6. Civil- IJCE- Effect of Patch Length Ratio of in-Plane - Jawadtalibadodi
Length of a Plane Curve
description
Transcript of Length of a Plane Curve
Length of a Plane Curve
Objective: To find the length of a plane curve.
Arc Length
• Our first objective is to define what we mean by length (or arc length) of a plane curve y = f(x) over an interval [a, b]. Once that is done we will be able to focus on the problem of computing arc lengths. We will have the requirement that f / be continuous on [a, b] and we say that y = f(x) is a smooth curve.
Arc Length Problem
• Suppose that y = f(x) is a smooth curve on the interval [a, b]. Define and find a formula for the arc length of L of the curve y = f(x) over the interval [a, b]
Arc Length
• To define the arc length of a curve we start by breaking the curve into small segments. Then we approximate the curve segments by line segments and add the lengths of the line segments to form a Riemann Sum. As we increase the number of segments, the approximation becomes better and better.
Arc Length
• To implement our idea, divide the interval [a, b] into n subintervals by inserting points between the values . Let be the points on the curve that join the line segments. These line segments form polygonal path that we can regard as an approximation to the curve y = f(x).
nxandbxa 0
121 ,...,, nxxx
nPPP ,...,, 10
Arc Length
• The length Lk of the kth line segment in the polygonal path is
21
222 )]()([)()()( kkkkkk xfxfxyxL
Arc Length
• The length Lk of the kth line segment in the polygonal path is
• If we now add the lengths of these line segments, we obtain the following approximation to the length L of the curve.
21
222 )]()([)()()( kkkkkk xfxfxyxL
21
2
11
)]()([)(
kkk
n
k
n
kk xfxfxLL
Arc Length
• To put this in the form of a Riemann Sum we will apply the Mean-Value Theorem. This Theorem implies that there is a point between and such that
or
1kx*kx kx
)()()( */
1
1k
kk
kk xfxx
xfxf
)()()( */1
kk
kk xfx
xfxf
kkkk xxfxfxf )()()( */1
Arc Length
• To put this in the form of a Riemann Sum we will apply the Mean-Value Theorem. This Theorem implies that there is a point between and such that
or
1kx*kx kx
)()()( */
1
1k
kk
kk xfxx
xfxf
21
2
11
)]()([)(
kkk
n
k
n
kk xfxfxLL
)()()( */1
kk
kk xfx
xfxf
kkkk xxfxfxf )()()( */1
2*/2
11
])([)( kkk
n
k
n
kk xxfxLL
Arc Length
• To put this in the form of a Riemann Sum we will apply the Mean-Value Theorem. This Theorem implies that there is a point between and such that
or
1kx*kx kx
)()()( */
1
1k
kk
kk xfxx
xfxf
kk
n
k
xxfL
2*/
1
)]([1
)()()( */1
kk
kk xfx
xfxf
2*/2
11
])([)( kkk
n
k
n
kk xxfxLL
Arc Length
• Thus, taking the limit as n increases and the widths of the subintervals approximate zero yields the following integral that defines the arc length L:
dxxfxxfLb
a
kk
n
kx
2/2*/
10max
)]([1)]([1lim
Definition
• 7.4.2 If y = f(x) is a smooth curve on the interval [a, b], then the arc length of L of this curve over [a, b] is defined as
dxxfLb
a 2/ )]([1
Definition
• 7.4.2 If x = g(y) is a smooth curve on the interval [c, d], then the arc length of L of this curve over [c, d] is defined as
dyygLd
c 2/ )]([1
Example 1
• Find the arc length of the curve from (1, 1) to (2, ) using both formulas. 22
2/3xy
Example 1
• Find the arc length of the curve from (1, 1) to (2, ) using both formulas. 22
2/3xy
2/1/
2
3)( xxf
09.22
31
2
1
22/1
dxxL
Example 1
• Find the arc length of the curve from (1, 1) to (2, ) using both formulas. 22
2/3xy
3/1/
3
2)( yyg
09.23
21
22
1
23/1
dyyL
3/2yx
Homework
• Page 469
• 3, 5
Average Value of a Function
• We will use the idea of average/mean and extend the concept so that we can compute not only the arithmetic average of finitely many function values but an average of all values of f(x) as x varies over a closed interval [a, b].
Average Value of a Function
• We will use the Mean-Value Theorem for Integrals, which states that if f is continuous on the interval [a, b], then there is at least one point in this interval such that
*x
))(()( * abxfdxxfb
a
Average Value of a Function
• We will use the Mean-Value Theorem for Integrals, which states that if f is continuous on the interval [a, b], then there is at least one point in this interval such that
• We will look at the equation in this form as our candidate for the average value of f over the interval [a, b].
*x
))(()( * abxfdxxfb
a
ave
b
a
fdxxfab
)(1
Average Value of a Function
• To explain what motivates this idea, divide the interval [a, b] into n subintervals of equal length
and choosing arbitrary points in successive subintervals. Then the arithmetic average of the values is
n
abx
**2
*1 ,...,, nxxx
)(),...,(),( **2
*1 nxfxfxf
)](...)()([1 **
2*1 nxfxfxf
nave
Average Value of a Function
• Using the fact that • we can say that
• and substitute to get this equation:
n
abx
n
kkn xxf
abxxfxxfxxf
abave
1
***2
*1 )(
1])(...)()([
1
nab
x 1
)](...)()([1 **
2*1 nxfxfxf
nave
Average Value of a Function
• Taking the limit as yields
b
a
n
kk
ndxxf
abxxf
ab)(
1)(
1lim
1
*
n
Average Value of a Function
• Definition 7.6.1 If f is continuous on [a, b], then the average value (or mean value) of f on [a, b] is defined to be
b
a
ave dxxfab
f )(1
Average Value of a Function
• If we look at the Mean-Value Theorem for Integrals together with the equation for average value, we can see the relationship.
b
a
ave dxxfab
f )(1
b
a
dxxfabxf )())(( *
Average Value of a Function
• The Mean-Value Theorem for Integrals guarantees the point where the rectangle has the right height. The average value is the height.
b
a
dxxfabxf )())(( *
b
a
ave dxxfab
f )(1
*x
Example 1
• Find the average value of the function over the interval [1, 4], and find all points in the interval at which the value of f is the same as the average.
xxf )(
Example 1
• Find the average value of the function over the interval [1, 4], and find all points in the interval at which the value of f is the same as the average.
• The average value of the functions is:
9
14
3
2
3
1
14
1)(
14
1
2/34
1
xdxxdxxfab
fb
a
ave
xxf )(
Example 1
• Find the average value of the function over the interval [1, 4], and find all points in the interval at which the value of f is the same as the average.
• The second question is the Mean-Value Theorem for Integrals.
81
1969
14)(
*
**
x
xxf
xxf )(
Average Velocity Revisited
• When we first looked at Rectilinear Motion, we defined the average velocity of the particle over a time interval to be its displacement over the time interval divided by the time elapsed. Thus, if the particle has position s(t), then its average velocity over a time interval [t0 , t1 ] is
01
01 )()(
tt
tstsvave
avev
Average Velocity Revisited
• However, the displacement is the integral of velocity over the given time interval. We can now look at average velocity as:
01
01 )()(
tt
tstsvave
1
0
)(1
01
t
t
ave dttvtt
v
Homework
• Pages 479-480
• 1-9 odd
• Section 5.8
• Pages 388-389
• 1-11 odd