74 CHAPTER 2. FUNCTIONS
Note that cosh(t)+ sinh(t) = et and cosh(t) = cosh(−t) while sinh(−t) =−sinh(t).These are the even and odd parts of the function t → et . This has just given all essentialfeatures of the hyperbolic functions.
2.7 Applications2.7.1 Interest Compounded Continuously
It is possible to compound interest continuously. If time is measured in years and if theinterest rate is r per year, compounded n times a year, then it can be shown that the amountafter t years is P
(1+ r
n
)nt . The idea is to let n get larger and larger. From material presentedlater, the amount becomes increasingly close to Pexp(rt) . This explains the followingprocedure for compounding interest continuously.
Procedure 2.7.1 If the interest rate is r and the interest is compounded continu-ously, to find the future value after t years you compute Pexp(rt) .
Example 2.7.2 The interest rate is 10% and the payment is $1000. What is the future valueafter 5 years if interest is compounded daily and if interest is compounded continuously.
To compound daily, you would have 365 payment periods in each year so the fu-ture value is 1000
(1+ .1
360
)5×360= $1648.60. Now compounding it continuously you get
1000exp(5× .1) = $1648.70. You see, compounding continuously is better than com-pounding daily. If you wait 5 years you get an extra 10 cents. Well, every little bit helps.
2.7.2 Exponential Growth and Decay
Suppose you have a bacteria culture and you feed it all it needs and there is no restriction onits growth due to crowding for example. Then in this case, the rate of growth is proportionalto the amount of bacteria present. This is because the more you have, the more bacteriathere are to divide and make new bacteria. Consider equally spaced intervals of time suchthat n of them equal one unit of time where n is large. The unit might be years, days, etc.Also let Ak denote the amount of whatever is growing at the end of the kth increment of time.In exponential growth Ak+1−Ak ≈ rAk (1/n) , A0 = P where r is a proportionality constantcalled the growth rate and the above difference equation should give a good description ofthe amount provided n is large enough. The symbol ≈ indicates approximately equal. Nowthis is easy to solve. Ak+1 ≈
(1+ r
n
)Ak, A0 = P and you look for Ak = αk and find α. This
is easily seen to be α =(1+ r
n
)and so Ak ≈ P
(1+ r
n
)k. Now let A(t) denote the amount
at time t. What is it? There are tn time intervals so k goes up to tn and you get
A(t)≈ P(
1+rn
)nt≈ P
(1+
rtnt
)nt
the approximation for A(t) getting better as n → ∞. As will be shown later, this becomesvery close to A(t) = Pexp(rt) when n is large which is the formula for exponential growth.As an example, consider P= 10 and r = .01. Then the following is the graph of the functiony = 10e.01t .