On-line Math 21

5.3  Indefinite integration

Example 3 Assume that a thingie is moving along a line so that its acceleration is 4feet/sec² . If at t = 0 the thingie is moving at 5feet/sec and is 7feet from origin, what is its position at any time t ?

Solution

Those conditions, contained in ``If at t = 0 the thingie is moving at 5feet/sec and is 7feet from origin'', are called initial conditions, for reasons that should be clear. Initial conditions allow you to solve for and determine the constants of integration, so that the answer we get will not involve some unknown C .

Since the acceleration a = 4 , and acceleration is the derivative of velocity,

v = ó õ a dt = ó õ 4 dt = 4t+C.

Note that t is the independent variable, representing time. Now, since at time t = 0 , v = 5, we have a string of equalities:

5 = v(0) = 4·0+C = C,

so that C = 5, or

v = 4t+5.

Furthermore, since the position s(t) is the integral of the velocity,

s(t) = ó õ v(t) dt,

we have

s(t) = ó õ v(t) dt = ó õ (4t+5)dt = 2t2+5t+c.

Again, the initial conditions can resolve the constant of integration, since s(0) = 7 :

7 = s(0) = 2·02+5·0+c = c,

so this constant is c = 7 , and

s(t) = 2t2+5t+7.

This is the position of the thingie at time t , which is what we needed to find.

Copyright (c) 2000 by David L. Johnson.