1.3. INITIAL VALUE PROBLEMS 31
1.3 INITIAL VALUE PROBLEMS
In this section we come to grips with the constant of integration and figure out what its value is.
is requires that we have a bit of additional information. Our motivating example is to build
up the position function s.t/ from the acceleration function a.t/ in steps, with the velocity
function v.t / as an intermediate object. e mathematical model of motion in one dimension
under constant acceleration is:
s.t/ D
1
2
at
2
C v
0
t C s
0
In English, the distance an object is from a reference point is equal to half the acceleration
times the time squared plus the initial velocity times the time plus the initial distance from that
reference point. If we break this into integrals we get:
v.t/ D
Z
t
t
0
a dx
D a .t t
0
/ CC
v.t
0
/ D a 0 CC
v
0
D C
So the constant of integration when we transform acceleration into velocity is the initial velocity.
Similarly:
s.t/ D
Z
t
t
0
v dx
D v .t t
0
/ CC
s.t
0
/ D v 0 C C
s
0
D C
e constant of integration for velocity is initial distance. is shows how the constant of inte-
gration can be solved for if we know the initial value of the quantity we are calculating. In fact,
all we need is the value of the thing we are calculating anywhere in the interval we are integrating
on. e initial value just has neater algebra.
32 1. INTEGRATION, AREA, AND INITIAL VALUE PROBLEMS
Example 1.50 Suppose we fire a cannonball directly upward with a velocity of 120 m/sec with
a gravitational acceleration of 10 m/sec
2
. If the cannon is at a height of 160 m above sea level,
find an expression for the distance above sea level of the cannonball after the cannon is fired at
t D 0.
Solution:
Plug into the equation of motion given above.
s.t/ D
1
2
10t
2
C 120t C 160 D 160 C 120t 5t
2
Since we have a model for this situation, the calculus is all done. Let’s look at a situation where
we need calculus.
˙
Example 1.51 Suppose a missile has an acceleration that builds gradually so that
a.t/ D 100 C 0:1t. If it is launched from a fixed position with a charge that gives it an
initial velocity of 20m/sec, find an expression for the distance the missile has traveled t seconds
after launch and find its position and velocity after t D 20 sec when its fuel runs out.
Solution:
First we find the velocity function. Assume that we start at t D 0.
v.t/ D
Z
a.t / dt
D
Z
.
100 C0:1t
/
dt
D 100t C 0:05t
2
C C
1.3. INITIAL VALUE PROBLEMS 33
Now solve for C :
20 D v.0/
20 D 100.0/ C0:05.0/ C C
20 D C
And we obtain:
v.t/ D 0:05t
2
C 100t C 20
is gives us the velocity after t D 20 sec: V .20/ D 2040 m/sec. Now we find the distance func-
tion.
s.t/ D
Z
v.t/ dt
D
Z
0:05t
2
C 100t C 20
dt
D
1
60
t
3
C 50t
2
C 20t C C
As before...
s.0/ D 0 C0 C 0 CC D 0
C D 0
So...
s.t/ D
1
60
t
3
C 50t
2
C 20t
is is the expression for the the distance the missile has traveled t seconds after launch, and its
position after t D 20 sec is s.20/ Š 20; 533 m.
˙
34 1. INTEGRATION, AREA, AND INITIAL VALUE PROBLEMS
Example 1.52 Suppose that
f .x/ D
Z
3x
2
C 1
dx
and we know that f .2/ D 3. Find an expression for f .x/ with no unknown constants.
Solution:
f .x/ D
Z
3x
2
C 1
dx
D x
3
C x CC
Now use the added information.
f .2/ D 2
3
C 2 C C
3 D 10 CC
7 D C
Combine
f .x/ D x
3
C x 7
˙
Knowledge Box 1.13
Solving initial value problems
When a function resulting from integration has the form
f .x/ CC;
an additional piece of information is needed to determine a value
for C .
1.3. INITIAL VALUE PROBLEMS 35
Example 1.53 Suppose that g.x/ D
Z
e
x
dx. For g.1/ D 12:2, find an expression for g.x/
that does not involve any unknown constants.
Solution:
e integral is trivial – e
x
is its own integral – and so
g.x/ D e
x
C C
Plug in the additional information and solve.
g.1/ D 12:2
e
1
C C D 12:2
C D 12:2 e
C Š 9:48
˙
If we need to do more than one integral, we will need one piece of added information per
unknown constant that arises.
Example 1.54 Suppose that the second derivative of h.x/ is h
00
.x/ D 1:2x 1. For h.0/ D 4
and h
0
.2/ D 2, find an expression for h.x/ that is free of unknown constants.
First we find the first derivative of h.x/:
h
0
.x/ D
Z
.1:2 x 1/dx
D 0:6x
2
x CC
1
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