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: 
First, we need to understand what $2^{-1}\mod 11$ stands for. Raising to the power $-1$ is usually used to denote the *multiplicative inverse*. That is, $b$ is a multiplicative inverse of $a$ if and only if $a\cdot b=1$. Thus, $2^{-1}\mod 11$ denotes the number such that $$2^{-1}\cdot2\equiv1\mod11$$ Since $2\cdot 6\equiv 12\equiv1\mod11$, we see $2^{-1}\equiv 6\mod11$. We can do the same to show $6^{-1}\equiv 2\mod 11$.

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For a reliable approach (that is, one that doesn't rely on simply testing $2\cdot 1,2\cdot 2,\cdots,2\cdot 10$ and see which one is $1$), we can use (where we're using that $\gcd(2,11)=1$) $$2^{\phi(11)}\equiv 1\mod 11$$ by [Euler's Totient Function](https://en.wikipedia.org/wiki/Euler%27s_totient_function) (or you could use [Fermat's Little Theorem](https://en.wikipedia.org/wiki/Fermat%27s_little_theorem) in this case, since $11$ is prime), and since $\phi(11)=10$, we have \begin{align}
2^{-1}&\equiv 2^{-1}\cdot 1\\
&\equiv 2^{-1}\cdot2^{\phi(11)}\\
&\equiv 2^{-1}\cdot2^{10}\\
&\equiv 2^{-1+10}\\
&\equiv 2^9\\
&\equiv 512\\
&\equiv 6\mod 11
\end{align}
You have $2\times 6=12$ which is $1$ mod $11$. Hence, we see that that $2\times 6=1$ mod $11$.

Now the set of congruence numbers modulo $11$ (often noted $\mathbb{Z}/11$) is a ring with addition and multiplication coming from $\mathbb{Z}$. 

Saying that $6^{-1}\text{ mod }11=2$ is saying that $6$ mod $11$ is invertible as an element of the ring $(\mathbb{Z}/11,+,\times)$ and that its inverse is $2$ mod $11$.