Transcript Ch.2 Limits and derivatives

Geometric application: arc length
Problem: Find the length s of the arc defined by the curve
y=f(x) from a to b.
 Solution: Use differential element method, take any
element [x,x+dx], the length of the sub-arc corresponding to
[x,x+dx] is:

s  (x)2  (y )2  (dx) 2  (dy ) 2  1  [ f ( x)]2 dx  ds,
Therefore the total arc length is
b
b
a
a
s   ds   1  [ f ( x)]2 dx.
Arc length formula

If a curve has the equation x=g(y), then the length of the
arc corresponding c  y  d is
d
d
c
c
s   ds  


(dx)  (dy)  
2
2
d
c
1  [ g ( y)]2 dy.
Ex. Find the length of the arc of the parabola y 2  x from
(0,0) to (1,1).
1
1
2
Sol. s   1  [ g ( y)] dy   1  4 y 2 dy
0
0
1 2
1

sec t  sec tdt  (sec t tan t  ln | sec t  tan t |) 0arctan 2
0
2
4
1
5 ln( 5  2)
 {sec(arctan 2)  2  ln[sec(arctan 2)  2]} 

.
4
2
4
arctan 2
Differential of arc length function
A smooth curve is defined by y=f(x) and let s(x) be the
length of the arc corresponding to the part from a to x. Then
s(x) is called the arc length function, and we have formula

x
x
a
a
s( x)   ds   1  [ f (t )]2 dt.

The differential of arc length is
ds  (dx) 2  (dy ) 2 .
Geometric application: surface area

A surface of revolution is generated by rotating a planar
curve about a line.

Problem: find the area of the surface obtained by rotating
the curve y  f ( x) (a  x  b) about the x-axis.
Surface area formula
Use differential element method, take an element [x,x+dx],
the small surface is approximately a portion of a circular cone,
thus its surface area is

y  ( y  dy )
S  2
ds  2 yds
2
 2 f ( x) 1  [ f ( x)]2 dx  dS ,
and therefore the total surface area is
b
b
b
a
a
a
S   dS   2 yds   2 f ( x) 1  [ f ( x)]2 dx.
Surface area formula

If the curve is given by x  g ( y)(c  y  d ), and rotate
the curve about y-axis, then the surface area formula is
d
d
d
c
c
c
S   dS   2 xds   2 g ( y) 1  [ g ( y)]2 dy.


Ex. The curve y  4  x 2 ,(1  x  1) about x-axis,
find the surface area.
Sol. S  1 2 f ( x) 1  [ f ( x)]2 dx

1
 2 
1
1
2
1
x
2
4  x  1
dx  4  dx  8
2
1
4 x
Surface area formula
If the curve is given by x  g ( y) (c  y  d ), and rotate it
about x-axis, the surface area formula is

d
d
c
c
S   2 yds   2 y 1  [ g ( y)]2 dy.
If the curve is y  f ( x)(a  x  b) and rotate about y-axis,
the formula is

b
b
b
a
a
a
S   dS   2 xds   2 x 1  [ f ( x)]2 dx.
Example
Ex. Find the area of the surface generated by rotating the
2
curve segment y  x from (1,1) to (2,4), about the y-axis.
2
2

2
2

xds

2

x
1

(2
x
)
dx

(17 17  5 5).
 Sol I. S 
1
1
6


Sol II.
4
4
S   2 yds   2 y  1  (
1
 2 
4
1
1
4
y  dy 
4
3
1
1
2 y
) 2 dy
1 3/ 2  

( y  4 )   6 (17 17  5 5)
1
4
Physical application: hydrostatic pressure
Background: when an object is submerged in a fluid, the
fluid will exert a force upon the object. The hydrostatic
pressure is defined to be the force per unit area.
 Suppose the object is a thin horizontal plate. Its area is A
and is submerged into depth d of a fluid with density  . Then
we have V  Ad  m  V   Ad
F
 F  mg   gAd  P    gd .
A
 By the principle of fluid pressure, at any point in a liquid
the pressure is the same in all directions, we have P   gd .

Hydrostatic force
Ex. A cylindrical drum with radius 3m is submerged in
water 10m deep. Find the hydrostatic force on one end of it.
 Sol. Use differential element method. First build up the
coordinate frame with origin placed at the center of the
drum. Take any infinitesimal element [x,x+dx], the pressure
on this part is P   gd  9800(7  x), and the force is

F  P  A  9800(7  x)  2 9  x2 dx,
Therefore the total force is
3
F   9800(7  x)2 9  x2 dx  617400 N .
3
Moments and centers of mass


Center of mass: a point on which a thin plate balances
horizontally
A system of two masses m1 and m2 , which lie at x1 and
x2 respectively, the center of mass of the system is x :
m1 x1  m2 x2
m1 ( x  x1 )  m2 ( x  x2 )  0  x 
.
m1  m2
m1 x1 and m2 x2 are called the moments of the masses m1
and m2 (with respect to the origin) respectively.
Moment and center of mass

A system of n particles m1 , m2 ,
x1 , x2 , , xn on the x-axis, then
m1 ( x  x1 ) 
, mn located at the points
 mn ( x  xn )  0  x 
n
n
m x m x
i 1
n
i i
 mi

i i
i 1
m
i 1
.
x is the center of mass of the system, m is the total mass
of the system, and the sum of individual moments
M   mi xi
is called the moments of the system about the origin.
Moment and center of mass

A system of n particles m1 , m2 , , mn located at the points
( x1, y1 ),( x2 , y2 ), ,( xn , yn ) in the xy-plane, then we define
the moment of the system about the y-axis to be
M y   mi xi
and the moment of the system about the x-axis as
M x   mi yi
the coordinates ( x, y ) of the center of mass of the system
x  M y / m, y  M x / m
where m   mi is the total mass.
Moment and center of mass





Consider a flat plate (called a lamina) with uniform density
 that occupies a region R of the plane.
Center of mass is called the centroid
The symmetry principle says that if R is symmetric about
a line l, then the centroid of R lies on l.
The moments are defined so that if the entire mass of a
region is concentrated at the center of mass, then its
moments remain unchanged.
The moment of the union of two nonoverlapping regions is
the sum of the moments of the individual regions.
Moment of a planar region


Suppose the region R lies between the lines x=a and x=b,
above the x-axis and below the curve y=f (x).
Use differential element method: the moment of the
subregion corresponding to [x,x+dx], about y-axis, is
M y   f ( x)dx  x   xf ( x)dx  dM y

So the moment of R about y-axis is
b
M y    xf ( x)dx
a
similarly, the moment of R about x-axis is
b
M x    f 2 ( x)dx
1
2
a
Center of mass of a planar region

The center of mass of R is defined by
x  M y / m, y  M x / m
the total mass is
b
m   A    f ( x)dx
a
so the coordinates of the centroid are
b
xf ( x)dx 1

x

 
m
A
f
(
x
)
dx

f ( x)dx 1
M

y

 
m
A
f
(
x
)
dx

My
a
b
b
a
xf ( x)dx
a
b
x
1
a 2
b
a
2
b
1
a 2
f 2 ( x)dx
Example
Ex. Find the centroid of the region bounded by the curves
y=cosx, y=0, x=0, and x   / 2.


Sol.
A
 /2
0
cos xdx  1
 /2
1 b

x   xf ( x)dx   x cos xdx   1
0
A a
2
1 b1 2
1  /2 2

y   2 f ( x)dx   cos xdx 
A a
2 0
8
Center of mass of a planar region


Suppose now the region R lies between the lines x=a and
x=b, above the curve y=g(x) and below the curve y=f (x)
where f ( x)  g ( x).
The coordinates of the centroid are
1 b
x   x[ f ( x)  g ( x)]dx
A a
1 b1 2
y   2 [ f ( x)  g 2 ( x)]dx
A a
Example

Ex. Find the centroid of the region bounded by the line y=x
2
and y  x .

Sol.
1
A   ( x  x )dx 
0
6
1
2
1
1 1
1
2
x   x[ f ( x)  g ( x)]dx  6 ( x  x )dx 
0
A 0
2
1
1 11 2
2
2
2
4
y   2 [ f ( x)  g ( x)]dx  3 ( x  x )dx 
0
A 0
5
Homework 20

Section 7.8: 25, 37, 40, 54, 55, 56, 58, 77

Section 8.1: 20, 37