Calculus: An Integrated Approach to Functions and their Rates of Change, Preliminary Edition Part 52 pdf

Recall that we have deposited $10,000 into a bank with a nominal annual interest rate of 100% and left it for one year.. Therefore, if a bank with nominal annual interest rate r compound

Let’s return to Example 15.1 Recall that we have deposited $10,000 into a bank with a nominal annual interest rate of 100% and left it for one year If the interest were compounded

ntimes a year we would have $10,000(1 +1n)n The question is whether (1 +1n)nincreases without bound or approaches some limiting value (When looked at in this context, it seems unreasonable that the limit would be 1.) Let’s experiment by returning to the numerical approach suggested by Example 15.1 and evaluating (1 +n1)nfor large n The largest value

we looked at in Example 15.1 was n = 525,600 Below are the results of evaluating (1 +1n)n for various values of n using a TI-83 All the digits displayed by this calculator are recorded here

For n = 525,600, the TI-83 gives 2.718279215

For n = 1,000,000, the TI-83 gives 2.718280469

For n = 1010, the TI-83 gives 2.718281828

For n = 1015, the TI-83 gives 1

What are we to make of this? For starters, look at the very last result Do you honestly think that (1 +10115)1015= 1? No The result must be larger than 1 As n increases, (1 +1n)n increases; we know this from the context of the problem What in fact is happening is that the calculator has treated (1 +10115)as 1 and then computed 11015and arrived at 1 Therefore, when considering the numerical results from the calculator, we need to disregard this one If

we evaluate (1 +n1)nfor n larger than 1010, the TI-83 will keep giving us 2.718281828 until

ngets so large that the TI-83 throws up its little calculator hands and gives us the number 1 The results of our numerical investigations might lead you to wonder whether the number 2.718281828 has some significance Where have you seen this number before? If you make your calculator display (to the best of its ability) the number e, it will match up, decimal for decimal, with 2.718281828.4This might lead you to conjecture that limn→∞(1 +1n)n= e

What Happens to the Limit as n Grows Without Bound?

Does the Limit Equal e?

Looking at numerical data and conjecturing that limn→∞(1 +n1)n

= e is great, but if we want to verify this conjecture5we cannot simply try to match up decimal places There are two logical difficulties First of all, e is an irrational number; e cannot be written as a finite decimal or as an infinite repeating one, so we cannot ever hope to match all the decimal places for e Second, and even more fundamental, is the question of where all of these

decimal places for e are coming from to begin with We have simply defined e to be the

number “a” such that the derivative of the function axis ax.6We have to rely exclusively on

4 If you get hold of a few more digits for e, you’ll get 2.718281828459 e is an irrational number; its decimal expansion

is nonrepeating.

5 We will show limx→∞( 1 +1) x = e and then conclude that lim n→∞ ( 1 +1) n = e We must be careful If lim x→∞ f (x) = L then limn→∞f (n) = L, but the converse is not necessarily true If lim n→∞ f (n) = L it is possible that lim x→∞ f (x) = L See the figure below.

1

y

6The Swiss mathematician Leonhard Euler, who first introduced the number e in the mid-1700s, in fact defined e to be the

limit in question (although a rigorous foundation for limits was not laid down until the 1800s) In this text we have not followed

this definition of e While the calculator has been instrumental in suggesting the conjecture,

it has no role in actually proving it.7 Knowing that the derivative of ex is ex, we concluded, after some work, that the derivative of its inverse function ln x is 1/x; this information will turn out to be useful

in verifying our conjecture If we can show limx→∞1 + 1xx= e we can conclude that limn→∞1 +1n

n

= e

Is limx→∞( 1 +x1)xReally e?

Let’s use the longstanding tactic of naming what we are looking for (What follows is valid

if we assume this limit exists and is finite.)

Let B = lim

x→∞

x

x

We have a variable in the exponent; we’d like to “bring it down” in order to make the expression on the right more tractable, so we’ll take the natural logarithm of both sides of the equation:

ln B = ln

 lim x→∞

1 + 1x

x Because the logarithm is a continuous function, this is equivalent to writing8

ln B = lim x→∞

 ln

1 +x1

x

= lim x→∞x ln

1 + 1x



As x → ∞ we have the product of two numbers, one of which is going toward zero (because ln 1 = 0), and the other which is growing without bound.9ln(1 +1x)is approaching

ln 1 = 0, but it is being multiplied by x, which is approaching ∞ It is still hard to decipher what is going on Our area of expertise is more along the lines of taking the limit as something tends toward zero, and this may tie in with our defining characteristic of e Let’s try a substitution in hopes of figuring out this limit Substitution is a tool for transforming the unfamiliar into the familiar

Let h =x1 As x → ∞, h → 0

With this substitution, limx→∞xln(1 + 1x)becomes limh→0 1hln(1 + h) = limh→0 ln(1+h)

h

ln B = lim h→0

ln(1 + h) h This latter limit is easier to evaluate In fact, you may recognize that it is the definition

of the derivative of ln x at x = 1 Verify, using Figure 15.3, that the slope of the secant line

7At this point, we don’t know how the calculator has arrived at its decimal expansion of e Numerically all we have shown is

that e lies between 2.7 and 2.8 If we can prove our conjecture, then we have a way of numerically approximating e.

8 If f is a continuous function, then f (limx→ag(x)) = lim x→a f (g(x)) This is limit principle (4).

9

Here’s another “indeterminate form”: ∞ · 0 On the one hand, any finite number multiplied by zero is zero On the other hand, if a number is growing without bound and we multiply it by any positive number, the product ought to grow without bound (Of course, the second factor is not identically zero; it is just tending toward zero Neither is the first factor “∞”; it is just growing

through (1, 0) and (1 + h, ln(1 + h)) is

ln(1 + h) − ln 1

ln(1 + h)

x

y

y = ln x

1 1 + h (1 + h, ln(1 + h))

Figure 15.3

So lim h→0

ln(1 + h)

x=1

The derivative of ln x is 1x, and evaluating at x = 1 gives 1 Therefore, ln B = 1 We were looking for B, not ln B; exponentiating shows that B = e1= e Eureka!

lim x→∞

x

x

= e

Not only have we gained a whole new perspective on e, but we now have a method of approximating e numerically

Having computed limx→∞(1 + 1x)x= e, we can compute variations on this limit

EXAMPLE 15.4 Show that for any constant r, limn→∞(1 +nr)n= er

will rename our variables (Changing variables is a standard mathematical technique for converting something that looks a bit unfamiliar to something well known and having familiar structure.) We would like to replacenr bym1 so we can utilize our previous result If 1

m=nr, then mr = n, so m =nr Therefore, we use the substitution m =nr We can assume that r = 0, because the case r = 0 can be easily handled independently

r → ∞ as well, because r is a constant

Thus,

lim n→∞

1 +nr

n

= lim m→∞

1 + m1

mr

= lim m→∞



1 + m1

mr But r is a constant, so this is equivalent to10

10 We’re using limit principle (4) again If f is continuous, then lim f (g(x)) = f (lim g(x)).

 lim m→∞

1 + m1

mr

= er

We conclude that

lim n→∞

1 +nr

n

EXAMPLE 15.5 Find limn→∞(1 +n3)4nusing the result of Example 15.4

SOLUTION CAUTIONThis limit business is subtle We don’t want to ad lib Instead, we need to use

exactly the results we have worked so hard to get

lim n→∞

1 + 3n

4n

= lim n→∞



1 +n3

n4

= [e3]4= e12

Implications of the Fact that limn→∞( 1 + rn)n= er

Recall:If you put $M0in a bank account with nominal annual interest rate r compounded

ntimes per year, then M(t), the amount of money in the account after t years, is given by

M(t ) = M0

1 + rn

nt

If we let the number of compounding periods increase without bound, we obtain

M(t ) = M0 lim

n→∞

1 + rn

nt

n→∞



1 + rn

nt Having shown that limn→∞1 + r

n

n

= er, we obtain M(t) = M0[er]t

= M0ert Therefore,

if a bank with nominal annual interest rate r compounds interest continuously,11then the money grows according to

M(t ) = M0ert

EXERCISE 15.2 You plan to deposit a fixed sum of money into one of two bank accounts and leave it there

for several years Which is a better choice of accounts, an account with a nominal interest

11 Do banks really do this? Certainly some banks compute interest on savings accounts every day If the nominal annual interest rate is 5% and interest is compounded daily, then

M(t ) = M 0

1 +0.05 365

 365t

≈ M 0 (1.051267) t

If we modeled this situation using continuous compounding, we would have

M(t ) = M 0 e 0.05t

≈ M 0 (1.051271) t Depending on the situation, the latter model may be quite reasonable and simpler to use (Even using the equation M(t) =

M 0



1 + 0.05 365

 365t

distorts reality a bit, because in reality if interest is compounded at the end of each day, M(t) ought to be

rate of 6% per year compounded monthly or an account with a nominal interest rate of 512% compounded continuously?

P R O B L E M S F O R S E C T I O N 1 5 1

1 Suppose you put $100,000 in a bank account with 6% interest and leave it for one year How much money will there be in the account if the interest is compounded

2 In 1996, inflation in Russia was 22% This was a decline from the 131% inflation rate in

1995 and the 2600% inflation rate in 1994 By contrast, the inflation rate in the United

States in 1996 was about 3% (Boston Globe, November 2, 1996.)

Compute the amount of time it would take for prices to double under each of the four inflation rates listed

3 A Boston Globe article on January 1, 1997, said that the best stock of 1996 was

Information Analysis, Incorporated, which closed the year at a price of $63 per share,

an increase of 1525% during the year The worst stock of 1996 was Mobilemedia Corporation, which closed the year at $7/16 per share, a decrease of 97.6%

What was the price of each of these stocks at the beginning of the year?

4 Compute the following limits In each case stop to think of a strategy, and use whatever strategy seems simplest to you For several of these limits there are different approaches

(x−3) 2

(c) limx→∞(1 − 3x1)7x (d) limt →0+(1 − 2t)1/t

5 Evaluate the following limits

(a) limx→∞ e−x

6 Suppose you invest $10,000 in an account with a nominal annual interest rate of 5% How much money will you have 10 years later if the interest is compounded

7 (a) A certain amount of money is put in an account with a fixed nominal annual interest rate, and interest is compounded continuously If 70 years later the money in the account has doubled, what is the nominal annual interest rate?

(b) Answer the same question if the interest is compounded only once a year

8 (a) Kevin has deposited money in a bank account that compounds interest quarterly

If the nominal interest rate is 5%, what is the effective interest rate?

(b) Ama has deposited money in a bank account that compounds interest quarterly If the effective interest rate is 5% per year, what is the nominal rate of interest?

9 Suppose that a person invests $10,000 in a venture that pays interest at a nominal rate of 8% per year compounded quarterly for the first 5 years and 3% per year compounded quarterly for the next 5 years

(a) How much does the $10,000 grow to after 10 years?

(b) Suppose there were another investment option that paid interest quarterly at a constant interest rate r What would r have to be for the two plans to be equivalent, ignoring taxes?

(c) If an investment scheme paid 3% interest compounded quarterly for the first 5 years and 8% interest compounded quarterly for the next 5 years, would it be better than, worse than, or equivalent to the first scheme?

10 Evaluate the following Substitution may be helpful; these problems are variations on the theme limn→∞1 + r

n

n

(a) lim

w→∞

w + 2 w

w

(c) lim x→∞

x − 1 x

2n

(d) lim n→∞

n + 1

n

(e) lim x→0 +(1 + 2x)3/(2x)

11 Suppose you put $6000 in a bank account at 5% (nominal) annual interest compounded continuously

(a) How much money do you have at the end of 7 years?

(b) How much money do you have at the end of t years?

(c) What is the instantaneous rate of change of money in the account with respect to time? (Find dMdt )

(d) True or False:dMdt = 0.05M Explain your reasoning!

(e) Write your answer to part (b) in the form M = Cat and use your calculator to approximate the value of “a” numerically

(f ) Each year, by what percent does your money grow? (This is called the effective annual yield and, if interest is compounded more than once a year, it is always bigger than the nominal annual interest rate.)

12 Evaluate the following limits Keep in mind that limit calculations can be subtle—don’t

ad lib, but instead keep the limits we looked at firmly in your mind and use substitution

in order to make the transfer to the problems here You can determine whether or not your answer is in the ballpark by using your calculator

(a) limx→∞(1 + 1x)x (b) lims→∞(1 +1s)3s (c) limr→∞(1 + 0.3r )r (d) limw→∞(1 + 2w−1)w (e) limw→∞(1 + (2w)−1)w

13 Which is a better deal, an account offering 4% annual interest compounded continu-ously or an account offering 4.2% interest compounded annually? What is the effective annual yield of the former account?

14 If M(t) = M0ert, finddMdt and show thatdMdt = rM

(dMdt = rM is called a differential equation because it is an equation with a derivative

in it You have just shown that M(t) = M0ertis a solution to this differential equation.)

15.2 INTRODUCING DIFFERENTIAL EQUATIONS

We have been characterizing exponential functions in two ways: If a quantity Q = Q(t) grows (or decays) exponentially, then we have said either

i Q can be written in the form Q(t) = Cbt, where C and b are constants; or

ii Q changes at a rate proportional to itself: dQdt = kQ for some constant k

The first statement gives an explicit formula for the amount, the second gives the rate of change Let’s express the statements in a form that makes their equivalence more transparent

In doing so we’ll sidle up to the subject of differential equations, equations involving a rate (or rates) of change

REMARK In the statements above we write Q and Q(t) interchangeably, using the latter to emphasize that Q is a function of time

We know that statement (i) implies statement (ii); if Q(t) = Cbt, then

dQ

dt = C · ln b · bt

= ln b · Cbt

= ln b · Q(t)

The constant “k” indQdt = kQ is equal to ln b

For a more aesthetically pleasing result, instead of writing Q(t) as Cbt, we can replace

bby ekand write Q(t) = Cekt Now dQdt = ln ek· Q = kQ

We can rewrite statements (i) and (ii) as follows If a quantity Q = Q(t) grows (or decays) exponentially, then there is a constant k (k > 0 for growth, k < 0 for decay) such that

i Q(t) = Cektand

ii dQdt = kQ

Let’s put this in the framework of a bank account If money in a bank account is growing

at a nominal rate of 10% per year compounded continuously, then the instantaneous rate of change of money in the account is 10% of the amount in the account

 (i) M(t) = M0e0.1t (ii) dMdt = 0.1M Equation (ii) really captures the situation in a simple way The rate of change of the amount of money is 10% of the amount of money

Frequently scientists have knowledge about the rate of change of a quantity and from this (and one piece of data about amount—like the initial amount) they try to find an amount equation If you read scientific journals, you will find that the form Q(t) = Cekt is often used to describe exponential growth or decay If the scientist began with an equation of the form dQdt = kQ describing the rate of change of the quantity in question, then that is the most natural form for the amount equation The equationdQ= kQ is called a differential

equation It is a differential equation that arises in many different disciplines, so we will take this opportunity to look at it more closely

Differential Equations and Their Solutions

D e f i n i t i o n

An equation that contains a derivative (or derivatives) is called a differential equation

Some examples of differential equations are:

dy

A function is a solution to a differential equation if it satisfies the differential equation; by

this we mean that when the function and its derivative(s) are substituted in the appropriate places in the differential equation, the two sides of the equation are equal A differential equation will actually have a whole family of solutions A solution whose parameters have

been determined is called a particular solution.

For example, the differential equation dM

dt = 0.1 has solutions M(t) = Ce0.1t, where C can be any constant at all M(t) = Ce0.1tgives a family of solutions In fact, it can be proven

that this is the general solution to the differential equation, meaning that any solution to

the differential equation can be written in this form M(t) = 350e0.1tis a particular solution

to the differential equation

Checking a solution to a differential equation is analogous to checking a solution to

an algebraic equation in that if we are solving for y, then to check we must replace all

occurrences of y (whether in the form y, y, or y ) by the proposed solution

For example, to check whether y = −1 is a solution to the algebraic equation y2− 3 = 2y, we replace all occurrences of y by −1 (whether in the form y or y2or ) We simply evaluate both sides of the equation at y = −1

(−1)2− 3= 2(−1)?

= −2

Analogously, to check whether y(t) = e−t is a solution to the differential equation

d 2 y

dt 2 − 3y = 2dydt, we evaluate both sides of the equation with y = e−t

d2

dt2[e−t] − 3[e−t] ?

= 2dtd [e−t] d

dt[−e−t] − 3e−t = −2e? −t

e−t− 3e−t = −2e? −t

EXAMPLE 15.6 Look for a solution to each of the differential equations given:

(a)dydx = 3 (b)dydt = 2t (c) dydt = y

Then look for a family of solutions Verify that this family of functions actually solves the differential equation

SOLUTION

(a) dydx= 3 We’re looking for a function y(x) whose derivative is 3

y = 3x will do It’s a line with slope 3

y = 3x + 2 will work as well

More generally, we have y = 3x + C for any constant C

y

x

Figure 15.4 A family of lines with slope 3

Check:The differential equation is dydx = 3 Is y = 3x + C a solution? Replace y by 3x + C and see if the two sides are equal

d

dx[3x + C] ?

= 3

(b) dydt = 2t We’re looking for a function y(t) whose derivative is 2t

y = t2will do Its derivative is 2t

More generally, we can use y = t2+ C for any constant C

y

x

Figure 15.5 A family of parabolas with slope 2t

Check:The differential equation is dydt = 2t Is y = t2+ C a solution? Replace y by

t2+ C and see if both sides are equal

d

dt[t2+ C]= 2t?

(c)dydt = y We’re looking for a function y(t) whose derivative is itself We know of one function whose derivative is itself: ex But we want a function of t, so y(t) = et Thinking back to what we were doing right before introducing the term “differen-tial equation,” we see that more generally y = Cetfor any constant C

Check:The differential equation isdxdy= y Is y = Cet a solution? Replace y by Cet and see if both sides are equal

d

dt[Cet] ?

= Cet

Cet= Cet √

y

x

y = Cet for C > 0

y = Cet for C < 0

Figure 15.6

Suppose that, carried away by the success of adding C in parts (a) and (b), we thought that y = et

+ C might be a solution Let’s check it

Check:dydt = y Is y = et+ C a solution? Replace y by et+ C and see if both sides are equal

d

dt[et+ C]= e? t+ C

et = e? t+ C No (unless C = 0)

So y = et

+ C is not a solution to the differential equation

We have verified that if Q(t) = Cekt, then dQdt = kQ; we know that the family of functions Q(t) = Cekt is a solution to the differential equation dQdt = kQ In fact, any solution must be of this form, so we refer to Q(t) = Cekt as the general solution to the

differential equation

REMARKWhen we study differential equations in greater depth we will see that if dydt =

f (t ), then, having found one solution, we can find the general solution by adding a constant

C, whereas ifdy= f (y), this is not the case See if you can make sense of this graphically.