1--- 2geometry: margin=2cm 3columns: 2 4graphics: yes 5--- 6 7# Graphing techniques 8 9## Reciprocal continuous functions 10 11If $y=f(x)$, the reciprocal function is: 12 13$$y={1 \over f(x)}$$ 14 15As $\quad f(x) \rightarrow \pm \infty,\quad {1 \over f(x)} \rightarrow 0^\pm$ (vert asymptote at $f(x)=0$) 16 17<!-- As $\quad x \rightarrow \pm \infty,\quad {-1 \over x}$ --> 18 19\includegraphics[width=0.25\textwidth]{./graphics/recip-parabola.png} 20\includegraphics[width=0.25\textwidth]{./graphics/recip-sin-cos.png} 21 22- reciprocal functions are always on the same side of $x=0$ 23- if $y=f(x)$ has a local max|min at $x=1$, then $y={1 \over f(x)}$ has a local max|min at $x=a$ 24- point of inflection at $P(1,1)$ 25 26## Locus of points 27 28- set of points that satisfy a given condition 29- path traced by a point that moves according to a condition 30- graph on CAS - **conics** 31 32### Circular loci 33 34point $P(x,y)$ has a constant distance $r$ from point $C(a,b)$ (centre) 35 36 37$$PC = r$$ 38 39$$(x-a)^2 + (y-b)^2 = r^2$$ 40 41 42 43### Linear loci 44 45$$QP = RP $$ 46$$\sqrt{(x_Q-q_P)^2+(y_Q-y_P)^2} = \sqrt{(x_R-x_P)^2+(y_R-y_P)^2}$$ 47 48points $Q$ and $R$ are fixed and have a perpendicular bisector $QR$. Therefore, any point on line $y=mx+c$ is equidistant from $QP$ and $RP$. 49 50Since the bisector of the line joining points $Q$ and $R$ is perpendicular to $QR $: 51 52$$m( QR ) \times m( RP ) = -1$$ 53 54### Parabolic loci 55 56$$PD = PF $$ 57$$|y-z|=\sqrt{(x-x_F)^2+(y-y_F)^2}$$ 58$$(y-z)^2=(x-x_F)^2+(y-y_F)^2$$ 59 60Distance of point $P(x,y)$ from fixed point $F(a,b)$ is equal to the distance of $P$ from $y=z \perp$. 61 62Fixed point $F$ is the **focus** (halfway between $y=z$ and $y=y_P$) 63 64Fixed line $x=z$ is the **directrix** 65 66### Elliptical loci 67 68Point $P$ moves so that the sum of its distances from two fixed points $F_1$ and $F_2$ is a constant $k$. 69 70$${F_1 P} + F_2 P =k$$ 71 72**Two** foci at $F_1$ and $F_2$ 73 74Cartesian equation for ellipses: 75$${(x-h)^2 \over a^2} + {(y-k)^2 \over b^2} = 1$$ 76centered at $(h,k)$. Width is $2a$, height is $2b$. 77 78### Transformations 79$$(x,y) \rightarrow (x \prime, y \prime)$$ 80 81where $x \prime$ and $y \prime$ are the transformation factors (dilation away from $x$-axis means coefficient of $y$ increases in $y \prime$, and vice versa). 82 83Transformed equation is the same as initial equation with each term divided by its dilation coefficients (must be in terms of $x\prime$ and $y\prime$). 84 85e.g. 86 87$x^2 + y^2 = 1$ is dilated $3$ from $x$, $5$ from $y$. 88Transformation rule is $(x\prime,y\prime) = (5x,3y)$ 89$x={x\prime \over 5},\quad y={y\prime \over 3}$ 90 91Equation $x^2 + y^2=1$ becomes 92 93$${(x\prime)^2 \over 25}+ {(y\prime)^2 \over 9}=1$$ 94 95 96### Hyperbolic loci 97 98$$|(F_2P - F_1P )| = k$$ 99 100Cartesian equation for hyperbolas ($a$ and $b$ are dilation factors): 101$${(x-h)^2 \over a^2} - {(y-k)^2 \over b^2} = 1$$ 102 103Distance between vertices is $2a$ 104Vertices given by $(h \pm a, k)$ 105 106Asymptotes at $y=\pm {b \over a}(x-h)+k$ 107To make hyperbola up/down rather than left/right, swap $x$ and $y$ 108 109$y^2-x^2=1$ produces hyperbola shifted 90 $^\circ$ (top and bottom of asymptotes) 110 111## Parametric equations 112 113Parametric curve: 114 115$$x=f(t), \quad y=g(t)$$ 116 117$t$ is the parameter 118 119To convert to cartesian, solve like simultaneous equations 120 121## Polar coordinates 122 123$$x = r\cos\theta, \quad y = r\sin\theta$$ 124 125### Spirals 126$$r={\theta \over n\pi}$$ 127- solve intercepts for multiples of $\pi \over 2$ 128- or draw table of values for $r$ and $\theta$ for each $n\pi \over 2$ 129 130### Circles 131$$r=a$$ 132 133### Lines 134 135Horizontal: $r={n \over \sin \theta}$ 136Vertical: $r={n \over \cos \theta}$ 137 138### Cardioids 139 140$$r=a(n+ \cos\theta)$$ 141 142### Roses 143 144$$r=\cos(k\theta)$$ 145 146If $k$ is odd, half of the petals will overlap (hence there are $n$ petals) 147 148If $k$ is even, petals will not overlap (hence $2n$ petals) 149 150\includegraphics[width=0.5\textwidth]{./graphics/rose.png} 151 152 153### Solving polar graphs 154 155solve in terms of $r$ 156 157e.g. $x=4$ 158 159$r\cos\theta = 4$ 160 161$r={4 \over \cos\theta}$ 162 163 164--- 165 166e.g. $y=x^2$ 167 168$r\sin\theta = r^2 \cos^2\theta$ 169 170$\sin \theta = r \cos^2\theta$ 171 172$r = {\sin \theta \over \cos^2\theta} = \tan\theta \sec\theta$ 173 174--- 175 176e.g. $r=6\cos \theta\quad$ *(multiply by $r$)* 177 178$r^2=6r\cos\theta$ 179 180$x^2+y^2=6x$ 181 182complete the square