What is g(x) in Terms of f(x): Distance Path Path

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In summary, the conversation discusses a function g(x) that contains all points that are at a constant distance from f(x). The function g(x) can be described parametrically by X=t\mp\frac{Df'(t)}{\sqrt{1+\left(f'(t)\right)^2}} and Y=f(t)\pm\frac{D}{\sqrt{1+\left(f'(t)\right)^2}}. It is easier to find the function g(x) if the original function is a parametrized function at unit speed. The conversation also includes a couple of plots showing the function g(x) at different distances from f(x).
  • #1
wheepep
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g(x) is a function which contains all points that are at a constant distance from f(x).

What is g(x) in terms of f(x)?
 

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  • #2
Let's begin by using a trivial parametrization of $f$ as follows:

\(\displaystyle x=t\)

\(\displaystyle y=f(t)\)

Now, a point on $g$ can be found to lie on the line:

\(\displaystyle y=-\frac{1}{f'(t)}(x-t)+f(t)\)

And the points on this line a distance $D$ from $(t,f(t))$ can be found from:

\(\displaystyle D=\sqrt{(x-t)^2+\left(y-f(t)\right)^2}=\left|y-f(t)\right|\sqrt{1+\left(f'(t)\right)^2}\)

Hence:

\(\displaystyle y=f(t)\pm\frac{D}{\sqrt{1+\left(f'(t)\right)^2}}\)

This then implies

\(\displaystyle -\frac{1}{f'(t)}(x-t)+f(t)=f(t)\pm\frac{D}{\sqrt{1+\left(f'(t)\right)^2}}\)

Which reduces to:

\(\displaystyle x=t\mp\frac{Df'(t)}{\sqrt{1+\left(f'(t)\right)^2}}\)

Thus, $g$ can be described parametrically by:

\(\displaystyle X=t\mp\frac{Df'(t)}{\sqrt{1+\left(f'(t)\right)^2}}\)

\(\displaystyle Y=f(t)\pm\frac{D}{\sqrt{1+\left(f'(t)\right)^2}}\)
 
  • #3
Re: Path

wheepep said:
g(x) is a function which contains all points that are at a constant distance from f(x).

What is g(x) in terms of f(x)?

Hi wheepep! ;)

Suppose we look at the parametrized function $\gamma(t) = (t, f(t))$.
The tangent of a parametrized function is $\gamma'(t)$.
The normal is $\gamma'^\perp(t)$.
And the normal at unit length is $N(t) = \frac{\gamma'^\perp(t)}{\| \gamma'(t) \|}$.
So the function that is at constant distance $c$ is:
$$\gamma_c(t) = \gamma(t) + cN(t) = \gamma(t) + \frac{c\gamma'^\perp(t)}{\| \gamma'(t) \|}$$In your problem statement we have:
$$\gamma(x) = (x,f(x)) \\
\gamma'(x) = (1,f'(x)) \\
\gamma'^\perp(x) = (-f'(x),1) \\
N(t) = \left(-\frac{f'(x)}{\sqrt{1+f'(x)^2}}, \frac{1}{\sqrt{1+f'(x)^2}}\right) \\
\gamma_c(x) = \left(x-\frac{cf'(x)}{\sqrt{1+f'(x)^2}}, f(x) + \frac{c}{\sqrt{1+f'(x)^2}}\right)
$$

EDIT: Which is the same result MarkFL just gave! (Wink)If we (can) reparametrize the function $\gamma$ to $\tilde\gamma$ such that it has unit speed, that is, has $\| \tilde\gamma'(s) \| = 1$, we have:
$$\tilde N(s) = \tilde\gamma'^\perp(s) \\
\tilde\gamma_c(s)=\tilde\gamma(s) + c\tilde\gamma'^\perp(s)
$$Let's pick an example.
Suppose we pick $\tilde\gamma(s) = (\cos s, \sin s)$, which corresponds to $f(x)=\sqrt{1-x^2}$, then it simplifies a bit:
$$\tilde \gamma'(s) = (-\sin s, \cos s) \\
\tilde N(s) = \tilde\gamma'^\perp(s) = (\cos s, \sin s)\\
\tilde\gamma_c(s) = \tilde\gamma s + c\tilde N(s) = (\cos s + c \cos s, \sin s + c \sin s) = (1+c)(\cos s, \sin s)
$$
which is a circle with radius $(1+c)$ as expected.

If we do the same thing with $f(x)=\sqrt{1-x^2}$ and $\gamma_c(x)=\left(x-\frac{cf'(x)}{\sqrt{1+f'(x)^2}}, f(x) + \frac{c}{\sqrt{1+f'(x)^2}}\right)$, we'll find the same result - it will just take much longer.As you can see, the problem becomes much simpler if the function is a parametrized function at unit speed.
Of course it all depends on the function we want to shift.
Did you have any particular function in mind? (Wondering)
 
  • #4
y = x2
 
  • #5
wheepep said:
y = x2

Let's see what it looks like when filling in the formula:
[LATEXS]
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[ymax=10, samples=51]
\addplot[blue, ultra thick] (x,x^2);
\addplot[red, ultra thick] ({x - 2*x/(sqrt(1+4*x^2))},{x^2 + 1/(sqrt(1+4*x^2))});
\end{axis}
\end{tikzpicture}
[/LATEXS]
[TIKZ]
%preamble \usepackage{pgfplots}
\begin{axis}[ymax=10, samples=51]
\addplot[blue, ultra thick] (x,x^2);
\addplot[red, ultra thick] ({x - 2*x/(sqrt(1+4*x^2))},{x^2 + 1/(sqrt(1+4*x^2))});
\end{axis}
[/TIKZ]
 
  • #6
Interesting
 
  • #7
Here's another plot with a couple more graphs at different distances:
\begin{tikzpicture}
%preamble \usepackage{pgfplots}
\begin{axis}[ymax=10, samples=151, grid=both,
axis lines=middle,
axis equal,
minor tick num=4,
xtick={-8,-6,...,8},
ytick={-4,-2,...,10},
grid style={line width=.1pt, draw=gray!30},
major grid style={line width=.2pt,draw=gray!70}
]
\addplot[green, ultra thick] ({x - 8*x/(sqrt(1+4*x^2))}, {x^2 + 4/(sqrt(1+4*x^2))});
\addplot[red, ultra thick] ({x - 4*x/(sqrt(1+4*x^2))}, {x^2 + 2/(sqrt(1+4*x^2))});
\addplot[blue, ultra thick] (x,x^2);
\addplot[red, ultra thick] ({x + 4*x/(sqrt(1+4*x^2))}, {x^2 - 2/(sqrt(1+4*x^2))});
\addplot[green, ultra thick] ({x + 8*x/(sqrt(1+4*x^2))}, {x^2 - 4/(sqrt(1+4*x^2))});
\end{axis}
\end{tikzpicture}
 
  • #8
Thanks
 

FAQ: What is g(x) in Terms of f(x): Distance Path Path

What is g(x) in Terms of f(x)?

g(x) is a function that represents the distance along a path from the starting point to a specific point. It is dependent on f(x), which is a function that represents the path taken.

Can g(x) and f(x) be used interchangeably?

No, g(x) and f(x) are two separate functions with different meanings. While f(x) represents the path taken, g(x) represents the distance along that path.

How is g(x) calculated?

To calculate g(x), you would need to use an integral or a derivative of f(x). The specific method would depend on the type of path and the information given.

What is the relationship between g(x) and distance?

g(x) is directly related to distance. It represents the distance traveled along a path from the starting point to a specific point. As the path changes, the value of g(x) will also change.

What are some real-life applications of g(x) in terms of f(x)?

g(x) and f(x) have many real-life applications, such as calculating the distance traveled by a moving object, determining the displacement of an object, and finding the area under a curve. They are also used in fields like physics, engineering, and economics.

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