\usepackage{pgfplots}
\usepackage{pst-plot}
\usepackage{rotating}
+%\usepackage{showframe} % debugging only
\usepackage{subfiles}
\usepackage{tabularx}
\usepackage{tcolorbox}
\hline
\end{tabularx}
+ \begin{theorembox}{title=Factor theorem}
+ If \(\beta z + \alpha\) is a factor of \(P(z)\), \\
+ \-\hspace{1em}then \(P(-\dfrac{\alpha}{\beta})=0\).
+ \end{theorembox}
+
\subsection*{\(n\)th roots}
\(n\)th roots of \(z=r\operatorname{cis}\theta\) are:
\addplot[gray, dotted, thick, domain=-35:35] {-1.5708} node [black, font=\footnotesize, above left, pos=1] {\(y=-\frac{\pi}{2}\)};
\end{axis}
\end{tikzpicture}
-\columnbreak
+
+ \subsection*{Mensuration}
+
+ \begin{tikzpicture}[draw=blue!70,thick]
+ \filldraw[fill=lblue] circle (2cm);
+ \filldraw[fill=white]
+ (320:2cm) node[right] {}
+ -- (220:2cm) node[left] {}
+ arc[start angle=220, end angle=320, radius=2cm]
+ -- cycle;
+ \node {Major Segment};
+ \node at (-90:2) {Minor Segment};
+
+ \begin{scope}[xshift=4.5cm]
+ \draw circle (2cm);
+ \filldraw[fill=lblue]
+ (320:2cm) node[right] {}
+ -- (0,0) node[above] {}
+ -- (220:2cm) node[left] {}
+ arc[start angle=220, end angle=320, radius=2cm]
+ -- cycle;
+ \node at (90:1cm) {Major Sector};
+ \node at (-90:1.5) {Minor Sector};
+ \end{scope}
+ \end{tikzpicture}
+
+
+ \begin{align*}
+ \textbf{Sectors: } A &= \pi r^2 \dfrac{\theta}{2\pi} \\
+ &= \dfrac{r^2 \theta}{2}
+ \end{align*}
+
+ \[ \textbf{Segments: } A = \dfrac{r^2}{2} \left(\theta - \sin \theta \right) \]
+
+ \begin{align*}
+ \textbf{Chords: } \operatorname{crd} \theta &= \sqrt{(1 - \cos\theta)^2 + \sin^2 \theta} \\
+ &= \sqrt{2 - 2\cos\theta} \\
+ &= 2 \sin \left(\dfrac{\theta}{2}\right)
+ \end{align*}
+
\section{Differential calculus}
+ \[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={\frac{dy}{dx}}\]
+
\subsection*{Limits}
\[\lim_{x \rightarrow a}f(x)\]
\(f(x)\) is continuous \(\iff L^-=L^+=f(x) \> \forall x\)
\end{enumerate}
- \subsection*{Gradients of secants and tangents}
+ \subsection*{Gradients}
\textbf{Secant (chord)} - line joining two points on curve\\
\textbf{Tangent} - line that intersects curve at one point
- \subsection*{First principles derivative}
-
- \[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={\frac{dy}{dx}}\]
-
- \subsubsection*{Logarithmic identities}
-
- \(\log_b (xy)=\log_b x + \log_b y\)\\
- \(\log_b x^n = n \log_b x\)\\
- \(\log_b y^{x^n} = x^n \log_b y\)
-
- \subsubsection*{Index identities}
-
- \(b^{m+n}=b^m \cdot b^n\)\\
- \((b^m)^n=b^{m \cdot n}\)\\
- \((b \cdot c)^n = b^n \cdot c^n\)\\
- \({a^m \div a^n} = {a^{m-n}}\)
-
- \subsection*{Reciprocal derivatives}
-
- \[\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}\]
-
- \subsection*{Differentiating \(x=f(y)\)}
- Find \(\dfrac{dx}{dy}\), then \(\dfrac{dy}{dx} = \dfrac{1}{\left(\dfrac{dx}{dy}\right)}\)
-
- \subsection*{Second derivative}
- \begin{align*}f(x) \longrightarrow &f^\prime (x) \longrightarrow f^{\prime\prime}(x)\\
- \implies y \longrightarrow &\frac{dy}{dx} \longrightarrow \frac{d^2 y}{dx^2}\end{align*}
-
- \noindent Order of polynomial \(n\)th derivative decrements each time the derivative is taken
-
\subsubsection*{Points of Inflection}
\emph{Stationary point} - i.e.
\end{warning}
+ \subsection*{Second derivative}
+ \begin{align*}f(x) \longrightarrow &f^\prime (x) \longrightarrow f^{\prime\prime}(x)\\
+ \implies y \longrightarrow &\frac{dy}{dx} \longrightarrow \frac{d^2 y}{dx^2}\end{align*}
+
+ \noindent Order of polynomial \(n\)th derivative decrements each time the derivative is taken
+
+
+ \subsection*{Slope fields}
+
+ \begin{tikzpicture}[declare function={diff(\x,\y) = \x+\y;}]
+ \begin{axis}[axis equal, ymin=-4, ymax=4, xmin=-4, xmax=4, ticks=none, enlargelimits=true, ]
+ \addplot[thick, orange, domain=-4:2] {e^(x)-x-1};
+ \pgfplotsinvokeforeach{-4,...,4}{%
+ \draw[gray] ( {#1 -0.1}, {4 - diff(#1, 4) *0.1}) -- ( {#1 +0.1}, {4 + diff(#1, 4) *0.1});
+ \draw[gray] ( {#1 -0.1}, {3 - diff(#1, 3) *0.1}) -- ( {#1 +0.1}, {3 + diff(#1, 3) *0.1});
+ \draw[gray] ( {#1 -0.1}, {2 - diff(#1, 2) *0.1}) -- ( {#1 +0.1}, {2 + diff(#1, 2) *0.1});
+ \draw[gray] ( {#1 -0.1}, {1 - diff(#1, 1) *0.1}) -- ( {#1 +0.1}, {1 + diff(#1, 1) *0.1});
+ \draw[gray] ( {#1 -0.1}, {0 - diff(#1, 0) *0.1}) -- ( {#1 +0.1}, {0 + diff(#1, 0) *0.1});
+ \draw[gray] ( {#1 -0.1}, {-1 - diff(#1, -1) *0.1}) -- ( {#1 +0.1}, {-1 + diff(#1, -1) *0.1});
+ \draw[gray] ( {#1 -0.1}, {-2 - diff(#1, -2) *0.1}) -- ( {#1 +0.1}, {-2 + diff(#1, -2) *0.1});
+ \draw[gray] ( {#1 -0.1}, {-3 - diff(#1, -3) *0.1}) -- ( {#1 +0.1}, {-3 + diff(#1, -3) *0.1});
+ \draw[gray] ( {#1 -0.1}, {-4 - diff(#1, -4) *0.1}) -- ( {#1 +0.1}, {-4 + diff(#1, -4) *0.1});
+ }
+ \end{axis}
+ \end{tikzpicture}
+
\begin{table*}[ht]
\centering
\begin{tabularx}{\textwidth}{|r|Y|Y|Y|}
\-\hspace{1em}\texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}
\end{cas}
- \subsection*{Slope fields}
+ \subsection*{Function of the dependent
+ variable}
- \begin{tikzpicture}[declare function={diff(\x,\y) = \x+\y;}]
- \begin{axis}[axis equal, ymin=-4, ymax=4, xmin=-4, xmax=4, ticks=none, enlargelimits=true, ]
- \addplot[thick, orange, domain=-4:2] {e^(x)-x-1};
- \pgfplotsinvokeforeach{-4,...,4}{%
- \draw[gray] ( {#1 -0.1}, {4 - diff(#1, 4) *0.1}) -- ( {#1 +0.1}, {4 + diff(#1, 4) *0.1});
- \draw[gray] ( {#1 -0.1}, {3 - diff(#1, 3) *0.1}) -- ( {#1 +0.1}, {3 + diff(#1, 3) *0.1});
- \draw[gray] ( {#1 -0.1}, {2 - diff(#1, 2) *0.1}) -- ( {#1 +0.1}, {2 + diff(#1, 2) *0.1});
- \draw[gray] ( {#1 -0.1}, {1 - diff(#1, 1) *0.1}) -- ( {#1 +0.1}, {1 + diff(#1, 1) *0.1});
- \draw[gray] ( {#1 -0.1}, {0 - diff(#1, 0) *0.1}) -- ( {#1 +0.1}, {0 + diff(#1, 0) *0.1});
- \draw[gray] ( {#1 -0.1}, {-1 - diff(#1, -1) *0.1}) -- ( {#1 +0.1}, {-1 + diff(#1, -1) *0.1});
- \draw[gray] ( {#1 -0.1}, {-2 - diff(#1, -2) *0.1}) -- ( {#1 +0.1}, {-2 + diff(#1, -2) *0.1});
- \draw[gray] ( {#1 -0.1}, {-3 - diff(#1, -3) *0.1}) -- ( {#1 +0.1}, {-3 + diff(#1, -3) *0.1});
- \draw[gray] ( {#1 -0.1}, {-4 - diff(#1, -4) *0.1}) -- ( {#1 +0.1}, {-4 + diff(#1, -4) *0.1});
- }
- \end{axis}
- \end{tikzpicture}
+ If \({\frac{dy}{dx}}=g(y)\), then
+ \(\frac{dx}{dy} = 1 \div \frac{dy}{dx} = \frac{1}{g(y)}\). Integrate both sides to solve equation. Only add \(c\) on one side. Express
+ \(e^c\) as \(A\).
+
+ \subsection*{Reciprocal derivatives}
+
+ \[\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}\]
+
+ \subsection*{Differentiating \(x=f(y)\)}
+ Find \(\dfrac{dx}{dy}\), then \(\dfrac{dy}{dx} = \dfrac{1}{\left(\dfrac{dx}{dy}\right)}\)
\subsection*{Parametric equations}
\[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\]
- \subsubsection*{Definite integrals}
-
- \[\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)\]
-
- \begin{itemize}
-
- \item
- Signed area enclosed by\\
- \(\> y=f(x), \quad y=0, \quad x=a, \quad x=b\).
- \item
- \emph{Integrand} is \(f\).
- \end{itemize}
-
\subsubsection*{Properties}
\begin{align*}
\(\sin 2x = 2 \sin x \cos x\)
\end{itemize}
+ \subsection*{Separation of variables}
+
+ If \({\frac{dy}{dx}}=f(x)g(y)\), then:
+
+ \[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]
+
\subsection*{Partial fractions}
To factorise \(f(x) = \frac{\delta}{\alpha \cdot \beta}\):
To reverse, use \texttt{combine(...)}
\end{cas}
+ \subsection*{Integrating \(\boldsymbol{\dfrac{dy}{dx} = g(y)}\)}
+
+ \[ \text{if } \dfrac{dy}{dx} = g(y), \text{ then } x = \int{\dfrac{1}{g(y)}} \> dy \]
+
\subsection*{Graphing integrals on CAS}
\begin{cas}
For restrictions, \texttt{Define\ f(x)=...} then \texttt{f(x)\textbar{}x\textgreater{}...}
\end{cas}
- \subsection*{Applications of antidifferentiation}
-
- \begin{itemize}
-
- \item
- \(x\)-intercepts of \(y=f(x)\) identify \(x\)-coordinates of
- stationary points on \(y=F(x)\)
- \item
- nature of stationary points is determined by sign of \(y=f(x)\) on
- either side of its \(x\)-intercepts
- \item
- if \(f(x)\) is a polynomial of degree \(n\), then \(F(x)\) has degree
- \(n+1\)
- \end{itemize}
-
- To find stationary points of a function, substitute \(x\) value of given
- point into derivative. Solve for \({\frac{dy}{dx}}=0\). Integrate to find
- original function.
-
\subsection*{Solids of revolution}
Approximate as sum of infinitesimally-thick cylinders
\end{enumerate}
\end{cas}
+ \subsection*{Applications of antidifferentiation}
+
+ \begin{itemize}
+
+ \item
+ \(x\)-intercepts of \(y=f(x)\) identify \(x\)-coordinates of
+ stationary points on \(y=F(x)\)
+ \item
+ nature of stationary points is determined by sign of \(y=f(x)\) on
+ either side of its \(x\)-intercepts
+ \item
+ if \(f(x)\) is a polynomial of degree \(n\), then \(F(x)\) has degree
+ \(n+1\)
+ \end{itemize}
+
+ To find stationary points of a function, substitute \(x\) value of given
+ point into derivative. Solve for \({\frac{dy}{dx}}=0\). Integrate to find
+ original function.
+
\subsection*{Rates}
\subsubsection*{Gradient at a point on parametric curve}
To verify solutions, find \(\frac{dy}{dx}\) from \(y\) and substitute into original
\end{warning}
- \subsubsection*{Function of the dependent
- variable}
-
- If \({\frac{dy}{dx}}=g(y)\), then
- \(\frac{dx}{dy} = 1 \div \frac{dy}{dx} = \frac{1}{g(y)}\). Integrate both sides to solve equation. Only add \(c\) on one side. Express
- \(e^c\) as \(A\).
-
\subsubsection*{Mixing problems}
\[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\]
- \subsubsection*{Separation of variables}
+ \subsection*{Euler's method}
- If \({\frac{dy}{dx}}=f(x)g(y)\), then:
+ \[\dfrac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
- \[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]
+ \[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
- \subsubsection*{Euler's method for solving DEs}
+ \begin{theorembox}{}
+ If \(\dfrac{dy}{dx} = g(x)\) with \(x_0 = a\) and \(y_0 = b\), then:
+ \begin{align*}
+ x_{n+1} &= x_n + h \\
+ y_{n+1} &= y_n + hg(x_n)
+ \end{align*}
+ \end{theorembox}
- \[\frac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
- \[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
\include{calculus-rules}
\end{align*}
\noindent \textbf{Distance travelled between \(t=a \rightarrow t=b\):}
- \[= \int^b_a \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \cdot dt \]
+ \begin{align*}
+ &= \int^{b}_{a}{\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}} \> dt \tag{2D} \\
+ &= \int^{t=b}_{t=a}{\dfrac{dx}{dt}} \> dt \tag{linear}
+ \end{align*}
\noindent \textbf{Shortest distance between \(\boldsymbol{r}(t_0)\) and \(\boldsymbol{r}(t_1)\):}
\[ = |\boldsymbol{r}(t_1) - \boldsymbol{r}(t_2)| \]