\rowcolor{shade2}
& \centering\(\dfrac{d^2 y}{dx^2} > 0\) & \centering \(\dfrac{d^2y}{dx^2}<0\) & \(\dfrac{d^2y}{dx^2}=0\) (inflection) \\
\hline
- \(\frac{dy}{dx}>0\) &
+ \(\dfrac{dy}{dx}>0\) &
\makecell{\\\begin{tikzpicture}\begin{axis}[xmin=-3, xmax=0.8, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(e^(x))}; \addplot[red] {x/2.5+0.75}; \end{axis}\end{tikzpicture} \\Rising (concave up)}&
\makecell{\\\begin{tikzpicture}\begin{axis}[xmin=0.1, xmax=4, scale=0.2, samples=50, unbounded coords=jump] \addplot[blue] {(ln(x))}; \addplot[red] {x/1.5-0.56}; \end{axis}\end{tikzpicture} \\Rising (concave down)}&
\makecell{\\\begin{tikzpicture}\begin{axis}[xmin=-1.5, xmax=1.5, scale=0.2, samples=100] \addplot[blue] {(sin((deg x)))}; \addplot[red] {x}; \end{axis}\end{tikzpicture} \\Rising inflection point}\\
\subsubsection*{Gradient at a point on parametric curve}
- \[{\frac{dy}{dx}} = {{\frac{dy}{dt}} \div {\frac{dx}{dt}}} \> \vert \> {\frac{dx}{dt}} \ne 0\]
+ \[{\frac{dy}{dx}} = {{\frac{dy}{dt}} \div {\frac{dx}{dt}}} \> \vert \> {\frac{dx}{dt}} \ne 0 \text{ (chain rule)}\]
\[\frac{d^2}{dx^2} = \frac{d(y^\prime)}{dx} = {\frac{dy^\prime}{dt} \div {\frac{dx}{dt}}} \> \vert \> y^\prime = {\frac{dy}{dx}}\]
\[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
- \end{multicols}
- \end{document}
+
+ \section{Kinematics \& Mechanics}
+
+ \subsection*{Constant acceleration}
+
+ \begin{itemize}
+ \item \textbf{Position} - relative to origin
+ \item \textbf{Displacement} - relative to starting point
+ \end{itemize}
+
+ \subsubsection*{Velocity-time graphs}
+
+ \begin{itemize}
+ \item Displacement: \textit{signed} area between graph and \(t\) axis
+ \item Distance travelled: \textit{total} area between graph and \(t\) axis
+ \end{itemize}
+
+ \[ \text{acceleration} = \frac{d^2x}{dt^2} = \frac{dv}{dt} = v\frac{dv}{dx} = \frac{d}{dx}\left(\frac{1}{2}v^2\right) \]
+
+ \begin{center}
+ \renewcommand{\arraystretch}{1}
+ \begin{tabular}{ l r }
+ \hline & no \\ \hline
+ \(v=u+at\) & \(x\) \\
+ \(v^2 = u^2+2as\) & \(t\) \\
+ \(s = \frac{1}{2} (v+u)t\) & \(a\) \\
+ \(s = ut + \frac{1}{2} at^2\) & \(v\) \\
+ \(s = vt- \frac{1}{2} at^2\) & \(u\) \\ \hline
+ \end{tabular}
+ \end{center}
+
+ \[ v_{\text{avg}} = \frac{\Delta\text{position}}{\Delta t} \]
+ \begin{align*}
+ \text{speed} &= |{\text{velocity}}| \\
+ &= \sqrt{v_x^2 + v_y^2 + v_z^2}
+ \end{align*}
+
+ \noindent \textbf{Distance travelled between \(t=a \rightarrow t=b\):}
+ \[= \int^b_a \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \cdot dt \]
+
+ \noindent \textbf{Shortest distance between \(\boldsymbol{r}(t_0)\) and \(\boldsymbol{r}(t_1)\):}
+ \[ = |\boldsymbol{r}(t_1) - \boldsymbol{r}(t_2)| \]
+
+ \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}