Question: I'm given a curve $g = 0$ in 2D specified by g(x,y) = f(x) - y. The normal to the curve is the gradient of $g$ - $(f', -1)$. Now I want express the change in the angle $\theta$ of the normal as I move along the curve. However, I'm not sure how to express this quantity. (For instance if $f$ is linear, this means that $d\theta = 0$ as the normal does not change direction$.

Answer: 
It's because $2\cdot 6 \equiv 1 \mod 11$
The notation $2^{-1} \pmod{11}$ means the *multiplicative inverse* of $2$ modulo $11$, that is, a number $x$ satisfying the equation $2x \equiv 1 \pmod{11}$. 

Since $2 \cdot 6 \equiv 12 \equiv 1 \pmod{11}$, $2^{-1} \equiv 6 \pmod{11}$. The same calculation shows that $6^{-1} \equiv 2 \pmod{11}$. 

The multiplicative inverse of $6$ exists modulo $11$ since $\gcd(6, 11) = 1$.

We can find the multiplicative inverse of $6$ modulo $11$ using the [extended Euclidean algorithm](https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm).
\begin{align\*}
11 & = 1 \cdot 6 + 5\\
6 & = 1 \cdot 5 + 1\\
5 & = 5 \cdot 1
\end{align\*}
We solve for $1$ as a linear combination of $6$ and $11$.
\begin{align\*}
1 & = 6 - 5\\
 & = 6 - (11 - 6)\\
 & = 2 \cdot 6 - 11
\end{align\*}
Hence, $2 \cdot 6 \equiv 1 \pmod{11} \iff 6^{-1} \equiv 2 \pmod{11}$.

Let's see what happens when we use the extended Euclidean algorithm to solve for $2^{-1}$ modulo $11$.

The multiplicative inverse of $2$ exists modulo $11$ since $\gcd(2, 11) = 1$.
\begin{align\*}
11 & = 5 \cdot 2 + 1\\
2 & = 2 \cdot 1
\end{align\*}
Solving for $1$ as a linear combination of $2$ and $11$ yields
$$1 = 11 - 5 \cdot 2$$
Thus, $$-5 \cdot 2 \equiv 1 \pmod{11} \iff 2^{-1} \equiv -5 \pmod{11}$$
What does this mean? It means that for each $t \in \mathbb{Z}$, 
$$2(-5 + 11t) \equiv 1 \pmod{11}$$
In particular, if $t = 1$, we obtain 
$$2 \cdot 6 \equiv 1 \pmod{11} \iff 2^{-1} \equiv 6 \pmod{11}$$
Why is $2(-5 + 11t) \equiv 1 \pmod{11}$ for $t \in \mathbb{Z}$? Observe that 
$$2(-5 + 11t) \equiv -10 + 22t \equiv 1 - 11 + 22t \equiv 1 + 11(2t - 1) \equiv 1 \pmod{11}$$