chapter 6 unit 6 定积分的物理应用定积分的物理应用. new words work 功 pressure...
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Chapter 6Chapter 6
Unit 6Unit 6Unit 6Unit 6
定积分的物理应用定积分的物理应用定积分的物理应用定积分的物理应用
New WordsWork 功 Pressure 压力
The universal gravitational constant
万有引力常数
Horizontal component 水平分力
Well-proportioned 均匀的
Perpendicular bisector 中垂线
Orthogonal triangle 直角三角形
The definite integral has wide applications in
mathematics, the physical science and engineering.
Here, we will introduce some simple applications in
the physical science.
At first, we will use the definite integral to compute
the work done by a varying force that moves an
object along a straight line.
Then we use the definite integral to compute the
pressure of water and the force between two objects.
To calculate a quantities by setting up a definite integral, the general procedure may be outlined as follows:
Step 1: Chop up the desired quantity into very small parts.
Step 2: Within each small part, calculate an approximation to the desired quantity.
Step 3: Add up the results of all of the small parts approximation from step 2
Step 4: Obtain a definite integral by taking the limit of
the sum in step 3 as the part get smaller and smaller.
1. Work1. Work
In this subsection we compute the work done by a
varying force that moves an object along a straight
line.
sFW
W
s
F
or
Distance Forcek Wor
simply is edaccomplish work the
then force, theofdirection in the distance aover
operatesit and alueconstant v thehas force theIf
But, if we are faced with a problem in which the force
is variable, we cannot use the above formula. Instead,
the definite integral is required.
Example 1Example 1
A rocket has mass 2000kg. Find the work done in
launching the rockets from the earth’s surface to an
altitude of 100 km. (Use 6400km as the radius of the
earth and ignore the weight of the fuel burned during
the flight)
SolutionSolution
According to the Newton’s law of gravitation. The
force F(x) needed to overcome the force of gravity
is 2x
GMmxF
body a of mass theis earth, theofthecenter
from distance theis and constant, nalgravitatio
universal theis )kg
Nkm10674.6( earth,
theof mass theis kg)10993.5( where
2
217
24
m
x
G
M
6500,6400d,,d
elyapproximat
is d to from liftingin done work The
xxxxxF
xxx
J
xxGMmxxfW
9
2417
6500
6400
26500
6400
10923.1
6500
1
6400
1200010993.510674.6
dd
is work total theHence
Example 2Example 2
A spring is stretched 0.5 meter longer than its rest len
gth. The force required to keep it at that length is 3 ne
wtons. Find the total work accomplished in stretching
the spring 0.5 meter from its rest length
SolutionSolution
Since the force is proportional to x, it is of the form kx
for some constant k. We know that F=3 when x=0.5, s
o6,5.03, kkkxF
To estimate the work involved in stretching the spring
from x to x+dx
The distance dx is small. As the end of the spring is str
etched from x to x+dx, the force is almost constant.
The work accomplished in stretching the spring from x
to x+dx is then approximately
xxxkxW d6dd
The element of work
Hence, the total work is
joule 75.065.0
0 xdxW
Example 3Example 3
SolutionSolution
A cylinder with base radius 3 meters and height 5
meters is full of work. How much work is required to
pump out all water in the cylinder?
See figure
variableof integral the
as ]5,0[ Choosing x
x
ox
dxx
5
ely approximat is
d,on cylinder small
of water of mass The
xxx
.d2.88d
ely approximat is d to
fromout water pumpingin done work The
xxw
xx
x
xd38.9 2
The element of work
Hence, the total work is
Jx
xxw 34622
2.88d2.885
0
25
0
Example 4Example 4
Solution:Solution:
?
,,
20N/s, 3m/s,
N,2000 50N, N,400
m,30 figure See
, ,
耳的功问克服重力需作多少焦
升至井口现将抓起污泥的抓斗提从抓斗缝隙中漏掉
的速率污泥以在提升过程中提升速度为
抓斗抓起的污泥重缆绳每米重
抓斗自重已知井深污泥后提出井口
抓起用缆绳将抓斗放入井底为清除井底的污泥
See figure
x
x
x+dx
0
30
321 WWWW
口需作的功
井抓起污泥的抓斗提升至
由题意知作的功
为提升污泥所所作的功
为克服缆绳重力作的功
为克服抓斗自重所其中
.
;
;
3
2
1
W
W
W
12000304001 W
xxW d3050d 2
22500d305030
02 xxW
作的功为
克服缆绳重力所处处提升到将抓斗由 , d xxx
提升污泥所作的功为在时间间隔 d, ttt
ttW d2020003d 3
s103
30时间为将污泥提升至井口共需
57000d202000310
03 ttW
JW 91500570002250012000
is work total theThus,
2. The pressure force of water2. The pressure force of water
Physics tells us the intensity of pressure of water
with depth h is p=rh, where r is the weight of one
cubic unit of water.
If a board with area A is placed horizontally in
water at a depth h, then the force on this board is
F=pA. But, if the board is submerged vertically in
water, we can not use the above formula. Instead,
the definite integral is required.
Example 5Example 5
Solution:Solution:
A cylindrical water tank has radius R (See figure), and it is full of water by half. Find the force against one end of the tank, where r is the weight of one cubic unit water.
See figure
x
ox
dxx
.d2 22 xxR
],0[],[
for and variable,of integral
theas ],0[ Choosing
Rdxxx
Rx
The area of small strip is
xxRxP d2d
ely approximat is strip small for the force typicalThe
22
xxRxPR
d2
is force total theThus,
22
0
.3
2
3
2
)(d
3
0
322
22
0
22
RxR
xRxR
R
R
Example 6Example 6
A orthogonal triangle board with side a and 2a is
SolutionSolution
The typical area of the small piece is
,d)(2 xxa
submerged vertically in water (see figure). Find
the force against the board
xxaaxP d1)(2)2(d
ely approximat is piece
small for the force typicalThe
x
oa2
a2a
The element of force
.3
7d))(2(2 3
0axxaaxP
a
3. Gravitation3. Gravitation
constant nalgravitatio universla theis
, mass and mass of distance theis where
is mass of
another and mass of particle onebetween
attraction nalgravitatio the:us tellsPhysics
21
221
2
1
k
mmrr
mmkF
m
m
If we want to calculate the gravitational attraction
between a mass and a piece of stick,
the definite integral is required, because the
distance between them is not constant.
Example 7Example 7
particle theandstick ebetween th
attraction nalgravitatio theFind bisector.
lar perpendicu itsin stick thefrom distance
thepoint with at the located is masswith
particle a is thereand , is length stick with
edproportion- wellof piece a ofdensity The
a
mM
l
SolutionSolution See figure
2l
2l
x
y
oMa
ry
dyy
]2
,2
[]d,[ interval smallany takeand
variable,of integral theas ]2
,2
[ Choosing
llyyy
lly
y
dyyy
d masswith
particle theas ],[ piece small theThinking
,d
is them
between attraction nalgravitatio typicalThe
,
is particle
theand piece small ebetween th distance The
22
22
ya
ymkF
yar
,)(
dd
iscomponent horizontal typical theAnd
23
22 ya
yamkFx
,)4(
2
)(
d
iscomponent horizontal the
21
23
2
2 2222 laa
lkm
ya
yamkF
l
lx
.0
iscomponent
vertical thesymmetry, the toAccording
yF
This section presented various physical applications o
f the definite integral. We showed that work, pressure
force and gravitational attraction can be represented a
s definite integrals.
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