Calculus: Analytic Geometry and Calculus, with Vectors

9.2 Derivatives and integrals of exponentials and logarithms 497

t (time) in such a way that its rate of change with respect tot is proportional to
it. There are in fact many situations in which it is assumed that y is a function
oft such that, for some constant k,

(1) dydt= yk

at each time t. We have thought about this matter before, and we shall think
about it again. The pedestrian way to start to extract information from (1)
is to divide by y. Without assuming that y 0 0, show (with complete attention
to each detail) that if (1) is true, then we can transpose ky and muliply by ekt
to obtain the first of the formulas

(2) d e kty = 0, e kty = A, y = 4ek'

and then the remaining ones in which A is a constant. Continue the work to

show that if y satisfies (1) and the boundary condition

(3)

then

y = yo when t=0,

(4) y = yoekt.

Remark: This problem provides a major reason why exponential functions are

important. In many cases, y decreases with passage of time and k is negative.

In such cases there is a positive number T such that

(5) vyoekt = ypek(t+T)

for each t. This number T, which is called the half-life of y, is the number of

units of time required for half of y to fade away. From (5) we find that - = ekr

or ekT = 2 or -kT = log 2 or

(6)

log 2 0.69315

T = - k k

This formula can be used to determine the half-life T when we know k, and it

can be used to determine k when we know the half-life.

8 Supposing that 0 =< x 5 A, observe that

(1) 05ex 5eA.

Replace x by t and integrate over the interval from 0 to x to obtain

(2) 0<=ex-1<=eAx.

Replace x by t and integrate over the interval from 0 to x to obtain

(3)

z

0ex-1 -x5eA2i-

Repeat the process to obtain

(4)

(5)

x2 x$

05 ex-1-x-Zi:e''31

xs x3 x4

0 ex -1-x-i T-1=eA4i