\[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
- \end{multicols}
- \end{document}
+
+ \section{Kinematics \& Mechanics}
+
+ \subsection*{Constant acceleration}
+ {\centering \begin{tabular}{ l r } % TODO need to fix centering here
+ \hline & no \\ \hline
+ $v=u+at$ & $x$ \\
+ $s = {1 \over 2}(v+u)t$ & $a$ \\
+ $s=ut+{1 \over 2}at^2$ & $v$ \\
+ $s=vt-{1 \over 2}at^2$ & $u$ \\
+ $v^2=u^2+2as$ & $t$ \\ \hline
+ \end{tabular}}
+
+ \[ v_{\text{avg}} = \frac{\Delta\text{position}}{\Delta t} \]
+ \begin{align*}
+ \text{speed} &= |{\text{velocity}}| \\
+ &= \sqrt{v_x(t)^2 + v_y(t)^2 + v_z(t)^2} \tag{for vector \(v\)}
+ \end{align*}
+ \textbf{Distance travelled between\(t=a \rightarrow t=b\):}
+ \[= \int^b_a \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \cdot dt \]
+
+ \subsection*{Vector functions}
+
+ \[ \boldsymbol{r}(t) = x \boldsymbol{i} + y \boldsymbol{j} + z \boldsymbol{k} \]
+
+ \begin{itemize}
+ \item If \(\boldsymbol{r}(t) \equiv\) position with time, then the graph of endpoints of \(\boldsymbol{r}(t) \equiv\) Cartesian path
+ \item Domain of \(\boldsymbol{r}(t)\) is the range of \(x(t)\)
+ \item Range of \(\boldsymbol{r}(t)\) is the range of \(y(t)\)
+ \end{itemize}
+
+ \subsection*{Vector calculus}
+
+ \subsubsection*{Derivative}
+
+ Let \(\boldsymbol{r}(t)=x(t)\boldsymbol{i} + y(t)\boldsymbol(j)\). If both \(x(t)\) and \(y(t)\) are differentiable, then:
+ \[ \boldsymbol{r}(t)=x(t)\boldsymbol{i}+y(t)\boldsymbol{j} \]
+
+ \end{multicols}
+\end{document}