# Show that $\kappa_{\delta} = \frac{\kappa}{1 - r\kappa}$

$\delta_r(t) = \gamma(t) + r U(t)$ is a parallel curve to a parametric curve $\gamma(t): I \rightarrow \mathbb{R}^2$ at distance $r$. I have already shown that $\delta_r' = (1 - r \kappa) \cdot | \gamma'| \cdot T$, where $\kappa$ is the curvature and $T$ is the unit tangent to $\gamma$. I have also shown that $U_{\delta} = U$ and $T_{\delta} = T$. I have also been told that $\kappa(t) < \frac{1}{r}$. I now need to show that $\kappa_{\delta} = \frac{\kappa}{1 - r\kappa}$. What I have done is this:

$\kappa = \frac{T' \circ U}{|\gamma'|}$ where $T' \circ U$ is the dot product between $T$ and $U$. So from here, we can say that

$$\kappa_{\delta} = \frac{T_{\delta}' \circ U_{\delta}}{|\delta'|}.$$

We have already shown that $T_{\delta}' = T, U_{\delta} = U$ and $\delta' = (1 - r \kappa) \cdot | \gamma'| \cdot T$. So we can now re-write our formula as

$$\kappa_{\delta} = \frac{T' \circ U}{|(1 - r \kappa) \cdot | \gamma'| \cdot T|}$$

As $T$ is a unit vector, we get $|T| = 1$ and so we get the formula to be

$$\kappa_{\delta} = \frac{T' \circ U}{|(1 - r \kappa) \cdot | \gamma'||}$$

Using the condition for $\kappa(t)$, we get that $|(1 - r\kappa)| = (1 - r \kappa)$ and so I end up with $(1 - r \kappa) |\gamma '|$ on the denominator, which is not the right answer. How do I get rid of the $|\gamma '|$?

$$\kappa_{\delta} = \frac{T' \circ U}{|\gamma'|} \cdot \frac{1}{1 - r \kappa} = \frac{\kappa}{1 - r\kappa}$$