take each one to reach 0.5 km tall? Solution: Shuttle: 50,000 cm = 2.5(2x)
20,000 = 2x
x = 14.29 days. This would be at
approximately 6:54 pm on the 14th day of the mission.
Exponentia: 50,000 cm = 761(1.5x)
65.7 = 1.5x
x = 10.32 days. This would be at
approximately 1:43 am on the night of the 10th day that the plant was on Exponentia (10.32
days translates to 7 hours, 43 minutes after the 10th complete day on the planet. Since the plant
was placed on the planet at 6 pm, 7 hours later makes the time 1:00am.)

5. Orbital time and orbital radius: German astronomer Johannes Kepler discovered in
1619 that the square of a planet’s orbital period is proportional to the cube of its orbital semi-major axis. This is known as Kepler’s 3rd Law of Motion. Another way to state this law is that the mean orbital radius, measured in astronomical units (AU), is equal to the time of one complete orbit around the sun, measured in years, raised to the power 2/3.8 (One AU is approx. 1.5 x 108 km or 9.3 x 107 miles.) 8 Problem 5 is adapted from Integrated Mathematics 3 by McDougal-Littell, 2002. Kathy Cox, State Superintendent of Schools Unit 3: Page 64 of 66
a. Write the equation that relates mean orbital radius, r, to the orbital time, t. Solution: r = t2/3
b. If Neptune requires 165.02 Earth years to orbit the sun, what is Neptune’s orbital radius? How many km is this? How many miles is this? Solution: 30.09 AU = 4,512,830,074 km = 27,979,554,646 miles
c. Saturn has an orbital radius of 9.542 AU. How many Earth years does it take for Solution: 9.542 = t2/3
9.5423/2 = t
t = 29.47536 years
Mercury Venus
(AU)

Time
(years)

Solution:
Mercury Venus
.7232
1
1.523
5.201
30.086
(AU)

.2408
1
29.475
84.008
Time
(years)

e. After observing Exponentia and its sun, Napier, for a number of years, the NASA astronomers determined that, even though the radius and time are still proportional, the exponent in the above equation would be different for their solar system. The astronomers determined that if Saturn was in Exponentia’s solar system, it would take 70 Earth years to orbit the sun. Kathy Cox, State Superintendent of Schools Unit 3: Page 65 of 66
Use the orbital radius above and determine the equation for the relationship between a planets orbital radius and time in Exponentia’s solar system. Solution: 9.542 = (70)x
.5309416578 0.53; r = t.53
f. One of the NASA astronomers, Delia, proposed using the following equation for the relationship between orbital radius and time in Exponentia’s solar system: t =(tm)5/4, where tm is the orbital time if the planet was in our solar system, the Milky Way. Evaluate Delia’s proposed equation. How does this relate to your equation in (e)? Solution: Delia’s equation is equivalent to t = (r3/2)5/4 = r15/8. This is the inverse of the equation
in (e). (e) could be written as t = r1/.53 = r1.89. 15/8 = 1.875, so the exponent values are quite
close.

6. Design a Viètian (vee-et-ee-an). The inhabitants of Exponentia are Vietians. Use the graphs
of exponential and logarithmic functions to design these creatures encountered by the astronauts on their visit to Exponentia. Your design must meet the following requirements: The design must use the basic exponential and logarithmic graphs as well as the natural exponential, e, and the natural logarithm. Use combinations of vertical and horizontal shifts, vertical and horizontal stretches or shrinks, and reflections through the x-axis or y-axis of the basic functions. Horizontal and vertical lines, as well as any other functions learned in previous courses and units, may also be used. For each equation used, label it in your picture (numbered). On a supplemental page, list the equations (with their numbers), the domains, ranges, asymptotes, and intercepts. Be sure to show any associated work. Solution: Answers will vary. Make sure that all requirements are met and are mathematically
correct. Alter this activity to fit your students. An extension of this activity is to have the
students write a program in their graphing calculators that will draw their picture.

Another extension: Students could also animate their picture to simulate waving, walking, etc.
Kathy Cox, State Superintendent of Schools Unit 3: Page 66 of 66