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a b | f ' - p ' | = i = 1 n x i - 1 x i | f ' - p ' | = a x 1 | f ' | + i = 2 n - 1 | f ' - p ' | + x n - 1 b | f ' | a a + δ ' | f ' | + b - δ ' b | f ' | + ϵ 4 ( b - a ) x 1 x n - 1 1 < ϵ .

The proof is now complete.

REMARK It should be evident that the preceding theorem can easily be generalized to a piecewise smooth function f , i.e., a function that is continuous on [ a , b ] , continuously differentiable on each subinterval ( t i - 1 , t i ) of a partition { t 0 < t 1 < ... < t n } , and whose derivative f ' is absolutely integrable on ( a , b ) . Indeed, just apply the theorem to each of the subintervals ( t i - 1 , t i ) , and then carefully piece together the piecewise linear functions on those subintervals.

Now we are ready to define what a smooth curve is.

By a smooth curve from a point z 1 to a different point z 2 in the plane, we mean a set C C that is the range of a 1-1, smooth, function φ : [ a , b ] C , where [ a , b ] is a bounded closed interval in R , where z 1 = φ ( a ) and z 2 = φ ( b ) , and satisfying φ ' ( t ) 0 for all t ( a , b ) .

More generally, if φ : [ a , b ] R 2 is 1-1 and piecewise smooth on [ a , b ] , and if { t 0 < t 1 < ... < t n } is a partition of [ a , b ] such that φ ' ( t ) 0 for all t ( t i - 1 , t i ) , then the range C of φ is called a piecewise smooth curve from z 1 = φ ( a ) to z 2 = φ ( b ) .

In either of these cases, φ is called a parameterization of the curve C .

Note that we do not assume that | φ ' | is improperly-integrable, though the preceding theorem might have made you think we would.

REMARK Throughout this chapter we will be continually faced with the fact that a given curve can have many different parameterizations.Indeed, if φ 1 : [ a , b ] C is a parameterization, and if g : [ c , d ] [ a , b ] is a smooth function having a nonzero derivative, then φ 2 ( s ) = φ 1 ( g ( s ) ) is another parameterization of C . Since our definitions and proofs about curves often involve a parametrization, we will frequently need to prove that the results we obtain are independent of the parameterization.The next theorem will help; it shows that any two parameterizations of C are connected exactly as above, i.e., there always is such a function g relating φ 1 and φ 2 .

Let φ 1 : [ a , b ] C and φ 2 : [ c , d ] C be two parameterizations of a piecewise smooth curve C joining z 1 to z 2 . Then there exists a piecewise smooth function g : [ c , d ] [ a , b ] such that φ 2 ( s ) = φ 1 ( g ( s ) ) for all s [ c , d ] . Moreover, the derivative g ' of g is nonzero for all but a finite number of points in [ c , d ] .

Because both φ 1 and φ 2 are continuous and 1-1, it follows from [link] that the function g = φ 1 - 1 φ 2 is continuous and 1-1 from [ c , d ] onto [ a , b ] . Moreover, from [link] , it must also be that g is strictly increasing or strictly decreasing. Write φ 1 ( t ) = u 1 ( t ) + i v 1 ( t ) ( u 1 ( t ) , v 1 ( t ) ) , and φ 2 ( s ) = u 2 ( s ) + i v 2 ( s ) ( u 2 ( s ) , v 2 ( s ) ) . Let { x 0 < x 1 < ... < x p } be a partition of [ a , b ] for which φ 1 ' is continuous and nonzeroon the subintervals ( x j - 1 , x j ) , and let { y 0 < y 1 < ... < y q } be a partition of [ c , d ] for which φ 2 ' is continuous and nonzero on the subintervals ( y k - 1 , y k ) . Then let { s 0 < s 1 < ... < s n } be the partition of [ c , d ] determined by the finitely many points { y k } { g - 1 ( x j ) } . We will show that g is continuously differentiable at each point s in the subintervals ( s i - 1 , s i ) .

Fix an s in one of the intervals ( s i - 1 , s i ) , and let t = φ 1 - 1 ( φ 2 ( s ) ) = g ( s ) . Of course this means that φ 1 ( t ) = φ 2 ( s ) , or u 1 ( t ) = u 2 ( s ) and v 1 ( t ) = v 2 ( s ) . Then t is in some one of the intervals ( x j - 1 , x j ) , so that we know that φ 1 ' ( t ) 0 . Therefore, we must have that at least one of u 1 ' ( t ) or v 1 ' ( t ) is nonzero. Suppose it is v 1 ' ( t ) that is nonzero. The argument, in case it is u 1 ' ( t ) that is nonzero, is completely analogous. Now, because v 1 ' is continuous at t and v 1 ' ( t ) 0 , it follows that v 1 is strictly monotonic in some neighborhood ( t - δ , t + δ ) of t and therefore is 1-1 on that interval. Then v 1 - 1 is continuous by [link] , and is differentiable at the point v 1 ( t ) by the Inverse Function Theorem. We will show that on this small interval g = v 1 - 1 v 2 , and this will prove that g is continuously differentiable at s .

Questions & Answers

how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
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Sherica
im all ears I need to learn
Sherica
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Tamia
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Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
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Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
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China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
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Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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