If we have a function $\phi:\Omega\subseteq\mathbb{R}^2\to\mathbb{R}$, $\phi\in C^2(\Omega)$ is there a way to find a function $d:\Omega\to\mathbb{R}, d\in C^2(\Omega)$ so that:

$|\nabla d(x,y)|=1,\ \forall\ (x,y)\in\Omega$ and $\phi(x)=0\Longleftrightarrow\ d(x)=0$.

For example, this is true when $\phi(x,y)=1-x^2-y^2$, and we can find $d(x,y)=1-\sqrt{x^2+y^2}$, with:

$|\nabla d|=\sqrt{\left (\frac{-x}{\sqrt{x^2+y^2}}\right )^2+\left (\frac{-y}{\sqrt{x^2+y^2}}\right )^2}=1$

If we have a function $\phi:\Omega\subseteq\mathbb{R}^2\to\mathbb{R}$, $\phi\in C^2(\Omega)$ is there a way to find a function $d:\Omega\to\mathbb{R}, d\in C^2(\Omega)$ so that:

$|\nabla d(x,y)|=1,\ \forall\ (x,y)\in\Omega$ and $\phi(x)=0\Longleftrightarrow\ d(x)=0$.

For example, this is true when $\phi(x,y)=1-x^2-y^2$, and we can find $d(x,y)=1-\sqrt{x^2+y^2}$, with:

$|\nabla d|=\sqrt{\left (\frac{-x}{\sqrt{x^2+y^2}}\right )^2+\left (\frac{-y}{\sqrt{x^2+y^2}}\right )^2}=1$

and of course $d(x,y)=0\ \Longleftrightarrow\ \phi(x,y)=0$.

If we have o function $\phi:\Omega\subseteq\mathbb{R}^2\to\mathbb{R}$, $\phi\in C^2(\Omega)$ is there a way to find a function $d:\Omega\to\mathbb{R}, d\in C^2(\Omega)$ so that:

$|\nabla d(x,y)|=1,\ \forall\ (x,y)\in\Omega$ and $\phi(x)=0\Longleftrightarrow\ d(x)=0$.

For example, this is true when $\phi(x,y)=1-x^2-y^2$, and we can find $d(x,y)=1-\sqrt{x^2+y^2}$, with:

$|\nabla d|=\sqrt{\left (\frac{-x}{\sqrt{x^2+y^2}}\right )^2+\left (\frac{-y}{\sqrt{x^2+y^2}}\right )^2}=1$

If we have a function $\phi:\Omega\subseteq\mathbb{R}^2\to\mathbb{R}$, $\phi\in C^2(\Omega)$ is there a way to find a function $d:\Omega\to\mathbb{R}, d\in C^2(\Omega)$ so that:

$|\nabla d(x,y)|=1,\ \forall\ (x,y)\in\Omega$ and $\phi(x)=0\Longleftrightarrow\ d(x)=0$.

For example, this is true when $\phi(x,y)=1-x^2-y^2$, and we can find $d(x,y)=1-\sqrt{x^2+y^2}$, with:

$|\nabla d|=\sqrt{\left (\frac{-x}{\sqrt{x^2+y^2}}\right )^2+\left (\frac{-y}{\sqrt{x^2+y^2}}\right )^2}=1$