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.
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.