# Parallel functions.

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:

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.

-
It would seem to me that you've managed to find the equations for a parallel curve 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. – J. M. May 18 '12 at 12:02

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:

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.

-
Very good examples Thanks a lot for graphs.Now I can see more general view of parallel curve reality. I wonder how we can find the limit distance to base function ($f(x)$) that parallel curves will be function too? Is It true that only line and circle have all parallel curves are also functions. Can we find other examples? Thanks – Mathlover May 19 '12 at 19:56
Sounds like that's worth a separate question. – J. M. May 20 '12 at 1:16
I opened a new separate question. math.stackexchange.com/questions/147857/… – Mathlover May 22 '12 at 6:31
The limit distance should simply be the minimum radius of curvature of the curve, right? - at that point, you'd get a cusp. After that, you get cross-overs. – kram1032 Mar 9 '13 at 3:21
the only way that a cusp is reached at the very same time uniformly everywhere is to have a curve of constant curvature, e.g. a circle or line. All others will not form equal cusps. (The circle forms a point eventually. The line will only produce more lines or, if you prefer, it'll form cusps at infinity. – kram1032 Mar 9 '13 at 3:24

You are looking for the parallel curves to a given curve. The link shows you how to obtain these curves for curves given parametrically.

-