Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 35 potx

The balance after t years is given by the function In particular, if $5000 is put in a bank paying 4% interest per year compounded annually, this formula says that Mt = 5000 1.04t, which

i Approximate the x-values where the graph of y = f (x) intersects the horizontal line y = k Equivalently, trace along the graph until the y-coordinate is k

ii Approximate the zeros of y = f (x) − k

Let’s use the second method to approximate the solution to 2 = (1.04)t Estimate the zeros of the function (1.04)t

− 2 You should come up with approximately 17.67; the money doubles approximately every 17.67 years.9

Generalization

Suppose we put M0dollars in a bank at interest rate r per year compounded annually (If the interest rate is 5%, then r = 0.05.) Assume the money is put in the bank at the beginning

of the year and interest is compounded at the end of the year Then each year the balance is multiplied by 1 + r The balance after t years is given by the function

In particular, if $5000 is put in a bank paying 4% interest per year compounded annually, this formula says that M(t) = 5000 (1.04)t, which agrees with our answer from Example 9.6 Notice that r is 0.04 and not 4; if we used r = 4 instead, then at the end of one year, the account would have grown from $5000 to $5000 · (1 + 4) = $25,000! (This is a nice deal

if you can get it, but not a reasonable answer to this problem.) COMMENT When we write the equation M(t) = 5000 (1.04)tand let the domain be t ≥ 0,

we are modeling a discontinuous phenomenon with a continuous model This equation will only mirror reality if t is an integer, i.e., if the bank has just paid the annual interest to the account For instance, if t ∈ (0, 1), then the interest has not yet been compounded so the balance should be exactly $5000 over this interval At t = 1 it should jump to $5200 However, it is convenient to model this discrete process with a continuous function, as we have seen done in examples in previous chapters

t B

5000 5408 5624.32

actual function (not continuous)

continuous model

Figure 9.4

EXAMPLE 9.7 Suppose we put the $5000 in a bank paying interest at an interest rate of 4% per year but

this time with interest compounded quarterly

This does not mean that you get 4 % interest each quarter.

9 This illustrates a very useful rule of thumb for estimating doubling time: the Rule of 70 This rule of thumb says that to estimate the number of years it will take something growing at a rate of A% per year to double, we calculate 70/A In Example 9.6, we calculate 70/4 = 17.5, and see that this is a very close estimate Keep in mind that this is not a hard and fast rule but rather only

Rather, your money earns 1%,4%4 , interest each quarter of a year The advantage is that interest is computed on the interest more frequently (four times per year, as opposed to just once per year in Example 9.6) The balance is growing by 1% every 1/4 of a year, so it grows according to

M(t ) = 5000 (1.01)4t (Refer to Example 9.3 if you need help with this.) Notice that the balance increases by 1% (is multiplied by 1.01) each quarter, so when t = 1, interest has been compounded four times, as expected M(1) = 5000(1.01)4≈ 5000(1.040604 ), so each year the interest is

effectively 4.0604 % This is called the effective annual interest rate.

4% is called a nominal annual interest rate because the interest is 4% “in name only.”

EXERCISE 9.6 Compare the amount of money you would have after 10 years in Example 9.6 with the

amount in Example 9.7

Answers are provided at the end of the chapter.

Generalization

Suppose you put $M0in a bank at interest rate r per year compounded n times per year Then the interest rate per compounding period is r

n Each year there are n compounding periods, so in t years there are nt compounding periods (The growth isnr every nth of a year.) Then

M(t ) = M0



1 +nr

nt

where M(t ) = the amount of money in the account at the end of t years,

M0= the original amount of money,

r = the annual interest rate (If the rate is 5%, then r = 0.05.),

n = the number of compounding periods per year

(If you have any trouble following this, make a table as was done in Example 9.2 Also, verify that Equation (9.1) is just a special case of the formula presented above.)

EXERCISE 9.7 Bank A gives a nominal interest rate of 6% per year compounded monthly Bank B gives

6.1% interest per year compounded annually Which is the preferable arrangement for an investor?

Radioactive Decay

Through experimentation scientists have found that unstable radioactive isotopes revert back

to nonradioactive form at a rate proportional to the amount of the isotope present The rate

of decay of radioactive substances is typically indicated by identifying the half-life of the

substance; this is the amount of time it takes for half of the original amount of radioactive substance to revert (decay) to its nonradioactive form

EXAMPLE 9.8 Tritium, the radioactive hydrogen isotope H3, has a half-life of 12.3 years (Tritium can be

used for determining the age of wines in a process similar to carbon-dating, which we will look at in Exercise 9.8.) Let H0denote the amount of tritium at time t = 0 Find a formula for H (t), the amount of tritium t years later

Approach 1

Use the ideas of Example 9.3 This idea of radioactive half-life is very similar to the idea

of doubling time we discussed when working with E coli When given information about

doubling time we used 2 as the base; in the case of tritium it makes sense to use 1/2 as our base

H (t ) = H0(1/2)t /12.3

Approach 2

Make a table of values

12.3 0.5 H0

24.6 (0.5)(0.5) H0

36.9 (0.5)3H0

t (0.5)t /12.3H0

So H (t) = H0(0.5)t /12.3

Approach 3

Radioactive material decays at a rate proportional to itself, so an exponential function models the process Therefore, H (t) = Cbtfor some constants C and b

The “initial condition,” the value of H when t = 0, helps us find C At t = 0, H = H0 Therefore, H0= Cb(0)= C and H (t) = H0bt

Now we must find b by using the fact that H (12.3) =21H0

0.5H0= H0b12.3, so 0.5 = b12.3 How do we solve for b?

Recall:If a3= 7, we can solve for a by taking the cube root of both sides That is,

(a3)1/3= 71/3, so a = 71/3

We can apply the same approach to solving for b in the equation 0.5 = b12.3

0.51/12.3= (b12.3)1/12.3 0.51/12.3= b

H (t ) = H0[0.51/12.3]t or H0(0.5)t /12.3

A person using Approach 3 could decide that 0.51/12.3is a clumsy way to express the value of b and therefore replace the expression by a numerical approximation 0.51/12.3≈ 0.945 The answer can be expressed as: H (t) ≈ H0(0.945)t H0(0.945)tis approximately the same as (H0)(0.5)t /12.3 Very large values of t will yield greater discrepancies than smaller ones It is advisable to work with exact values and save any rounding off until the end of the problem If you don’t do this, then you’ll need to think hard about error propagation

We have two different forms in which we can express H as a function of t, t in years:

 1 2

t /12.3

From the former, we can quickly determine that every year there is 94.5% of the amount

of tritium there was the previous year (Another way of putting this is that each year the amount of tritium decreases by 5.5%, because 100% − 94.5% = 5.5%.) From the latter, we can read the information that every 12.3 years half the tritium remains

We can take advantage of the fact that there are many (infinitely many, in fact) ways to write any given number For example, 64 = 26= 40961/2= 0.5−6= 43, and so on

EXERCISE 9.8 Carbon Dating Cosmic rays in the upper atmosphere convert some of the stable form

of carbon, C12, to the unstable radioactive isotope C14, a form with two extra electrons Through photosynthesis, plants take in carbon dioxide from the atmosphere and incorporate the carbon atoms; therefore plants contain the same ratio of C14to C12atoms as exists in the atmosphere Animals eat plants and so have the same ratio of C14to C12as do the plants and the atmosphere Thus, during the lifetime of a living organism the ratio of C14to C12

in the organism is fixed; it is the same as the ratio in the earth’s atmosphere

When an organism dies it stops incorporating carbon atoms The C14 that is in the organism decays exponentially (reverting to the stable C12form), and the C14supply is not replenished Assuming that the ratio of C12to C14has remained constant in the atmosphere, scientists are able to use the unstable C14isotope to date organisms

The half-life of carbon-14 is approximately 5730 years

(a) Let C0denote the amount of radioactive carbon in an organism right before it died Find a formula for C(t), the amount of radioactive carbon in this organism t years after

it has died

(b) A sample from a tree that perished in a volcanic eruption has been analyzed and found

to have 70% of the C14it would have had were it alive Approximately how long ago did the volcanic eruption occur?

Answers are provided at the end of the chapter.

Writing and Rewriting Exponential Functions; Cheap Information, Yours for the Taking

Doubling Time in Minutes; Doubling Time in Hours. We know that the number of

organisms in a colony of E coli under ideal reproductive conditions doubles in size every

20 minutes The number of E coli after t minutes is given by

B(t ) = B0· 2t /20, where B0= B(0) and t is given in minutes

Doubling every 20 minutes is equivalent to doubling every 1/3 hour (or doubling three times each hour) so we can write

B(h) = B0· 23h, where h is given in hours

Percentage Change Per Minute, Percentage Change Per Hour. If we rewrite B(t) =

B0· 2t /20in the form B0bt, we can readily read off the percent change in population each minute

B(t ) = B0· 2t /20= B0(21/20)t≈ B0(1.0353)t= B0(1 + 0.0353)t,

so every minute the E coli population increases by about 3.53%.10

For h measured in hours, B(h) = B023h= B0(23)h= B0(8)h= B0(1 + 7)h= B0



1 +700100

h

,

so every hour the E coli population increases by 700%.

Nominal Annual Interest and Effective Annual Interest. Suppose a bank has a nominal interest rate of 5.4% per year compounded monthly If M0dollars are deposited in the account at time t = 0 and left for t years, the balance in the account will be

M(t ) = M0



1 + 0.05412

12t

Effectively the annual interest will be slightly more than 5.4% due to the compounding Rewriting the function above lets us read off the effective annual interest rate:

M(t ) = M0



12

12t

= M0(1.0045)12t= M0[(1.0045)12]t≈ M0(1.05536)t

We see that a nominal annual interest rate of 5.4% compounded monthly is equivalent to

an effective annual growth rate of about 5.536%

The idea behind the last few examples is that an exponential equation can be written

in many different forms Different forms make different bits of information (annual growth rate, or quarterly growth rate, or doubling time, or tripling time, etc.) available at our fingertips

EXERCISE 9.9 A filtering system is used to purify water used in industrial processes The longer the

water remains in the filtering system the less contaminants remain in the water The system removes contaminants at a rate proportional to the amount of contaminants in the water Let C(t) be the number of milligrams of contaminants in the water t hours after it enters the system Suppose that when the water enters the system it contains C0 milligrams of contaminant

(a) Find C(t) if

i the system removes 20% of the contaminants every hour

10 Alternatively, we could have calculated B(1) and computed the percent change from B(0) to B(1):

B( 1) − B(0) B(0) ≈ 3.53%.

ii the system removes 10% of the contaminants every half hour.

iii the system removes 4.9% of the contaminants every quarter of an hour

iv the system removes 60% of the contaminants every 4 hours

(b) Write each of the functions from part (a) in the form C(t) = C0bt Use your answers

to determine which of the systems is most efficient for water purification Explain your reasoning

Answers are provided at the end of the chapter.

Variations on a Theme—A Difference That Decays Exponentially

EXAMPLE 9.9 You’ve just taken a hot apple cobbler out of the oven and placed it on the kitchen counter to

cool The cobbler comes out of the oven at a piping hot 275◦F Assume the kitchen maintains

a constant temperature of 65◦F How does the temperature of the cobbler change with time?

SOLUTION From experience we know that the cobbler will cool most rapidly when it is first taken from

the oven, when the difference between its temperature and the temperature of the kitchen air is greatest Newton’s Law of Cooling tells us that the temperature difference between the cobbler and the air will decrease exponentially In other words, if A(t) is the temperature

of the apple cobbler at time t, then the quantity A(t) − 65 decays exponentially with time

275

65

Temperature (°F)

time

Figure 9.5

Recall that if y decays exponentially with time, then y(t) = Cbt for some b ∈ (0, 1) Therefore our last statement translates to A(t) − 65 = Cbtfor some b ∈ (0, 1)

When t = 0, A(t) = 275 so at t = 0

275 − 65 = Cb0

210 = C C is the initial temperature difference

We know A(t) = 210bt

+ 65, where b ∈ (0, 1) The graph of y = btfor b ∈ (0, 1) looks like Figure 9.6(a) After doing Exercise 9.5, the graphs drawn in Figures 9.6(b) and (c) should make sense

t

y

210

y

t

y

t

65 275

y = 210b t + 65 for b (0, 1)

y = 210b t for b (0, 1)

y = b t for b (0, 1)

Figure 9.6

Notice that in this example A(t) does not decay exponentially, (A(t) = Cbt), but the temperature difference, (A(t) − 65), does decay exponentially with time

In order to find b we would need to put a thermometer into the apple cobbler and find its temperature at some specified time t This data point would allow us to solve for b For instance, suppose that after 10 minutes the temperature of the cobbler is 175◦F As an exercise you can show that A(t) = 2101121t /10+ 65

Exploratory Problems for Chapter 9

The Derivative of the Exponential Function

What is the derivative of an exponential function bx? We will look at case studies to investigate this derivative numerically and graphically

Case Studies. Let f (x) = 2x, g(x) = 3x, and h(x) = 10x You will investigate f, g, and hfrom both graphical and numer-ical perspectives

1.On the same set of axes, sketch the graphs of f (x) = 2x, g(x) =

3x, and h(x) = 10x Then, on another set of axes, draw a rough sketch of the graphs of f(x), g(x), and h(x) The graphs of f ,

g, and h all intersect the vertical axis at (0, 1) On the other hand, the graphs of f(x), g(x), and h(x)do not intersect at the same point Which has the largest vertical intercept? The smallest? This next problem is best done in groups Each group chooses one of the functions f , g, or h If there are many groups, other bases can be used in addition to 2, 3, and 10, but these constitute the minimum

2.. Complete a table like the one below for your function Ap-proximate the slope of the tangent at a point by the slope of

a secant line For instance, to approximate the derivative of

f (x) = 2x at x = 1, use the slope of the secant line between (1, f (1)) and (1 + h, f (1 + h)) for some very small h Letting

f (x)be 2x you get the two points (1, 21)and (1 + h, 21+h)

f(1) ≈ msec=21+hh−21 for h very small

x f (x) Estimate of f(x) 0

1 2 3 4 5

1 1+h x

f

1+h

2

(1+h, )

(1, 2 )

h

1+h 1+h

2

2

21

–

f (x) = 2x

Post these completed tables so that everyone in the class can see

them

. In light of the data you have collected, make a conjecture about a

formula for f(x)for your function Post your conjecture under

your completed table

If you and your group haven’t been able to come up with

a conjecture based on your data, look at other tables of data for

inspiration.11Or see if you can predict f(6) Or collect more

data until you see a pattern Or look at the conjectures of other

groups and see if you can adapt these conjectures to your own

data

3.After posting all the tables of data and all the conjectures from

each of the groups, try to come up with a conjecture that might

hold more generally, for any function of the form f (x) = bx

4.Do you think that there is a value of b for which the derivative of

bxis bx? If so, about how big do you think this value is?

11 Looking at the data for the function 10 x might help you out.

P R O B L E M S F O R S E C T I O N 9 3

1 Carbon Dating: Carbon-14, with a half-life of approximately 5730 years, can be used

to date organic remains on earth (See Exercise 9.8 for an explanation of carbon dating.) (a) Let C0denote the amount of radioactive carbon in an organism when it is alive Find a formula for C(t), the amount of radioactive carbon in this organism t years after it has died

(b) In the summer of 1993, while in Syria, I visited the ruins of Palmyra (Tadmor

in Arabic—the City of Dates) Palmyra was mentioned in tablets dating as far back as the nineteenth century b.c., but it reached its heyday in the time of Queen Zenobia around 137 a.d While exploring the site near the Funerary Towers, I came across something that looked disturbingly like that of a human leg bone lying in the sand A Harvard medical student I met at the site confirmed that this was indeed

a human bone—really Assuming that the relevant person died around 137 a.d., what percentage of the original C14remained in the bone?

(c) Rewrite the equation you got in part (a) so it is in the form f (t) = C0at From this, fill in the blank in the following statement: Each year the amount of carbon-14 in

2 On July 15, 1999, the Harvard University Gazette reported that workers gutting Holden

Chapel in the course of renovations at the university unexpectedly came across some human bones An archaeology concentrator, Rachel Sexton, who was observing the excavations, called in Professor Carole Mandryk of the Anthropology Department for her insight Since some of the bones were sawed in half, Mandryk first hypothesized a horrible murder—but further investigation by Sexton revealed that before 1850 Holden

Chapel had been used as a medical laboratory The Gazette reports that

Whether the bones would be considered evidence or artifacts came down to a question of age Under Massachusetts law, human remains less than 100 years old are the concern of the police and the forensics lab, while those more than 100 years old fall under the jurisdiction of the state archaeologist.12

The bones were deemed to date back to around 1850 What percent of the original C14 remained in the bones in 1999?

3 A population of bacteria is growing exponentially At 7:00 a.m the mass of the popu-lation is 12 mg Five hours later it is 14 mg

(a) What will be the mass of the bacteria after another 5 hours?

(b) At 7:00 p.m what do we expect the mass to be?

(c) What was the mass of the population at 8:00 a.m.? Given your answer, by what percent is the mass of the population increasing each hour? By what percent is it increasing each day?

4 The population in a certain area of the country is increasing In 1970 the population was 100,000, and by 1990 it was 200,000

12