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APPLIED PROJECT
7.6
CALCULUS AND BASEBALL
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APPLIED PROJECT: CALCULUS AND BASEBALL
This project can be completed
anytime after you have studied
Section 7.66 in the textbook.
In this project we explore three of the many applications of calculus to baseball. The physical
interactions of the game, especially the collision of ball and bat, are quite complex and their
models are discussed in detail in a book by Robert Adair, The Physics of Baseball, 3d ed.
(New York: HarperPerennial, 2002).
1. It may surprise you to learn that the collision of baseball and bat lasts only about a thou-
sandth of a second. Here we calculate the average force on the bat during this collision by
first computing the change in the ball’s momentum.
The momentum p of an object is the product of its mass m and its velocity v, that is,
p 苷 mv. Suppose an object, moving along a straight line, is acted on by a force F 苷 F共t兲
that is a continuous function of time.
(a) Show that the change in momentum over a time interval 关t0 , t1 兴 is equal to the integral
of F from t0 to t1; that is, show that
t1
p共t1 兲 ⫺ p共t0 兲 苷 y F共t兲 dt
t0
Batter’s box
An overhead view of the position of a
baseball bat, shown every fiftieth of
a second during a typical swing.
(Adapted from The Physics of Baseball)
This integral is called the impulse of the force over the time interval.
(b) A pitcher throws a 90-mi兾h fastball to a batter, who hits a line drive directly back
to the pitcher. The ball is in contact with the bat for 0.001 s and leaves the bat with
velocity 110 mi兾h . A baseball weighs 5 oz and, in US Customary units, its mass is
measured in slugs: m 苷 w兾t where t 苷 32 ft 兾s 2.
(i) Find the change in the ball’s momentum.
(ii) Find the average force on the bat.
2. In this problem we calculate the work required for a pitcher to throw a 90-mi兾h fastball
by first considering kinetic energy.
The kinetic energy K of an object of mass m and velocity v is given by K 苷 12 mv 2.
Suppose an object of mass m , moving in a straight line, is acted on by a force F 苷 F共s兲
that depends on its position s. According to Newton’s Second Law
F共s兲 苷 ma 苷 m
dv
dt
where a and v denote the acceleration and velocity of the object.
(a) Show that the work done in moving the object from a position s0 to a position s1 is
equal to the change in the object’s kinetic energy; that is, show that
s1
W 苷 y F共s兲 ds 苷 12 mv12 ⫺ 12 mv 02
s0
where v0 苷 v共s0 兲 and v1 苷 v共s1 兲 are the velocities of the object at the positions s0 and
s1. Hint: By the Chain Rule,
m
dv
dv ds
dv
苷m
苷 mv
dt
ds dt
ds
(b) How many foot-pounds of work does it take to throw a baseball at a speed of 90 mi兾h ?
Copyright © 2013, Cengage Learning. All rights reserved.
3. (a) An outfielder fields a baseball 280 ft away from home plate and throws it directly to
the catcher with an initial velocity of 100 ft兾s. Assume that the velocity v共t兲 of the
ball after t seconds satisfies the differential equation dv兾dt 苷 ⫺ 101 v because of air
;
resistance. How long does it take for the ball to reach home plate? (Ignore any vertical motion of the ball.)
(b) The manager of the team wonders whether the ball will reach home plate sooner if it
is relayed by an infielder. The shortstop can position himself directly between the
outfielder and home plate, catch the ball thrown by the outfielder, turn, and throw the
ball to the catcher with an initial velocity of 105 ft兾s. The manager clocks the relay
time of the shortstop (catching, turning, throwing) at half a second. How far from
home plate should the shortstop position himself to minimize the total time for the
ball to reach the plate? Should the manager encourage a direct throw or a relayed
throw? What if the shortstop can throw at 115 ft兾s?
(c) For what throwing velocity of the shortstop does a relayed throw take the same time
as a direct throw?
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APPLIED PROJECT
CALCULUS AND BASEBALL
SOLUTIONS
dv
, so by the Substitution Rule we have
dt
] v1
] t1 ] t1
dv
dt = m
F (t) dt =
m
dv = [mv]vv10 = mv1 − mv0 = p(t1 ) − p(t0 )
dt
t0
t0
v0
1. (a) F = ma = m
110(5280)
3600
(b) (i) We have v1 = 110 mi/h =
baseball is m =
w
=
g
5/16
32
=
5
512 .
ft/s = 161.3 ft/s, v0 = −90 mi/h = −132 ft/s, and the mass of the
So the change in momentum is
161.3 − (−132) ≈ 2.86 slug-ft/s.
U 0.001
F (t)dt = p(0.001) − p(0) ≈ 2.86, so the average force over
(ii) From part (a) and part (b)(i), we have 0
U
0.001
1
1
the interval [0, 0.001] is 0.001
F (t) dt ≈ 0.001
(2.86) = 2860 lb.
0
p(t1 ) − p(t0 ) = mv1 − mv0 =
2. (a) W =
]
s1
5
512
F (s) ds, where F (s) = m
s0
W =
]
s1
F (s) ds =
s0
]
s1
s0
mv
dv
dv ds
dv
=m
= mv
and so, by the Substitution Rule,
dt
ds dt
ds
dv
ds =
ds
]
v(s1 )
mv dv =
v(s0 )
1
2
mv 2
v1
v0
= 12 mv12 − 12 mv02
(b) From part (b)(i), 90 mi/h = 132 ft/s. Assume v0 = v(s0 ) = 0 and v1 = v(s1 ) = 132 ft/s (note
that s1 is the point of release of the baseball). m =
W =
1
mv12
2
−
1
mv02
2
=
1
2
·
5
512
5
,
512
so the work done is
2
· (132) ≈ 85 ft-lb.
3. (a) Here we have a differential equation of the form dv/dt = kv, so by Theorem 5 .5 .2, the solution is
1
and v(0) = 100 ft/s, so v(t) = 100e−t/10 . We are interested in the
v(t) = v(0)ekt . In this case k = − 10
time t that the ball takes to travel 280 ft, so we find the distance function
s(t) =
Ut
v(x) dx =
0
Ut
0
k
lt
100e−x/10 dx = 100 −10e−x/10 = −1000(e−t/10 − 1) = 1000(1 − e−t/10 )
0
Now we set s(t) = 280 and solve for t: 280 = 1000(1 − e−t/10 ) ⇒ 1 − e−t/10 =
1
7
⇒ t ≈ 3.285 seconds.
t = ln 1 − 25
− 10
7
25
⇒
(b) Let x be the distance of the shortstop from home plate. We calculate the time for the ball to reach home plate as a
function of x, then differentiate with respect to x to find the value of x which corresponds to the minimum time.
The total time that it takes the ball to reach home is the sum of the times of the two throws, plus the relay time
1 s . The distance from the fielder to the shortstop is 280 − x, so to find the time t1 taken by the first throw, we
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solve the equation
Copyright © 2013, Cengage Learning. All rights reserved.
s1 (t1 ) = 280 − x ⇔ 1 − e−t1 /10 =
280 − x
1000
⇔ t1 = −10 ln
720 + x
. We find the time t2 taken by
1000
the second throw if the shortstop throws with velocity w, since we see that this velocity varies in the rest of the
problem. We use v = we−t/10 and isolate t2 in the equation s(t2 ) = 10w 1 − e−t2 /10 = x
⇔ e−t2 /10 = 1 −
tw (x) =
x
10w
⇔ t2 = −10 ln
10w − x
, so the total time is
10w
720 + x
10w − x
1
− 10 ln
+ ln
. To find the minimum, we differentiate:
2
1000
10w
APPLIED PROJECT
CALCULUS AND BASEBALL
1
dtw
1
= −10
−
, which changes from negative to positive when 720 + x = 10w − x ⇔
dx
720 + x
10w − x
x = 5w − 360. By the First Derivative Test, tw has a minimum at this distance from the shortstop to home plate.
So if the shortstop throws at w = 105 ft/s from a point x = 5(105) − 360 = 165 ft from home plate, the
+ 165
− 165
≈ 3.431 seconds. This is longer than the
+ ln 1050
minimum time is t105 (165) = 12 − 10 ln 7201000
1050
time taken in part (a), so in this case the manager should encourage a direct
throw. If w = 115 ft/s, then x = 215 ft from home, and the minimum time is
+ 215
− 215
≈ 3.242 seconds. This is less than the time taken in part (a),
t115 (215) = 12 − 10 ln 7201000
+ ln 1150
1150
so in this case, the manager should encourage a relayed throw.
(c) In general, the minimum time is
tw (5w − 360) =
=
360 + 5w
360 + 5w
1
− 10 ln
+ ln
2
1000
10w
(w + 72)2
1
− 10 ln
2
400w
We want to find out when this is about 3.285 seconds, the same time as the direct throw. From the graph, we
estimate that this is the case for w ≈ 112.8 ft/s. So if the shortstop can throw the ball with this velocity, then a
Copyright © 2013, Cengage Learning. All rights reserved.
relayed throw takes the same time as a direct throw.
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