%%%%NOTE: this is ADD-MATH, not GCSE \input{../include.tex} \input{../units.tex} \usepackage{keystroke} \setmainfont{Junicode} \begin{document} \title{Revision Notes for OCR ADD-Math} \author{} \date{2008--2009} \maketitle \section{Algebra} You must be very proficient at GCSE level algebra. In addition you may meet the following: \subsection{Simplifying Algebraic Expressions} \paragraph{Example} \[\frac{2}{x + 2} + \frac{3}{x} - \frac{4}{x^2 + 2x}\] Factorise all denominators if possible: \[ = \frac{2}{x + 2} + \frac{3}{x} - \frac{4}{x(x + 2)} \] Multiply top and bottom of each term by something to make the denominators of each term the same. \[ = \frac{2x}{x(x + 2)} + \frac{3(x + 2)}{x(x + 2)} - \frac{4}{x(x + 2)} \] Write as one fraction: \[ = \frac{2x + 3(x + 2) - 4}{x(x + 2)} \] Simplify: \[ = \frac{5x + 2}{x^2 + 2x} \] \subsection{Solving Equations with Fractions} \paragraph{Example} \[\mathrm{Solve} \:\: \frac{3}{3x - 2} - \frac{4}{x + 1} = \frac{2}{(3x - 2)(x + 1)}\] Multiply each term by the Lowest Common Multiple (LCM) of the denominator: \begin{eqnarray*} \frac{3\cancel{(3x - 2)}(x + 1)}{\cancel{3x - 2}} - \frac{4(3x - 2)\cancel{(x + 1)}}{\cancel{x + 1}} & = & \frac{2\cancel{(3x - 2)}\cancel{(x + 1)}}{\cancel{(3x - 2)}\cancel{(x + 1)}} \\ 3(x + 1) - 4(3x - 2) & = & 2 \\ 3x + 3 - 12x + 8 & = & 2 \\ -9x & = & -9 \\ x & = & 1 \\ \end{eqnarray*} \note{In this case the equation leads to a single value of $x$. Be prepared also for a quadratic ending needing factorising, or the formula.} \subsection{Indices} \[ \begin{array}{rcl|rcl} \multicolumn{3}{c}{\textbf{\large Facts}} & \multicolumn{3}{c}{\textbf{\large Examples}} \\ a^m \times a^n & = & a^{m + n} & a^7 \times a^5 &=& a^{12} \\ a^m \div a^n & = & a^{m - n} & a^9 \div a^5 &=& a^4 \\ (a^m)^n & = & a^{mn} & (a^2)^5 &=& a^{10} \\ a^{\frac{1}{n}} & = & \sqrt[n]{a} & 8^\frac{1}{3} &=& 2 \\ a^0 & = & 1 & (6.8)^0 &=& 1 \\ a^{-m} & = & \frac{1}{a^m} & 8^{-2} &=& \frac{1}{64} \\ a^{\frac{p}{q}} & = & (\sqrt[q]{a})^p = \sqrt[q]{a^p} & 4^\frac{3}{2} &=& 8 \\ \end{array} \] A harder example: \[\bfrac{27}{8}^{-\frac{4}{3}} = \frac{1}{\sqrt[3]{\bfrac{27}{8}^4}} = \frac{16}{81}\] \subsection{Quadratic Equations} Try factorising: \begin{eqnarray*} 3x^2 - 11x &=& 0 \\ (3x - 2)(x - 3) &=& 0 \\ \end{eqnarray*} \[\mathrm{Either:} \:\: x = 3 \: \mathrm{or} \: x = \frac{2}{3}\] If it does not factorise or if significant figures/decimal places are mentioned then use the quadratic formula: \framemath{\[ ax^2 + bx + c = 0 \:\:\:\: \Leftrightarrow \:\:\:\: x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]} Note that for real roots $b^2 - 4ac \geq 0$. If you get the square root of a negative number then you have gone wrong -- probably a mistake with signs. \paragraph{Example} Solve the equation $2x^2 -7x - 5 = 0$, giving you answers correct to 2 decimal places. \begin{eqnarray*} x &=& \frac{7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} \\ x &=& \frac{7 \pm \sqrt{89}}{4} \\ x &=& 4.11 \:\: \mathrm{or} \:\: -0.61 \:\: \mathrm{(to\: 2\: d.p.)} \\ \end{eqnarray*} \subsection{Problems Leading to Equations} \paragraph{Example 1} The product of three consecutive positive odd whole numbers when added to $k$ times the middle number is equal to the cube of the middle number. Find the value of $k$. \[(2n + 1)(2n + 3)(2n + 5) + k(2n + 3) = (2n + 3)^3\] Divide each term by $(2n + 3)$, and expand and simplify: \begin{eqnarray*} 4n^2 + 12n + 5 + k &=& 4n^2 + 12n + 9 \\ 5 + k &=& 9 \\ k &=& 4 \\ \end{eqnarray*} \paragraph{Example 2} On a regular journey of 200km a car driver averages $x\mathrm{kmh}^{-1}$. When he buys a new car he finds that his average speed increases by $10\mathrm{kmh}^{-1}$ and he saves half an hour on the journey. Find his new average speed. \begin{eqnarray*} \mathrm{Old \: time} - \mathrm{New \: time} &=& \mathrm{Time \: saved} \\ \frac{200}{x} - \frac{200}{x + 10} &=& \frac{1}{2} \\ 400(x + 10) - 400x &=& x(x + 10) \\ x^2 + 10x - 4000 &=& 0 \\ \end{eqnarray*} Using the quadratic formula: \begin{eqnarray*} x &=& \frac{-10 \pm \sqrt{100 + 4 \cdot 1 \cdot 4000}}{2} \\ x &=& 58.4\mathrm{kmh}^{-1} \:\: \mathrm{or} \:\: -73.47 \\ \end{eqnarray*} His new average speed is $68.4\mathrm{kmh}^{-1}$ (we assume that he is driving forwards, rather than backawards, so the negative root is ignored). \subsection{Simultaneous Equations (one linear, one quadratic)} \paragraph{Method} Isolate either $x$ or $y$ in the linear equation, substitute it into the quadratic equation, solve, then substitute back into the linear equation to find the other unknown. \paragraph{Example} \[ \left\{ \begin{array}{lcrl} xy &=& -6 & \mathrm{this\: is\: quadratic} \\ 2x + 3y &=& -5 & \mathrm{this\: is\: linear} \\ \end{array} \right. \] From the linear equation $x = \frac{1}{2}(-5 - 3y)$. Substituting into $xy = -6$ gives: \begin{eqnarray*} \frac{1}{2}(-5 - 3y)y &=& -6 \\ -5y - 3y^2 &=& -12 \\ 3y^2 + 5y - 12 &=& 0 \\ (3y - 4)(y + 3) &=& 0 \\ \end{eqnarray*} Either $y = -3$ and $x = 2$ or $y = 1\frac{1}{3}$ and $x = -4\frac{1}{2}$. \subsection{Binomial Theorem} \framemath{\[ (a + b)^n = a^n + \binom{n}{1}a^{n - 1}b + \binom{n}{2}a^{n - 2}b^2 + \cdots + \binom{n}{r}a^{n - r}b^r \cdots + n^n \] \[ \binom{n}{r} = \frac{n!}{(n - r)!r!} \]} \paragraph{Example} Find the first 4 items in the expansion of $1 - 2x)^7$. \begin{eqnarray*} (1-2x)^7 &=& 1^7 + \binom{7}{1}1^6(-2x) + \binom{7}{2}1^5(-2x)^2 + \binom{7}{3}1^4(-2x)^3 + \cdots \\ &=& 1 - 14x + 84x^2 - 280x^3 + \cdots \\ \end{eqnarray*} \paragraph{Pascal's Triangle} could be used: \[\begin{array}{ccccccccccccccc} & & & & & & & 1 & & & & & & & \\ & & & & & & 1 & & 1 & & & & & & \\ & & & & & 1 & & 2 & & 1 & & & & & \\ & & & & 1 & & 3 & & 3 & & 1 & & & & \\ & & & 1 & & 4 & & 6 & & 4 & & 1 & & & \\ & & 1 & & 5 & & 10 & & 10 & & 5 & & 1 & & \\ & 1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1 & \\ 1 & & 7 & & 21 & & 35 & & 35 & & 21 & & 7 & & 1 \\ \end{array}\] The last line gives the values of $7\choose0$, $7\choose1$, $7\choose2$, etc. \note{Many calculators will give $\binom{n}{r}$ using the \keystroke{$_nC_r$} button. Check that \keystroke{7} \keystroke{$_nC_r$} \keystroke{2} $= {7\choose2} = 21$ on your calculator.} \paragraph{Numerical Applications -- Example} Expand $(1 - 2x)^7$ as far as the term in $x^3$. Use your expandsion to find $(0.998)^7$ correct to 7 significant figures. \begin{eqnarray*} \mathrm{From above:} \:\:\:\: (1 - 2x)^7 &=& 1 - 14x + 84x^2 - 280x^3 + \cdots \\ \mathrm{If} \:\: x = 0.001 \:\:\:\: (0.998)^7 &=& 1 - 0.014 + 0.000084 - 0.00000028 + \cdots \\ &=& 0.9860837 \cdots \end{eqnarray*} \note{Calculators are normally not allowed in answering this type of question.} \section{Plane Cartesian Geometry} \subsection{Straight Lines} \subsubsection{Gradient} \begin{tikzpicture} \fill[green!20!white] (-0.5,-0.5) rectangle (4.5cm,4.5cm); \draw (0,0) -- (1,1) node (second) [label=right:{$(x_2,y_2)$}] {} -- (3,3) node (first) [label=left:{$(x_1,y_1)$}] {} -- (4,4); \drawcross{first} \drawcross{second} \draw (6,2) node [right,text width=6cm] {\begin{minipage}{\textwidth} \begin{eqnarray*}% \mathrm{Gradient} &=& \frac{y_1-y_2}{x_1-x_2} \\% \mathrm{or\:\: Gradient} &=& \frac{y_2-y_1}{x_2-x_1} \\% \end{eqnarray*} \end{minipage}}; \end{tikzpicture} \subsubsection{Equation}\label{gradient1point} (given gradient and 1 point) \begin{tikzpicture} \fill[green!20!white] (-0.5,-0.5) rectangle (4.5cm,4.5cm); \draw (0,0) -- (2,2) node (first) [label=left:{$(x_1,y_1)$}] {} -- (4,4); \drawcross{first} \draw (0.5,4) node [right,text width=6cm] {$\mathrm{Gradient} = m$}; \draw (0.5,4) node [right,text width=6cm] {$\mathrm{Gradient} = m$}; \draw (4.75,2) node [right,text width=5cm] {\begin{minipage}{\textwidth} {\centering \textbf{Method 1}}\\ \[\frac{y-y_1}{x-x_1} = m\] Put in the values of $x_1$, $y_1$, and $m$ and rearrange the equation. \end{minipage}}; \draw (10,4.5) -- (10,-0.5); \draw (10.5,2) node [right,text width=5cm] {\begin{minipage}{\textwidth} {\centering \textbf{Method 2}}\\ \[\mathrm{Use} \:\: y = mx + c\] Put in values of $x$ and $y$ from the given point and the given value of $m$. Hence find value of $c$. Rewrite the equation with $m$ and $c$ instead. \end{minipage}}; \end{tikzpicture} \subsubsection{Equation} (given two points on the line) \begin{tikzpicture} \fill[green!20!white] (-0.5,-0.5) rectangle (4.5cm,4.5cm); \draw (0,0) -- (1,1) node (second) [label=right:{$(x_2,y_2)$}] {} -- (3,3) node (first) [label=left:{$(x_1,y_1)$}] {} -- (4,4); \drawcross{first} \drawcross{second} \draw (4.75,2) node [right,text width=5cm] {\begin{minipage}{\textwidth} {\centering \textbf{Method 1}}\\ Substitute into the formula: \[\frac{y-y_1}{x-x_1} = \frac{y_1-y_2}{x_1-x_2}\] then rearrange the equation. \end{minipage}}; \draw (10,4.5) -- (10,-0.5); \draw (10.5,2) node [right,text width=5cm] {\begin{minipage}{\textwidth} {\centering \textbf{Method 2}}\\ Find the gradient as before ($m$). \[\mathrm{Use} \:\:\:\: y = mx + c\] Put in values of $x$ and $y$ for one fo the points and the value of $m$. Hence find the value of $c$. Rewrite $y = mx + c$ with values of $m$ and $c$. \end{minipage}}; \end{tikzpicture} \subsubsection{Parallel Lines} \begin{tikzpicture} \fill[green!20!white] (-1.5,-0.5) rectangle (5.5cm,4.5cm); \draw[->] (0,0) -- (1.5,2) node[above left] {$\mathrm{Gradient} = m_1$}; \draw (1.5,2) -- (3,4); \draw[->] (1,0) -- (2.5,2) node[below right] {$\mathrm{Gradient} = m_2$}; \draw (2.5,2) -- (4,4); \draw (6,2) node [right,text width=5cm] {\begin{minipage}{\textwidth} Parallel lines have equal gradients i.e.\ $m_1 = m_2$. \end{minipage}}; \end{tikzpicture} \subsubsection{Perpendicular Lines} \begin{tikzpicture} \fill[green!20!white] (-1.5,-0.5) rectangle (5.5cm,4.75cm); \draw (4,0) -- (2,2) node (intersec) {} -- (0,4) node[above] {$\mathrm{Gradient} = m_1$}; \draw (0,0) -- (4,4) node[above] {$\mathrm{Gradient} = m_2$}; \draw[rotate=45] (intersec) rectangle +(8pt,8pt); \draw (6,2) node [right,text width=5cm] {\begin{minipage}{\textwidth} The product of the gradients is $-1$, i.e. \[m_1m_2 = -1\] If $m_1 = 3$ \[m_2 = \frac{1}{3} \:\:\mathrm{etc.}\] \end{minipage}}; \end{tikzpicture} \subsection{Distance Between Two Points} \begin{tikzpicture} \fill[green!20!white] (-0.5,-0.5) rectangle (4.5cm,4.5cm); \draw (0,0) -- (1,1) node (second) [label=right:{$(x_2,y_2)$},label=left:B] {} -- (3,3) node (first) [label=left:{$(x_1,y_1)$},label=right:A] {} -- (4,4); \drawcross{first} \drawcross{second} \draw (6,2) node [right,text width=6cm] {\begin{minipage}{\textwidth} \[\mathrm{Distance} \:\:\:\:AB = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\] \end{minipage}}; \end{tikzpicture} \subsubsection{Midpoints} See diagram above. The midpoint of AB has co-ordinates: \[ \left( \frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} \right) \] \section{Calculus} \subsection{Differentiation} \framemath{\[y = x^n \:\: \Rightarrow \:\: \frac{\diff y}{\diff x} = nx^{n - 1}\]} \paragraph{Note} If $y = c$ (any constant) then $\frac{\diff y}{\diff x} = 0$. If $y = ax^n$, $\frac{\diff y}{\diff x} = a \cdot nx^{n-1}$ %Function notation may be used, e.g.\ If $f(x) = x^2$, $f'(x) = 2x$, $f''(x) = 2$. \paragraph{Examples} \begin{eqnarray*} y = 3x^2 &\Rightarrow& \frac{\diff y}{\diff x} = 6x \Rightarrow \frac{\diff^2y}{\diff x^2} = 6 \\ %f(x) = 6x^2 - 5 &\Rightarrow& f'(x) = 12x \\ y = \frac{1}{x} = x^{-1} &\Rightarrow& \frac{\diff y}{\diff x} = -1 \cdot x^{-2} = \frac{-1}{x^2} \\ %f(x) = \sqrt{x} = x^\frac{1}{2} &\Rightarrow& f'(x) = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}} \\ y = \frac{1}{4x^2} = \frac{1}{4} \cdot \frac{1}{x^2} = \frac{1}{4}x^{-2} &\Rightarrow& \frac{\diff y}{\diff x} = \frac{1}{4} \cdot -2x^{-3} = \frac{-1}{2x^3} \\ \end{eqnarray*} \subsection{Integration} \framemath{\[ \int x^n\diff x = \frac{x^{n + 1}}{n + 1} + c \] \hfill $(n \neq -1)$} \paragraph{Note} \[\int a \diff x = ax + c \:\: \therefore \:\: \int ax^n \diff x = a \cdot \frac{x^{n + 1}}{n + 1} + c = \frac{ax^{n + 1}}{n + 1} + c\] \paragraph{Examples} \begin{eqnarray*} \int x^2 \diff x &=& \frac{x^3}{3} + c \\ \int \frac{1}{x^3} \diff x = \int x^{-3} \diff x &=& \frac{x^{-2}}{-2} + c = \frac{-1}{2x^2} + c\\ \int \sqrt[3]{x^4} \diff x = \int x^\frac{4}{3} \diff x &=& \frac{x^\frac{7}{3}}{\frac{7}{3}} + c = \frac{3}{7} \cdot \sqrt[3]{x^7} + c \\ \int \frac{1}{3x^5} \diff x = \int \frac{1}{3}x^-5 \diff x &=& \frac{1}{3} \cdot \frac{x^{-4}}{-4} + c = \frac{-1}{12x^4} + c \\ \int \left(3x^3 - 2 + \frac{5}{x^2}\right) \diff x = \int (3x^3 - 2 + 5x^{-2}) \diff x &=& \frac{3}{4}x^4 - 2x + \frac{5x^{-1}}{-1} + c = \frac{3}{4}x^4 - 2x - \frac{5}{x} + c \\ \end{eqnarray*} $\frac{\diff y}{\diff x} = 6x^2 - 10x + 7$ is the gradient function of a function passing through $(2,5)$. Then: \begin{eqnarray*} y &=& \frac{6x^3}{3} - \frac{10x^2}{2} + 7x + c \\ y &=& 2x^3 - 5x^2 + 7x + c \\ \therefore 5 &=& 2 \cdot 2^3 - 5 \cdot 2^2 + 7 \cdot 2 + c \\ \therefore c &=& -5 \\ \end{eqnarray*} \subsection{Gradients of Curves} The gradient of a curve at a particular point is the value of the derivative at that point. Hence, first differentiate the function and then put in the value of $x$ at that point. \paragraph{Example} Find the gradient of the curve $y = 3x^2 - 2x$ at the point $(2,8)$. \[ \frac{\diff y}{\diff x} = 6x - 2 \] \[ \mathrm{When} \:\: x = 2, \frac{\diff y}{x} = 6 \times 2 - 2 = gradient = 10 \] \subsection{Tangents to Curves} To find the equation of a tangent to a curve, first find the gradient of the curve at that point, then use the method in ``given Gradient and 1 Point'' on page \pageref{gradient1point}. \paragraph{Example} Find the equation of the tangent to the curve $y = 3x^2 + 5x - 2$ at the point $(1,6)$. \[ \mathrm{If} \:\: y = 3x^2 + 5x - 2 \] \[ \mathrm{then} \:\: \frac{\diff y}{\diff x} = 6x + 5\mathrm{.} \] \[ \mathrm{When} \:\: x = 1, \frac{\diff y}{\diff x} = 11 \] Therefore we need a line with gradient $11$ that goes through $(1,6)$. \[ \frac{y - 6}{x - 1} = 11 \] Equation of tangent is $y = 11x - 5$. \subsection{Normals to Curves} The normal to a curve at a point is perpendicular to the tangent at that point. First find the gradient of the curve and put in the value of $x$ at that point. Take the negative reciprocal of the gradient of the tangent to find the gradient of the normal. Then use the method in ``given Gradient and 1 Point'' on page \pageref{gradient1point}. \paragraph{Example} Find the equation of the normal to the curve $y = 3x^2 + 5x - 2$ at the point $(1,6)$. \[ \mathrm{If} \:\: y = 3x^2 + 5x - 2 \] \[ \mathrm{then} \:\: \frac{\diff y}{\diff x} = 6x + 5 \] \[ \mathrm{When} \:\: x = 1, \frac{\diff y}{\diff x} = 11 \] Gradient of tangent is then 11 so gradient of normal is $\frac{-1}{11}$. \[ \frac{y - 6}{x - 1} = \frac{-1}{11} \] Equation of normal is $y = -\frac{1}{11}x + \frac{67}{11}$ \subsection{Finding Curves, Given the Gradient Function} \paragraph{Example} For a particular curve it is known that $\frac{\diff y}{\diff x} = x^2 + 2x - 3$ and that the curve passes through the point $(3,11)$. Find the equation of the curve. \begin{eqnarray*} \frac{\diff y}{\diff x} &=& x^2 + 2x - 3 \\ y &=& \frac{1}{3}x^3 + x^2 - 3x + c \\ \end{eqnarray*} But it is known that the point $(3,11)$ lies on this curve. \begin{eqnarray*} 11 &=& 9 + 9 - 9 + c \\ c &=& 2 \\ \end{eqnarray*} Equation of the curve is then $y = \frac{1}{3}x^3 + x^2 - 3x + 2$ \subsection{Maxima \& Minima} \paragraph{Routine Procedure} \begin{enumerate} \item Express the quantity to be maximised/minimised $(y)$ in terms of a single variable $(x)$. \item Find $\frac{\diff y}{\diff x}$. \item Put $\frac{\diff y}{\diff x} = 0$, solve the equation to find value(s) of $x$. \item Find the corresponding values of $y$ from the original equation. \item Find the gradient on each side of each turning point found above. A gradient change from $+$ to $-$ as $x$ increases shows a maximum. A gradient change from $-$ to $+$ as $x$ increases shows a minimum. \begin{framed} Alternatively, this method may be used: \\ % If $f''(x)$ or $\frac{\diff^2y}{\diff x^2}$ is positive at the turning point $\Rightarrow$ minimum. \\ % If $f''(x)$ or $\frac{\diff^2y}{\diff x^2}$ is negative at the turning point $\Rightarrow$ maximum. \end{framed} \item Answer the question. \end{enumerate} \paragraph{Example 1} Sketch the curve $y = x(x^2 - 3)$. \[ y = x(x^2 - 3) = x^3 - 3x \] \[ \frac{\diff y}{\diff x} = 3x^2 - 3 \] \[ \left[\frac{\diff^2y}{\diff x^2} = 6x\right]\] For maximum or minimum values $\frac{\diff y}{\diff x} = 0$ so $3x^2 - 3 = 0$ \[ 3(x - 1)(x + 1) = 0 \] \[ x = 1 \:\: \mathrm{or} \:\: x = -1 \] Turning points are then $(1,-2)$ and $(-1,2)$. \bigskip \begin{minipage}{0.4\textwidth} {\centering \textbf{For the point $(1,-2)$}} \[\begin{array}{c|c|c|c} x & 0 & 1 & 2 \\ \hline \mathrm{Grad} & - & 0 & + \\ \end{array}\] \[\left[\frac{\diff^2y}{\diff x^2} = +6\right]\] $\therefore$ minimum. \end{minipage} \hfill \vline \hfill \begin{minipage}{0.4\textwidth} {\centering \textbf{For the point $(-1,2)$}} \[\begin{array}{c|c|c|c} x & -2 & -1 & 0 \\ \hline \mathrm{Grad} & + & 0 & - \\ \end{array}\] \[\left[\frac{\diff^2y}{\diff x^2} = -6\right]\] $\therefore$ maximum. \end{minipage} \bigskip \begin{centering} \begin{tikzpicture}[scale=1.5] \fill[blue!10!white] (-5,-2.5) rectangle (5,2.5); \draw[ystep=0.5cm,xstep=1cm,color=black!30!white] (-4.5,-2.25) grid (4.5,2.25); \draw[->] (0,-2.25) -- (0,-2) node[label=left:{$-2$}] {} -- (0,2) node[label=left:{$2$}] {} -- (0,2.25); \draw (0,-2) -- +(-4pt,0); \draw (0,2) -- +(-4pt,0); \draw[->] (-4.5,0) -- (-4,0) node[label=below:{$-2$}] {} -- (-2,0) node[label=below:{$-1$}] {} -- (2,0) node[label=below:{$1$}] {} -- (4,0) node[label=below:{$2$}] {} -- (4.5,0); \draw (-4,0) -- +(0,-4pt); \draw (-2,0) -- +(0,-4pt); \draw (2,0) -- +(0,-4pt); \draw (4,0) -- +(0,-4pt); \draw plot[domain=-2:2,samples=50] ({(\x)*2},{(\x*(\x^2 - 3))}); \draw (-3.464,0) node[label=above left:{$-\sqrt{3}$}] {} -- +(2pt,2pt) -- +(-2pt,-2pt) -- +(2pt,2pt) -- (-3.464,0) -- +(2pt,-2pt) -- +(-2pt,2pt) -- +(2pt,-2pt); \draw (3.464,0) node[label=above left:{$\sqrt{3}$}] {} -- +(2pt,2pt) -- +(-2pt,-2pt) -- +(2pt,2pt) -- (3.464,0) -- +(2pt,-2pt) -- +(-2pt,2pt) -- +(2pt,-2pt); \draw (-2,2) node[above] {\tiny maximum}; \draw (2,-2) node[below] {\tiny minimum}; \end{tikzpicture} \end{centering} \paragraph{Example 2} A cone is such that the sum of its diameter and height is 10cm. Find its volume. \begin{minipage}{0.25\textwidth} \begin{tikzpicture} \fill[blue!10!white] (-2.5,-1.5) rectangle (2.5,6.5); \draw[dashed] (-2,0) .. controls (-2,1.1045) and (2,1.1045) .. (2,0); \draw (-2,0) .. controls (-2,-1.1045) and (2,-1.1045) .. (2,0); \draw[dashed] (-2,0) -- (-1,0) node[below] {$r$ cm} -- (0,0) -- (0,1.5) node[right] {$10 - 2r$} -- (0,6); \draw (-8pt,0) -- +(0,8pt) -- +(8pt,8pt); \draw (-2,0) -- (0,6) -- (2,0); \end{tikzpicture} \end{minipage} \begin{minipage}{0.75\textwidth} \begin{eqnarray*} V &=& \frac{1}{3}\pi r^2(10 - 2r) \\ &=& \frac{10}{3}\pi r^2 - \frac{2}{3}\pi r^3 \\ \frac{dV}{dr} &=& \frac{10}{3}\pi 2r - \frac{2}{3}\pi 3r^2 \\ \left[\frac{\diff^2V}{dr^2}\right. &=& \left.\frac{20\pi}{3} - 4\pi r\right] \\ \end{eqnarray*} \hfill For maximum or minimum values $\frac{dV}{dr} = 0$. \hfill \hfill \begin{eqnarray*} \mathrm{Hence} \:\:\:\: \frac{20}{3}\pi r - \frac{2}{3}\pi 3r^2 &=& 0 \\ r &=& \frac{4}{3} \\ \end{eqnarray*} \[\begin{array}{c|c|c|c} r & 3 & \frac{4}{3} & 4 \\ \hline \mathrm{Grad} & + & 0 & - \\ \end{array} \:\:\:\: \left[\frac{\diff^2V}{dr^2} = \frac{20\pi}{3} - \frac{40\pi}{3} = -\frac{20\pi}{3}\right]\] \hfill This is a maximum. \hfill \hfill \end{minipage} Hence the volume of the cone is a maximum when $r = \frac{4}{3}$ and in this case $V = 38.8\mathrm{cm}^3$. \section{Trigonometric Functions} \subsection{$\cos$ and $\sin$} Graphs of $\cos \theta\degree$ and $\sin \theta\degree$, $0\leq\theta\leq 360$. \begin{eqnarray*} \sin 0 &=& 0\\ \sin 90 &=& 1\\ \sin 180 &=& 0\\ \sin 270 &=& -1\\ \sin 360 &=& 0\\ \cos 0 &=& 1\\ \cos 90 &=& 0\\ \cos 180 &=& -1\\ \cos 270 &=& 0\\ \cos 360 &=& 1\\ \end{eqnarray*} \begin{center} \begin{tikzpicture}[domain=0:2,scale=0.35,samples=100] \draw[->] (-0.2,0) -- (5,0) node[below] {$50$} -- (9,0) node[above] {$90$} -- (10,0) node[below] {$100$} -- (15,0) node[above] {$150$} -- (18,0) node[above] {$180$} -- (20,0) node[below] {$200$} -- (25,0) node[below] {$250$} -- (27,0) node[above] {$270$} -- (30,0) node[below] {$300$} -- (35,0) node[below] {$350$} -- (36,0) node[above] {$360$} -- (37,0) node[right] {$\theta\degree$}; \foreach \x in {5,9,10,15,18,20,25,27,30,36} \drawcrosses{\x,0}{6pt}; \draw[->] (0,-10.9) -- (0,-10) node[left] {$-1.0$} -- (0,-5) node[left] {$-0.5$} -- (0,0) node[left] {$0$} -- (0,5) node[left] {$0.5$} -- (0,10) node[left] {$1.0$} -- (0,10.9) node[above,blue] (cos) {$\cos \theta\degree$}; \foreach \x in {-10,-5,0,5,10} \drawcrosses{0,\x}{6pt}; \node at (cos.north) [above,red] (sin) {$\sin \theta\degree$}; \draw[red,xscale=18,yscale=10] plot (\x,{sin((\x*pi) r)}); \draw[blue,xscale=18,yscale=10] plot (\x,{cos((\x*pi) r)}); \end{tikzpicture} \end{center} \subsection{$\tan$} Graph of $\tan \theta\degree$, $0\leq\theta\leq 360$. \begin{eqnarray*} \tan 0 &=& 0\\ \tan 90 &\simeq& \infty\\ \tan 180 &=& 0\\ \tan 270 &\simeq& \infty\\ \tan 360 &=& 0\\ \end{eqnarray*} \begin{center} \begin{tikzpicture}[samples=100] \begin{axis}[enlargelimits=false,ymax=4.2,ymin=-4.2,xmin=0,xmax=364,width=\textwidth,axis x line=middle,axis y line=left,extra x ticks={90,180,270,360},extra x tick style={yshift=16},tick style={black!100}] \addplot[purple,mark={}] plot[domain=0:89] (\x,{(sin(\x))/(cos(\x))}); \addplot[purple,mark={}] plot[domain=91:269] (\x,{(sin(\x))/(cos(\x))}); \addplot[purple,mark={}] plot[domain=271:359] (\x,{(sin(\x))/(cos(\x))}); \foreach \foo in {90,180,270,360} \addplot[dashed,dash pattern=on 4pt off 2pt,thin,mark={}] plot (\foo,\x); \end{axis} \end{tikzpicture} \end{center} \end{document}