Mixing
How to prove the arc length of smooth curve converges to the straight length?
$begingroup$
Suppose there is a smooth curve $L(t)=begin{cases}x(t)\y(t)end{cases}, tin[a,b]subset R$ with continues first derivative.
To find the length of $L$ we separate $L$ into several small part $l_i$ .
Define the arc length of $l_i$ is $s_i$ and the length of the straight line connecting the 2 end points of $l_i$ is $d_i$ .
Now my question is how to prove that $s_i=d_i+o(d_i)$ where $lim_{d_ito0}o(d_i)=0$ .
In other words, $d_i$ converges to $s_i$ when it becomes infinitely small. Which is the base that allows me to use integration to find the length.
Furthermore, does it has to be smooth curve? What about other non-normal curves (like those everywhere continuous, nowhere derivable functions)?
Thanks a lot!
real-analysis calculus integration limits analysis
asked Jan 14 at 1:37
ShoreShore
183
$endgroup$
add a comment |
$begingroup$
Suppose there is a smooth curve $L(t)=begin{cases}x(t)\y(t)end{cases}, tin[a,b]subset R$ with continues first derivative.
To find the length of $L$ we separate $L$ into several small part $l_i$ .
Define the arc length of $l_i$ is $s_i$ and the length of the straight line connecting the 2 end points of $l_i$ is $d_i$ .
Now my question is how to prove that $s_i=d_i+o(d_i)$ where $lim_{d_ito0}o(d_i)=0$ .
In other words, $d_i$ converges to $s_i$ when it becomes infinitely small. Which is the base that allows me to use integration to find the length.
Furthermore, does it has to be smooth curve? What about other non-normal curves (like those everywhere continuous, nowhere derivable functions)?
Thanks a lot!
real-analysis calculus integration limits analysis
asked Jan 14 at 1:37
ShoreShore
183
$endgroup$
$begingroup$
We usually express that as $s_i/d_ito 1$ as $d_ito 0$.
$endgroup$
– Paramanand Singh
Jan 14 at 3:53
add a comment |
$begingroup$
Suppose there is a smooth curve $L(t)=begin{cases}x(t)\y(t)end{cases}, tin[a,b]subset R$ with continues first derivative.
To find the length of $L$ we separate $L$ into several small part $l_i$ .
Define the arc length of $l_i$ is $s_i$ and the length of the straight line connecting the 2 end points of $l_i$ is $d_i$ .
Now my question is how to prove that $s_i=d_i+o(d_i)$ where $lim_{d_ito0}o(d_i)=0$ .
In other words, $d_i$ converges to $s_i$ when it becomes infinitely small. Which is the base that allows me to use integration to find the length.
Furthermore, does it has to be smooth curve? What about other non-normal curves (like those everywhere continuous, nowhere derivable functions)?
Thanks a lot!
real-analysis calculus integration limits analysis
asked Jan 14 at 1:37
ShoreShore
183
$endgroup$
Suppose there is a smooth curve $L(t)=begin{cases}x(t)\y(t)end{cases}, tin[a,b]subset R$ with continues first derivative.
To find the length of $L$ we separate $L$ into several small part $l_i$ .
Define the arc length of $l_i$ is $s_i$ and the length of the straight line connecting the 2 end points of $l_i$ is $d_i$ .
Now my question is how to prove that $s_i=d_i+o(d_i)$ where $lim_{d_ito0}o(d_i)=0$ .
In other words, $d_i$ converges to $s_i$ when it becomes infinitely small. Which is the base that allows me to use integration to find the length.
Furthermore, does it has to be smooth curve? What about other non-normal curves (like those everywhere continuous, nowhere derivable functions)?
Thanks a lot!
real-analysis calculus integration limits analysis
real-analysis calculus integration limits analysis
asked Jan 14 at 1:37
ShoreShore
183
asked Jan 14 at 1:37
ShoreShore
183
asked Jan 14 at 1:37
ShoreShore
183
asked Jan 14 at 1:37
ShoreShore
183
asked Jan 14 at 1:37
ShoreShore
183
183
$begingroup$
We usually express that as $s_i/d_ito 1$ as $d_ito 0$.
$endgroup$
– Paramanand Singh
Jan 14 at 3:53
add a comment |
$begingroup$
We usually express that as $s_i/d_ito 1$ as $d_ito 0$.
$endgroup$
– Paramanand Singh
Jan 14 at 3:53
$begingroup$
We usually express that as $s_i/d_ito 1$ as $d_ito 0$.
$endgroup$
– Paramanand Singh
Jan 14 at 3:53
$begingroup$
We usually express that as $s_i/d_ito 1$ as $d_ito 0$.
$endgroup$
– Paramanand Singh
Jan 14 at 3:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is best to setup the coordinate system in such a manner that the part of curve with length $l_i$ starts from origin and then we have $$l_i=int_{0}^{h}sqrt{{x'(t)}^2+{y'(t)}^2},dt$$ and $$d_i=sqrt{x^2(h)+y^2(h)}$$ and we need to show that $l_i/d_ito 1$ as $hto 0$. If we assume that one of the derivatives $x'(0),y'(0)$ is non-zero then we can see via Fundamental Theorem of Calculus that $$frac{l_i}{h}tosqrt{{x'(0)}^2+{y'(0)}^2}$$ and by definition of derivative $d_i/h$ also tends to the same value and thus our job is done.
The definition of a curve assumes that $x(t), y(t) $ are continuous and further the notion of arc-length requires that these functions be of bounded variation. The corresponding analysis under such general conditions is more difficult.
Based on feedback received in comments let me just mention the development of arc-length and its representation as an integral in brief.
We first begin with the notion of a partition. Let $[a, b] $ be a closed interval. A partition of $[a, b] $ is a finite set $$P={t_0,t_1,t_2,dots,t_n} $$ such that $$a=t_0<t_1<t_2<dots<t_n=b$$ Let the set of all possible partitions of $[a, b] $ be denoted by $mathcal {P} [a, b] $ ie $$mathcal{P} [a, b] ={Pmid Ptext{ is a partition of }[a, b] } $$
Let $f:[a, b] tomathbb {R} $ be a function. Let $$P={t_0,t_1,t_2,dots,t_n}$$ be a partition of $[a, b] $ and we form a sum $$V_{f} (P) =sum_{i=1}^{n}|f(t_i)-f(t_{i-1})|$$ The function $f$ is said to be of bounded variation if the set $${V_{f} (P) mid Pinmathcal{P} [a, b] } $$ of sums $V_{f} (P) $ for all partitions $P$ of $[a, b] $ is bounded. And in this case the total variation of $f$ on $[a, b] $ is defined to be the supremum $$V_{f} [a, b] =sup, {V_{f} (P) mid Pinmathcal{P} [a, b] } $$
It is easy to prove that any monotone function is of bounded variation and a bit more difficult to prove that a function is of bounded variation if and only if it can be expressed as a difference of two increasing functions.
Next we come to the topic of interest. Let $f, g$ be two functions from interval $[a, b] $ to $mathbb {R} $ and let's assume that they are continuous on $[a, b] $. A curve is a set of points $$mathcal{C} ={(x, y) mid x=f(t), y=g(t), tin[a, b] } $$ To define the arc-length of this curve $mathcal{C} $ we start with a partition $$P={t_0,t_1,t_2,dots,t_n} $$ of $[a, b] $ and form the sum $$L_{mathcal{C}} (P) =sum_{i=1}^{n}sqrt {{f(t_i)-f(t_{i-1})}^2+{g(t_i)-g(t_{i-1})}^2} $$ In other words corresponding to the partition $P$ of $[a, b] $ we have points $A_0,A_1,A_2dots,A_n$ on the curve $mathcal{C} $ with $A_i=(f(t_i), g(t_i)) $ and the above expression for $L_{mathcal{C}} (P) $ is the sum of lengths of these line segments $A_{i-1}A_{i}$.
If the set $${L_{mathcal{C}} (P) mid Pinmathcal{P} [a, b] } $$ of sums $L_{mathcal{C}} (P) $ is bounded for all partitions $P$ of $[a, b] $ then we say that the curve $mathcal{C} $ is rectifiable (ie possesses a well defined arc-length) and its arc-length $L_{mathcal{C}} $ is defined to be the supremum of all such sums ie $$L_{mathcal{C}} =sup, {L_{mathcal{C}} (P) mid Pinmathcal{P} [a, b] } $$ It can be proved with some effort that the curve $mathcal{C} $ is rectifiable if and only if both $f, g$ are of bounded variation on $[a, b] $.
Let us now assume that the functions $f, g$ used to define the curve $mathcal{C} $ are differentiable and further $f'(t) neq 0$ for all $tin[a, b] $. Then $f$ is one-one (by Rolle's theorem if $f$ takes same value at two points then its derivative vanishes somewhere in between) and for any partition $P={t_0,t_1,dots,t_n} $ of $[a, b] $ we have $f(t_i) - f(t_{i-1}) neq 0$. Hence we can write $$L_{mathcal{C}} (P) =sum_{i=1}^{n}|f(t_i)-f(t_{i-1})|sqrt{1+left(frac{g(t_i)-g(t_{i-1})}{f(t_i)-f(t_{i-1})}right)^2}$$ Using Cauchy mean value theorem the ratio inside square roots can be written as $g'(xi_i) /f'(xi_i) $ for some $xi_iin(t_{i-1}, t_i) $. Also using mean value theorem the expression outside square roots can be written as $|f'(eta_i)|(t_i-t_{i-1})$ for some $eta_iin(t_{i-1},t_i)$ and thus we have $$L_{mathcal{C}} (P) =sum_{i=1}^{n}|f'(eta_i)|sqrt{1+left(frac{g'(xi_i)}{f'(xi_i)}right)^2}(t_i-t_{i-1})$$ and the above looks like a Riemann sum for the integral $$int_{a} ^{b} |f'(t) |sqrt{1+left(frac{g'(t)}{f'(t)}right)^2},dt=int_{a}^{b}sqrt{{f'(t)}^2+{g'(t)}^2},dt$$ (there is a slight technicality involved here due to the different set of points $eta_i, xi_i $). As partitions $P$ become finer and finer the expression $L_{mathcal{C}} (P) $ tends to its supremum $L_{mathcal{C}} $ and the Riemann sums above tend to the integral above and thus we get the arc-length formula $$L_{mathcal{C}} =int_{a}^{b}sqrt{{f'(t)}^2+{g'(t)}^2},dt$$ This assumes that the integral on right exists.
In the above development we have nowhere use the fact $s_i/d_ito 1$. The integral formula for arc-length is a consequence of definition of arc-length and mean value theorems. And the result in your question is a consequence of this integral formula.
The restriction $f'(t) neq 0$ can be removed by using Langrange Mean Value Theorem on $f, g$ in the sum $L_{mathcal {C}} (P) $ and Duhamel Principle for integrals.
answered Jan 14 at 4:09
Paramanand SinghParamanand Singh
50.1k556163
$endgroup$
$begingroup$
Hi thanks very much, can I ask 2 more questions. Firstly could you be more specific on proving $frac{l_i}{h}tosqrt{{x'(0)}^2+{y'(0)}^2}$ ? Secondly, as far as I know the function $l_i=int_0^hsqrt{{x'(t)}^2+{y'(t)}^2}dt$ is proved from the results of $frac{l_i}{d_1}to1$ .......
$endgroup$
– Shore
Jan 14 at 6:20
$begingroup$
@Shore: the limit of $l_i/h$ is obtained by a direct use of fundamental theorem of calculus (the derivative of integral is the function itself) and you need to observe that limit of $l_i/h$ is the derivative of the integral here at point $h=0$.
$endgroup$
– Paramanand Singh
Jan 14 at 8:12
$begingroup$
@Shore : the fact that arc length can be represented by the integral depends on Cauchy mean value theorem and not on the limit you seek. I will update this part in my answer.
$endgroup$
– Paramanand Singh
Jan 14 at 8:14
$begingroup$
Thanks, my calculus text book proved it based on the infinity small straight line is the length of the curve. But this need to prove that on a small scale, the straight line is close enough to the arc length.
$endgroup$
– Shore
Jan 14 at 9:04
$begingroup$
Hi Paramanand Singh, I've read some references about mean value theorem, It seems still using the fact that arc length can be represented by connecting line length with infinite small intervals........
$endgroup$
– Shore
Jan 15 at 0:41
|
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$begingroup$
It is best to setup the coordinate system in such a manner that the part of curve with length $l_i$ starts from origin and then we have $$l_i=int_{0}^{h}sqrt{{x'(t)}^2+{y'(t)}^2},dt$$ and $$d_i=sqrt{x^2(h)+y^2(h)}$$ and we need to show that $l_i/d_ito 1$ as $hto 0$. If we assume that one of the derivatives $x'(0),y'(0)$ is non-zero then we can see via Fundamental Theorem of Calculus that $$frac{l_i}{h}tosqrt{{x'(0)}^2+{y'(0)}^2}$$ and by definition of derivative $d_i/h$ also tends to the same value and thus our job is done.
The definition of a curve assumes that $x(t), y(t) $ are continuous and further the notion of arc-length requires that these functions be of bounded variation. The corresponding analysis under such general conditions is more difficult.
Based on feedback received in comments let me just mention the development of arc-length and its representation as an integral in brief.
We first begin with the notion of a partition. Let $[a, b] $ be a closed interval. A partition of $[a, b] $ is a finite set $$P={t_0,t_1,t_2,dots,t_n} $$ such that $$a=t_0<t_1<t_2<dots<t_n=b$$ Let the set of all possible partitions of $[a, b] $ be denoted by $mathcal {P} [a, b] $ ie $$mathcal{P} [a, b] ={Pmid Ptext{ is a partition of }[a, b] } $$
Let $f:[a, b] tomathbb {R} $ be a function. Let $$P={t_0,t_1,t_2,dots,t_n}$$ be a partition of $[a, b] $ and we form a sum $$V_{f} (P) =sum_{i=1}^{n}|f(t_i)-f(t_{i-1})|$$ The function $f$ is said to be of bounded variation if the set $${V_{f} (P) mid Pinmathcal{P} [a, b] } $$ of sums $V_{f} (P) $ for all partitions $P$ of $[a, b] $ is bounded. And in this case the total variation of $f$ on $[a, b] $ is defined to be the supremum $$V_{f} [a, b] =sup, {V_{f} (P) mid Pinmathcal{P} [a, b] } $$
It is easy to prove that any monotone function is of bounded variation and a bit more difficult to prove that a function is of bounded variation if and only if it can be expressed as a difference of two increasing functions.
Next we come to the topic of interest. Let $f, g$ be two functions from interval $[a, b] $ to $mathbb {R} $ and let's assume that they are continuous on $[a, b] $. A curve is a set of points $$mathcal{C} ={(x, y) mid x=f(t), y=g(t), tin[a, b] } $$ To define the arc-length of this curve $mathcal{C} $ we start with a partition $$P={t_0,t_1,t_2,dots,t_n} $$ of $[a, b] $ and form the sum $$L_{mathcal{C}} (P) =sum_{i=1}^{n}sqrt {{f(t_i)-f(t_{i-1})}^2+{g(t_i)-g(t_{i-1})}^2} $$ In other words corresponding to the partition $P$ of $[a, b] $ we have points $A_0,A_1,A_2dots,A_n$ on the curve $mathcal{C} $ with $A_i=(f(t_i), g(t_i)) $ and the above expression for $L_{mathcal{C}} (P) $ is the sum of lengths of these line segments $A_{i-1}A_{i}$.
If the set $${L_{mathcal{C}} (P) mid Pinmathcal{P} [a, b] } $$ of sums $L_{mathcal{C}} (P) $ is bounded for all partitions $P$ of $[a, b] $ then we say that the curve $mathcal{C} $ is rectifiable (ie possesses a well defined arc-length) and its arc-length $L_{mathcal{C}} $ is defined to be the supremum of all such sums ie $$L_{mathcal{C}} =sup, {L_{mathcal{C}} (P) mid Pinmathcal{P} [a, b] } $$ It can be proved with some effort that the curve $mathcal{C} $ is rectifiable if and only if both $f, g$ are of bounded variation on $[a, b] $.
Let us now assume that the functions $f, g$ used to define the curve $mathcal{C} $ are differentiable and further $f'(t) neq 0$ for all $tin[a, b] $. Then $f$ is one-one (by Rolle's theorem if $f$ takes same value at two points then its derivative vanishes somewhere in between) and for any partition $P={t_0,t_1,dots,t_n} $ of $[a, b] $ we have $f(t_i) - f(t_{i-1}) neq 0$. Hence we can write $$L_{mathcal{C}} (P) =sum_{i=1}^{n}|f(t_i)-f(t_{i-1})|sqrt{1+left(frac{g(t_i)-g(t_{i-1})}{f(t_i)-f(t_{i-1})}right)^2}$$ Using Cauchy mean value theorem the ratio inside square roots can be written as $g'(xi_i) /f'(xi_i) $ for some $xi_iin(t_{i-1}, t_i) $. Also using mean value theorem the expression outside square roots can be written as $|f'(eta_i)|(t_i-t_{i-1})$ for some $eta_iin(t_{i-1},t_i)$ and thus we have $$L_{mathcal{C}} (P) =sum_{i=1}^{n}|f'(eta_i)|sqrt{1+left(frac{g'(xi_i)}{f'(xi_i)}right)^2}(t_i-t_{i-1})$$ and the above looks like a Riemann sum for the integral $$int_{a} ^{b} |f'(t) |sqrt{1+left(frac{g'(t)}{f'(t)}right)^2},dt=int_{a}^{b}sqrt{{f'(t)}^2+{g'(t)}^2},dt$$ (there is a slight technicality involved here due to the different set of points $eta_i, xi_i $). As partitions $P$ become finer and finer the expression $L_{mathcal{C}} (P) $ tends to its supremum $L_{mathcal{C}} $ and the Riemann sums above tend to the integral above and thus we get the arc-length formula $$L_{mathcal{C}} =int_{a}^{b}sqrt{{f'(t)}^2+{g'(t)}^2},dt$$ This assumes that the integral on right exists.
In the above development we have nowhere use the fact $s_i/d_ito 1$. The integral formula for arc-length is a consequence of definition of arc-length and mean value theorems. And the result in your question is a consequence of this integral formula.
The restriction $f'(t) neq 0$ can be removed by using Langrange Mean Value Theorem on $f, g$ in the sum $L_{mathcal {C}} (P) $ and Duhamel Principle for integrals.
answered Jan 14 at 4:09
Paramanand SinghParamanand Singh
50.1k556163
$endgroup$
$begingroup$
Hi thanks very much, can I ask 2 more questions. Firstly could you be more specific on proving $frac{l_i}{h}tosqrt{{x'(0)}^2+{y'(0)}^2}$ ? Secondly, as far as I know the function $l_i=int_0^hsqrt{{x'(t)}^2+{y'(t)}^2}dt$ is proved from the results of $frac{l_i}{d_1}to1$ .......
$endgroup$
– Shore
Jan 14 at 6:20
$begingroup$
@Shore: the limit of $l_i/h$ is obtained by a direct use of fundamental theorem of calculus (the derivative of integral is the function itself) and you need to observe that limit of $l_i/h$ is the derivative of the integral here at point $h=0$.
$endgroup$
– Paramanand Singh
Jan 14 at 8:12
$begingroup$
@Shore : the fact that arc length can be represented by the integral depends on Cauchy mean value theorem and not on the limit you seek. I will update this part in my answer.
$endgroup$
– Paramanand Singh
Jan 14 at 8:14
$begingroup$
Thanks, my calculus text book proved it based on the infinity small straight line is the length of the curve. But this need to prove that on a small scale, the straight line is close enough to the arc length.
$endgroup$
– Shore
Jan 14 at 9:04
$begingroup$
Hi Paramanand Singh, I've read some references about mean value theorem, It seems still using the fact that arc length can be represented by connecting line length with infinite small intervals........
$endgroup$
– Shore
Jan 15 at 0:41
|
show 4 more comments
$begingroup$
It is best to setup the coordinate system in such a manner that the part of curve with length $l_i$ starts from origin and then we have $$l_i=int_{0}^{h}sqrt{{x'(t)}^2+{y'(t)}^2},dt$$ and $$d_i=sqrt{x^2(h)+y^2(h)}$$ and we need to show that $l_i/d_ito 1$ as $hto 0$. If we assume that one of the derivatives $x'(0),y'(0)$ is non-zero then we can see via Fundamental Theorem of Calculus that $$frac{l_i}{h}tosqrt{{x'(0)}^2+{y'(0)}^2}$$ and by definition of derivative $d_i/h$ also tends to the same value and thus our job is done.
The definition of a curve assumes that $x(t), y(t) $ are continuous and further the notion of arc-length requires that these functions be of bounded variation. The corresponding analysis under such general conditions is more difficult.
Based on feedback received in comments let me just mention the development of arc-length and its representation as an integral in brief.
We first begin with the notion of a partition. Let $[a, b] $ be a closed interval. A partition of $[a, b] $ is a finite set $$P={t_0,t_1,t_2,dots,t_n} $$ such that $$a=t_0<t_1<t_2<dots<t_n=b$$ Let the set of all possible partitions of $[a, b] $ be denoted by $mathcal {P} [a, b] $ ie $$mathcal{P} [a, b] ={Pmid Ptext{ is a partition of }[a, b] } $$
Let $f:[a, b] tomathbb {R} $ be a function. Let $$P={t_0,t_1,t_2,dots,t_n}$$ be a partition of $[a, b] $ and we form a sum $$V_{f} (P) =sum_{i=1}^{n}|f(t_i)-f(t_{i-1})|$$ The function $f$ is said to be of bounded variation if the set $${V_{f} (P) mid Pinmathcal{P} [a, b] } $$ of sums $V_{f} (P) $ for all partitions $P$ of $[a, b] $ is bounded. And in this case the total variation of $f$ on $[a, b] $ is defined to be the supremum $$V_{f} [a, b] =sup, {V_{f} (P) mid Pinmathcal{P} [a, b] } $$
It is easy to prove that any monotone function is of bounded variation and a bit more difficult to prove that a function is of bounded variation if and only if it can be expressed as a difference of two increasing functions.
Next we come to the topic of interest. Let $f, g$ be two functions from interval $[a, b] $ to $mathbb {R} $ and let's assume that they are continuous on $[a, b] $. A curve is a set of points $$mathcal{C} ={(x, y) mid x=f(t), y=g(t), tin[a, b] } $$ To define the arc-length of this curve $mathcal{C} $ we start with a partition $$P={t_0,t_1,t_2,dots,t_n} $$ of $[a, b] $ and form the sum $$L_{mathcal{C}} (P) =sum_{i=1}^{n}sqrt {{f(t_i)-f(t_{i-1})}^2+{g(t_i)-g(t_{i-1})}^2} $$ In other words corresponding to the partition $P$ of $[a, b] $ we have points $A_0,A_1,A_2dots,A_n$ on the curve $mathcal{C} $ with $A_i=(f(t_i), g(t_i)) $ and the above expression for $L_{mathcal{C}} (P) $ is the sum of lengths of these line segments $A_{i-1}A_{i}$.
If the set $${L_{mathcal{C}} (P) mid Pinmathcal{P} [a, b] } $$ of sums $L_{mathcal{C}} (P) $ is bounded for all partitions $P$ of $[a, b] $ then we say that the curve $mathcal{C} $ is rectifiable (ie possesses a well defined arc-length) and its arc-length $L_{mathcal{C}} $ is defined to be the supremum of all such sums ie $$L_{mathcal{C}} =sup, {L_{mathcal{C}} (P) mid Pinmathcal{P} [a, b] } $$ It can be proved with some effort that the curve $mathcal{C} $ is rectifiable if and only if both $f, g$ are of bounded variation on $[a, b] $.
Let us now assume that the functions $f, g$ used to define the curve $mathcal{C} $ are differentiable and further $f'(t) neq 0$ for all $tin[a, b] $. Then $f$ is one-one (by Rolle's theorem if $f$ takes same value at two points then its derivative vanishes somewhere in between) and for any partition $P={t_0,t_1,dots,t_n} $ of $[a, b] $ we have $f(t_i) - f(t_{i-1}) neq 0$. Hence we can write $$L_{mathcal{C}} (P) =sum_{i=1}^{n}|f(t_i)-f(t_{i-1})|sqrt{1+left(frac{g(t_i)-g(t_{i-1})}{f(t_i)-f(t_{i-1})}right)^2}$$ Using Cauchy mean value theorem the ratio inside square roots can be written as $g'(xi_i) /f'(xi_i) $ for some $xi_iin(t_{i-1}, t_i) $. Also using mean value theorem the expression outside square roots can be written as $|f'(eta_i)|(t_i-t_{i-1})$ for some $eta_iin(t_{i-1},t_i)$ and thus we have $$L_{mathcal{C}} (P) =sum_{i=1}^{n}|f'(eta_i)|sqrt{1+left(frac{g'(xi_i)}{f'(xi_i)}right)^2}(t_i-t_{i-1})$$ and the above looks like a Riemann sum for the integral $$int_{a} ^{b} |f'(t) |sqrt{1+left(frac{g'(t)}{f'(t)}right)^2},dt=int_{a}^{b}sqrt{{f'(t)}^2+{g'(t)}^2},dt$$ (there is a slight technicality involved here due to the different set of points $eta_i, xi_i $). As partitions $P$ become finer and finer the expression $L_{mathcal{C}} (P) $ tends to its supremum $L_{mathcal{C}} $ and the Riemann sums above tend to the integral above and thus we get the arc-length formula $$L_{mathcal{C}} =int_{a}^{b}sqrt{{f'(t)}^2+{g'(t)}^2},dt$$ This assumes that the integral on right exists.
In the above development we have nowhere use the fact $s_i/d_ito 1$. The integral formula for arc-length is a consequence of definition of arc-length and mean value theorems. And the result in your question is a consequence of this integral formula.
The restriction $f'(t) neq 0$ can be removed by using Langrange Mean Value Theorem on $f, g$ in the sum $L_{mathcal {C}} (P) $ and Duhamel Principle for integrals.
answered Jan 14 at 4:09
Paramanand SinghParamanand Singh
50.1k556163
$endgroup$
$begingroup$
Hi thanks very much, can I ask 2 more questions. Firstly could you be more specific on proving $frac{l_i}{h}tosqrt{{x'(0)}^2+{y'(0)}^2}$ ? Secondly, as far as I know the function $l_i=int_0^hsqrt{{x'(t)}^2+{y'(t)}^2}dt$ is proved from the results of $frac{l_i}{d_1}to1$ .......
$endgroup$
– Shore
Jan 14 at 6:20
$begingroup$
@Shore: the limit of $l_i/h$ is obtained by a direct use of fundamental theorem of calculus (the derivative of integral is the function itself) and you need to observe that limit of $l_i/h$ is the derivative of the integral here at point $h=0$.
$endgroup$
– Paramanand Singh
Jan 14 at 8:12
$begingroup$
@Shore : the fact that arc length can be represented by the integral depends on Cauchy mean value theorem and not on the limit you seek. I will update this part in my answer.
$endgroup$
– Paramanand Singh
Jan 14 at 8:14
$begingroup$
Thanks, my calculus text book proved it based on the infinity small straight line is the length of the curve. But this need to prove that on a small scale, the straight line is close enough to the arc length.
$endgroup$
– Shore
Jan 14 at 9:04
$begingroup$
Hi Paramanand Singh, I've read some references about mean value theorem, It seems still using the fact that arc length can be represented by connecting line length with infinite small intervals........
$endgroup$
– Shore
Jan 15 at 0:41
|
show 4 more comments
$begingroup$
It is best to setup the coordinate system in such a manner that the part of curve with length $l_i$ starts from origin and then we have $$l_i=int_{0}^{h}sqrt{{x'(t)}^2+{y'(t)}^2},dt$$ and $$d_i=sqrt{x^2(h)+y^2(h)}$$ and we need to show that $l_i/d_ito 1$ as $hto 0$. If we assume that one of the derivatives $x'(0),y'(0)$ is non-zero then we can see via Fundamental Theorem of Calculus that $$frac{l_i}{h}tosqrt{{x'(0)}^2+{y'(0)}^2}$$ and by definition of derivative $d_i/h$ also tends to the same value and thus our job is done.
The definition of a curve assumes that $x(t), y(t) $ are continuous and further the notion of arc-length requires that these functions be of bounded variation. The corresponding analysis under such general conditions is more difficult.
Based on feedback received in comments let me just mention the development of arc-length and its representation as an integral in brief.
We first begin with the notion of a partition. Let $[a, b] $ be a closed interval. A partition of $[a, b] $ is a finite set $$P={t_0,t_1,t_2,dots,t_n} $$ such that $$a=t_0<t_1<t_2<dots<t_n=b$$ Let the set of all possible partitions of $[a, b] $ be denoted by $mathcal {P} [a, b] $ ie $$mathcal{P} [a, b] ={Pmid Ptext{ is a partition of }[a, b] } $$
Let $f:[a, b] tomathbb {R} $ be a function. Let $$P={t_0,t_1,t_2,dots,t_n}$$ be a partition of $[a, b] $ and we form a sum $$V_{f} (P) =sum_{i=1}^{n}|f(t_i)-f(t_{i-1})|$$ The function $f$ is said to be of bounded variation if the set $${V_{f} (P) mid Pinmathcal{P} [a, b] } $$ of sums $V_{f} (P) $ for all partitions $P$ of $[a, b] $ is bounded. And in this case the total variation of $f$ on $[a, b] $ is defined to be the supremum $$V_{f} [a, b] =sup, {V_{f} (P) mid Pinmathcal{P} [a, b] } $$
It is easy to prove that any monotone function is of bounded variation and a bit more difficult to prove that a function is of bounded variation if and only if it can be expressed as a difference of two increasing functions.
Next we come to the topic of interest. Let $f, g$ be two functions from interval $[a, b] $ to $mathbb {R} $ and let's assume that they are continuous on $[a, b] $. A curve is a set of points $$mathcal{C} ={(x, y) mid x=f(t), y=g(t), tin[a, b] } $$ To define the arc-length of this curve $mathcal{C} $ we start with a partition $$P={t_0,t_1,t_2,dots,t_n} $$ of $[a, b] $ and form the sum $$L_{mathcal{C}} (P) =sum_{i=1}^{n}sqrt {{f(t_i)-f(t_{i-1})}^2+{g(t_i)-g(t_{i-1})}^2} $$ In other words corresponding to the partition $P$ of $[a, b] $ we have points $A_0,A_1,A_2dots,A_n$ on the curve $mathcal{C} $ with $A_i=(f(t_i), g(t_i)) $ and the above expression for $L_{mathcal{C}} (P) $ is the sum of lengths of these line segments $A_{i-1}A_{i}$.
If the set $${L_{mathcal{C}} (P) mid Pinmathcal{P} [a, b] } $$ of sums $L_{mathcal{C}} (P) $ is bounded for all partitions $P$ of $[a, b] $ then we say that the curve $mathcal{C} $ is rectifiable (ie possesses a well defined arc-length) and its arc-length $L_{mathcal{C}} $ is defined to be the supremum of all such sums ie $$L_{mathcal{C}} =sup, {L_{mathcal{C}} (P) mid Pinmathcal{P} [a, b] } $$ It can be proved with some effort that the curve $mathcal{C} $ is rectifiable if and only if both $f, g$ are of bounded variation on $[a, b] $.
Let us now assume that the functions $f, g$ used to define the curve $mathcal{C} $ are differentiable and further $f'(t) neq 0$ for all $tin[a, b] $. Then $f$ is one-one (by Rolle's theorem if $f$ takes same value at two points then its derivative vanishes somewhere in between) and for any partition $P={t_0,t_1,dots,t_n} $ of $[a, b] $ we have $f(t_i) - f(t_{i-1}) neq 0$. Hence we can write $$L_{mathcal{C}} (P) =sum_{i=1}^{n}|f(t_i)-f(t_{i-1})|sqrt{1+left(frac{g(t_i)-g(t_{i-1})}{f(t_i)-f(t_{i-1})}right)^2}$$ Using Cauchy mean value theorem the ratio inside square roots can be written as $g'(xi_i) /f'(xi_i) $ for some $xi_iin(t_{i-1}, t_i) $. Also using mean value theorem the expression outside square roots can be written as $|f'(eta_i)|(t_i-t_{i-1})$ for some $eta_iin(t_{i-1},t_i)$ and thus we have $$L_{mathcal{C}} (P) =sum_{i=1}^{n}|f'(eta_i)|sqrt{1+left(frac{g'(xi_i)}{f'(xi_i)}right)^2}(t_i-t_{i-1})$$ and the above looks like a Riemann sum for the integral $$int_{a} ^{b} |f'(t) |sqrt{1+left(frac{g'(t)}{f'(t)}right)^2},dt=int_{a}^{b}sqrt{{f'(t)}^2+{g'(t)}^2},dt$$ (there is a slight technicality involved here due to the different set of points $eta_i, xi_i $). As partitions $P$ become finer and finer the expression $L_{mathcal{C}} (P) $ tends to its supremum $L_{mathcal{C}} $ and the Riemann sums above tend to the integral above and thus we get the arc-length formula $$L_{mathcal{C}} =int_{a}^{b}sqrt{{f'(t)}^2+{g'(t)}^2},dt$$ This assumes that the integral on right exists.
In the above development we have nowhere use the fact $s_i/d_ito 1$. The integral formula for arc-length is a consequence of definition of arc-length and mean value theorems. And the result in your question is a consequence of this integral formula.
The restriction $f'(t) neq 0$ can be removed by using Langrange Mean Value Theorem on $f, g$ in the sum $L_{mathcal {C}} (P) $ and Duhamel Principle for integrals.
answered Jan 14 at 4:09
Paramanand SinghParamanand Singh
50.1k556163
$endgroup$
It is best to setup the coordinate system in such a manner that the part of curve with length $l_i$ starts from origin and then we have $$l_i=int_{0}^{h}sqrt{{x'(t)}^2+{y'(t)}^2},dt$$ and $$d_i=sqrt{x^2(h)+y^2(h)}$$ and we need to show that $l_i/d_ito 1$ as $hto 0$. If we assume that one of the derivatives $x'(0),y'(0)$ is non-zero then we can see via Fundamental Theorem of Calculus that $$frac{l_i}{h}tosqrt{{x'(0)}^2+{y'(0)}^2}$$ and by definition of derivative $d_i/h$ also tends to the same value and thus our job is done.
The definition of a curve assumes that $x(t), y(t) $ are continuous and further the notion of arc-length requires that these functions be of bounded variation. The corresponding analysis under such general conditions is more difficult.
Based on feedback received in comments let me just mention the development of arc-length and its representation as an integral in brief.
We first begin with the notion of a partition. Let $[a, b] $ be a closed interval. A partition of $[a, b] $ is a finite set $$P={t_0,t_1,t_2,dots,t_n} $$ such that $$a=t_0<t_1<t_2<dots<t_n=b$$ Let the set of all possible partitions of $[a, b] $ be denoted by $mathcal {P} [a, b] $ ie $$mathcal{P} [a, b] ={Pmid Ptext{ is a partition of }[a, b] } $$
Let $f:[a, b] tomathbb {R} $ be a function. Let $$P={t_0,t_1,t_2,dots,t_n}$$ be a partition of $[a, b] $ and we form a sum $$V_{f} (P) =sum_{i=1}^{n}|f(t_i)-f(t_{i-1})|$$ The function $f$ is said to be of bounded variation if the set $${V_{f} (P) mid Pinmathcal{P} [a, b] } $$ of sums $V_{f} (P) $ for all partitions $P$ of $[a, b] $ is bounded. And in this case the total variation of $f$ on $[a, b] $ is defined to be the supremum $$V_{f} [a, b] =sup, {V_{f} (P) mid Pinmathcal{P} [a, b] } $$
It is easy to prove that any monotone function is of bounded variation and a bit more difficult to prove that a function is of bounded variation if and only if it can be expressed as a difference of two increasing functions.
Next we come to the topic of interest. Let $f, g$ be two functions from interval $[a, b] $ to $mathbb {R} $ and let's assume that they are continuous on $[a, b] $. A curve is a set of points $$mathcal{C} ={(x, y) mid x=f(t), y=g(t), tin[a, b] } $$ To define the arc-length of this curve $mathcal{C} $ we start with a partition $$P={t_0,t_1,t_2,dots,t_n} $$ of $[a, b] $ and form the sum $$L_{mathcal{C}} (P) =sum_{i=1}^{n}sqrt {{f(t_i)-f(t_{i-1})}^2+{g(t_i)-g(t_{i-1})}^2} $$ In other words corresponding to the partition $P$ of $[a, b] $ we have points $A_0,A_1,A_2dots,A_n$ on the curve $mathcal{C} $ with $A_i=(f(t_i), g(t_i)) $ and the above expression for $L_{mathcal{C}} (P) $ is the sum of lengths of these line segments $A_{i-1}A_{i}$.
If the set $${L_{mathcal{C}} (P) mid Pinmathcal{P} [a, b] } $$ of sums $L_{mathcal{C}} (P) $ is bounded for all partitions $P$ of $[a, b] $ then we say that the curve $mathcal{C} $ is rectifiable (ie possesses a well defined arc-length) and its arc-length $L_{mathcal{C}} $ is defined to be the supremum of all such sums ie $$L_{mathcal{C}} =sup, {L_{mathcal{C}} (P) mid Pinmathcal{P} [a, b] } $$ It can be proved with some effort that the curve $mathcal{C} $ is rectifiable if and only if both $f, g$ are of bounded variation on $[a, b] $.
Let us now assume that the functions $f, g$ used to define the curve $mathcal{C} $ are differentiable and further $f'(t) neq 0$ for all $tin[a, b] $. Then $f$ is one-one (by Rolle's theorem if $f$ takes same value at two points then its derivative vanishes somewhere in between) and for any partition $P={t_0,t_1,dots,t_n} $ of $[a, b] $ we have $f(t_i) - f(t_{i-1}) neq 0$. Hence we can write $$L_{mathcal{C}} (P) =sum_{i=1}^{n}|f(t_i)-f(t_{i-1})|sqrt{1+left(frac{g(t_i)-g(t_{i-1})}{f(t_i)-f(t_{i-1})}right)^2}$$ Using Cauchy mean value theorem the ratio inside square roots can be written as $g'(xi_i) /f'(xi_i) $ for some $xi_iin(t_{i-1}, t_i) $. Also using mean value theorem the expression outside square roots can be written as $|f'(eta_i)|(t_i-t_{i-1})$ for some $eta_iin(t_{i-1},t_i)$ and thus we have $$L_{mathcal{C}} (P) =sum_{i=1}^{n}|f'(eta_i)|sqrt{1+left(frac{g'(xi_i)}{f'(xi_i)}right)^2}(t_i-t_{i-1})$$ and the above looks like a Riemann sum for the integral $$int_{a} ^{b} |f'(t) |sqrt{1+left(frac{g'(t)}{f'(t)}right)^2},dt=int_{a}^{b}sqrt{{f'(t)}^2+{g'(t)}^2},dt$$ (there is a slight technicality involved here due to the different set of points $eta_i, xi_i $). As partitions $P$ become finer and finer the expression $L_{mathcal{C}} (P) $ tends to its supremum $L_{mathcal{C}} $ and the Riemann sums above tend to the integral above and thus we get the arc-length formula $$L_{mathcal{C}} =int_{a}^{b}sqrt{{f'(t)}^2+{g'(t)}^2},dt$$ This assumes that the integral on right exists.
In the above development we have nowhere use the fact $s_i/d_ito 1$. The integral formula for arc-length is a consequence of definition of arc-length and mean value theorems. And the result in your question is a consequence of this integral formula.
The restriction $f'(t) neq 0$ can be removed by using Langrange Mean Value Theorem on $f, g$ in the sum $L_{mathcal {C}} (P) $ and Duhamel Principle for integrals.
answered Jan 14 at 4:09
Paramanand SinghParamanand Singh
50.1k556163
edited Jan 21 at 0:40
answered Jan 14 at 4:09
Paramanand SinghParamanand Singh
50.1k556163
answered Jan 14 at 4:09
Paramanand SinghParamanand Singh
50.1k556163
answered Jan 14 at 4:09
Paramanand SinghParamanand Singh
50.1k556163
50.1k556163
$begingroup$
Hi thanks very much, can I ask 2 more questions. Firstly could you be more specific on proving $frac{l_i}{h}tosqrt{{x'(0)}^2+{y'(0)}^2}$ ? Secondly, as far as I know the function $l_i=int_0^hsqrt{{x'(t)}^2+{y'(t)}^2}dt$ is proved from the results of $frac{l_i}{d_1}to1$ .......
$endgroup$
– Shore
Jan 14 at 6:20
$begingroup$
@Shore: the limit of $l_i/h$ is obtained by a direct use of fundamental theorem of calculus (the derivative of integral is the function itself) and you need to observe that limit of $l_i/h$ is the derivative of the integral here at point $h=0$.
$endgroup$
– Paramanand Singh
Jan 14 at 8:12
$begingroup$
@Shore : the fact that arc length can be represented by the integral depends on Cauchy mean value theorem and not on the limit you seek. I will update this part in my answer.
$endgroup$
– Paramanand Singh
Jan 14 at 8:14
$begingroup$
Thanks, my calculus text book proved it based on the infinity small straight line is the length of the curve. But this need to prove that on a small scale, the straight line is close enough to the arc length.
$endgroup$
– Shore
Jan 14 at 9:04
$begingroup$
Hi Paramanand Singh, I've read some references about mean value theorem, It seems still using the fact that arc length can be represented by connecting line length with infinite small intervals........
$endgroup$
– Shore
Jan 15 at 0:41
|
show 4 more comments
$begingroup$
Hi thanks very much, can I ask 2 more questions. Firstly could you be more specific on proving $frac{l_i}{h}tosqrt{{x'(0)}^2+{y'(0)}^2}$ ? Secondly, as far as I know the function $l_i=int_0^hsqrt{{x'(t)}^2+{y'(t)}^2}dt$ is proved from the results of $frac{l_i}{d_1}to1$ .......
$endgroup$
– Shore
Jan 14 at 6:20
$begingroup$
@Shore: the limit of $l_i/h$ is obtained by a direct use of fundamental theorem of calculus (the derivative of integral is the function itself) and you need to observe that limit of $l_i/h$ is the derivative of the integral here at point $h=0$.
$endgroup$
– Paramanand Singh
Jan 14 at 8:12
$begingroup$
@Shore : the fact that arc length can be represented by the integral depends on Cauchy mean value theorem and not on the limit you seek. I will update this part in my answer.
$endgroup$
– Paramanand Singh
Jan 14 at 8:14
$begingroup$
Thanks, my calculus text book proved it based on the infinity small straight line is the length of the curve. But this need to prove that on a small scale, the straight line is close enough to the arc length.
$endgroup$
– Shore
Jan 14 at 9:04
$begingroup$
Hi Paramanand Singh, I've read some references about mean value theorem, It seems still using the fact that arc length can be represented by connecting line length with infinite small intervals........
$endgroup$
– Shore
Jan 15 at 0:41
$begingroup$
Hi thanks very much, can I ask 2 more questions. Firstly could you be more specific on proving $frac{l_i}{h}tosqrt{{x'(0)}^2+{y'(0)}^2}$ ? Secondly, as far as I know the function $l_i=int_0^hsqrt{{x'(t)}^2+{y'(t)}^2}dt$ is proved from the results of $frac{l_i}{d_1}to1$ .......
$endgroup$
– Shore
Jan 14 at 6:20
$begingroup$
Hi thanks very much, can I ask 2 more questions. Firstly could you be more specific on proving $frac{l_i}{h}tosqrt{{x'(0)}^2+{y'(0)}^2}$ ? Secondly, as far as I know the function $l_i=int_0^hsqrt{{x'(t)}^2+{y'(t)}^2}dt$ is proved from the results of $frac{l_i}{d_1}to1$ .......
$endgroup$
– Shore
Jan 14 at 6:20
$begingroup$
@Shore: the limit of $l_i/h$ is obtained by a direct use of fundamental theorem of calculus (the derivative of integral is the function itself) and you need to observe that limit of $l_i/h$ is the derivative of the integral here at point $h=0$.
$endgroup$
– Paramanand Singh
Jan 14 at 8:12
$begingroup$
@Shore: the limit of $l_i/h$ is obtained by a direct use of fundamental theorem of calculus (the derivative of integral is the function itself) and you need to observe that limit of $l_i/h$ is the derivative of the integral here at point $h=0$.
$endgroup$
– Paramanand Singh
Jan 14 at 8:12
$begingroup$
@Shore : the fact that arc length can be represented by the integral depends on Cauchy mean value theorem and not on the limit you seek. I will update this part in my answer.
$endgroup$
– Paramanand Singh
Jan 14 at 8:14
$begingroup$
@Shore : the fact that arc length can be represented by the integral depends on Cauchy mean value theorem and not on the limit you seek. I will update this part in my answer.
$endgroup$
– Paramanand Singh
Jan 14 at 8:14
$begingroup$
Thanks, my calculus text book proved it based on the infinity small straight line is the length of the curve. But this need to prove that on a small scale, the straight line is close enough to the arc length.
$endgroup$
– Shore
Jan 14 at 9:04
$begingroup$
Thanks, my calculus text book proved it based on the infinity small straight line is the length of the curve. But this need to prove that on a small scale, the straight line is close enough to the arc length.
$endgroup$
– Shore
Jan 14 at 9:04
$begingroup$
Hi Paramanand Singh, I've read some references about mean value theorem, It seems still using the fact that arc length can be represented by connecting line length with infinite small intervals........
$endgroup$
– Shore
Jan 15 at 0:41
$begingroup$
Hi Paramanand Singh, I've read some references about mean value theorem, It seems still using the fact that arc length can be represented by connecting line length with infinite small intervals........
$endgroup$
– Shore
Jan 15 at 0:41
|
show 4 more comments
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$begingroup$
We usually express that as $s_i/d_ito 1$ as $d_ito 0$.
$endgroup$
– Paramanand Singh
Jan 14 at 3:53