My son has been working on equations for finding length of the length of a curve without any resources (computers, calculators...) while in prison. He is wondering if he is on the right path. Sadly my understanding of mathematics is severely limited; I cannot help him. Dear mathematicians, could you please offer him some feedback? Is he on the right track? Any advice or assistance or even comments from someone knowledgeable would be greatly appreciated. Below is his information. Thank you, Rimel
The length of a curve by calculus incorporates limits to each point next to each other along the curve. MY method (Daniel) incorporates the ever expanding, contracting nature of a measure of $x$ and $y$(or $x^2,1-(\sqrt{1-x^2})$ etc.) forming a continuous curve(arc) as the representation of a linear function in Cartesian representations.
To find the length of an arc in terms of $x$ and $y$ (being a function of $x$)), start with a method I derived to find $\langle x \rangle$ and $\langle y \rangle$ ,$\langle . \rangle$ stands for average.
ex: length of $x,x^2$ from $(0,0) \rightarrow (1,1)$
$\langle x(x) \rangle$ derivative of $x^2 = 2x$.
$2x=\frac{(√2+1)}{2}$ $x= 0.603$($\langle x \rangle$ from $(0,0) \rightarrow (1,1)$ in $(x,x^2)$)
$y=x^2$ $2*\langle x \rangle$ (this $2*$ has nothing to do with the derivative in terms of $2x$ but rather $2*\text{(the average)}$)
$2*\langle x \rangle= 2^(\langle y \rangle+1)$
solve for $\langle y \rangle = -0.729$
next calculation: $(2-(2√((\langle x \rangle)^2 + (\langle y \rangle)^2))) + \sqrt{2} = 1.522$ (length of $x,x^2$ from $(0,0)$ to $(1,1)$ in terms of $\langle x \rangle$ and $\langle y \rangle$ diagonals NOT adjacent points making the curve(arc))
This method can be used to find linear curves BUT to find ALL lengths in terms of these ratio representations I am still working on.
another example: $1/4$ circumference a circle OR $x,1-(\sqrt(1-x^2))$ from $(0,0) \rightarrow (1,1)$ an underhand upside down circle derived from x^2 + y^2 = 1
use the chain rule to find the derivative of $1-\sqrt{1-x^2}$ $x$. in derivative = $0.77$
uh oh! This $\langle x \rangle$ is higher than the box of equal e $x,y$ $0.707$
you must subtract to find $0.644 = \langle x \rangle$
$2*\langle x \rangle = 2^{(\langle y \rangle + 1)}$ $\langle y \rangle= 0.633$
notice how thi exhibits itself in these 2 close values!
plug in $\langle x \rangle$ and $\langle y \rangle$ to $$ (2-(2*(\sqrt{(\langle x \rangle)^2 + (\langle y \rangle)^2)))) + \sqrt{2}= 1.608 $$ ($1/4$ a circle circumference with radius $1$)
other results $x^3 = 1.58$ $x^4 = 1.644$
Notice when partially plotted the circle length part lies in between $x^3$ and $x^4$ so correctly! done with pen,paper, calculators
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proper representation of x throughout the graph curve. mine=$1.608$ vs $\frac{\pi}{2}=1.57079 x,1-\sqrt{1-x^2} (0,0)\rightarrow (1,1)$
$x^2=1.522$ $x^3=1.58$ $1/4$circumference a circle= $1.608$ $x^4=1.64$(approx.)
