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In 2 dimensions, we can draw 2 parallel lines that have the same distance from a line.

I wanted to find parallel functions of a function and their distance is $d$ to the function for all inputs and tangents are equal as shown in the picture.

I assume we have $f(x)$ and we try to find parallel functions that named $g(x)$. $g(x)$ must have 2 solutions $g_1(x)$ and $g_2(x)$ as shown in the picture:

diagram for parallel functions

Equations to find g(x):

Equation $(1)$ :Parallel condition $f'(x_1)=g'(x_2)$

Equation $(2)$ : $A(x_1,f(x_1))$ and $B(x_2,g(x_2))$ they are in same line. $g(x_2)-f(x_1)=\frac{-1}{f'(x_1)}(x_2-x_1)$

Equation $(3)$ : d is between $A(x_1,f(x_1))$ and $B(x_2,g(x_2))$ $d^2=(x_2-x_1)^2+(g(x_2)-f(x_1))^2$

$(x_2-x_1)^2+(\frac{1}{(f'(x_1))^2}(x_2-x_1)^2=d^2$

$(x_2-x_1)^2+\frac{1}{(f'(x_1))^2}(x_2-x_1)^2=d^2$

$(x_2-x_1)^2==\frac{d^2(f'(x_1))^2}{1+(f'(x_1))^2} $

$x_2-x_1=+\frac{d.f'(x_1)}{\sqrt{1+(f'(x_1))^2}} $

$x_2-x_1=-\frac{d.f'(x_1)}{\sqrt{1+(f'(x_1))^2}} $

if we want to find first solution of $g(x)$

then need to take $x_2=x_1+\frac{df'(x_1)}{\sqrt{1+(f'(x_1))^2}} $ and put in Equation (1)

$f'(x_1)=g'(x_1+\frac{d.(f'(x_1))}{\sqrt{1+(f'(x_1))^2}})$ replace $x_1$ with $x$ means for all values . and I tried to find $g(x)$ $f'(x)=g'(x+\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})$

$f'(x)(1+(\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})')=(1+(\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})') g'(x+\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})$

$\int f'(x)(1+(\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})') dx= \int (1+(\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})'). g'(x+\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}}) dx$

$\int f'(x)(1+(\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})') dx= g(x+\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})$

$g(x+\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})=f(x)+\int f'(x)(\frac{d.(f'(x))}{\sqrt{1+(f'(x))^2}})') dx$

Am I in the right way to find $g(x)$? Can I find g(x) after integrations? Thanks in advance.

  • 3
    It would seem to me that you've managed to find the equations for a [parallel curve](http://mathworld.wolfram.com/ParallelCurves.html) in the special case $y=f(x)$. In general, however, parallel curves will almost always be difficult to recast in explicit $f(x,y)=0$ form.2012-05-18

2 Answers 2

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Now that André has already told you about parallel/offset curves, I am writing this post to drive a certain point home: the parallels of a function may not be functions themselves. To illustrate this point, I will show a family of parallel curves for four common functions:

parallels of functions

Note that each family of parallels has at least one member that possesses a cusp, and a point of self-intersection. Some of them might be functions themselves (though the likelihood of them possessing a simple $y=g(x)$ formula is not too high), but generally speaking, parallels are not functions.

  • 0
    I do not have a formal proof for this but that's what my intuition would say about the topic. I guess you could get some other piecewise cases, like a periodicly repeating half-circle which would result in a mirrored and shifted version of itself in the cusping limit. Such cases have to be considered more carefully, but even then, they won't never form a cusp. (double negative)2013-03-09
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You are looking for the parallel curves to a given curve. The link shows you how to obtain these curves for curves given parametrically.