Resources for Direction Fields
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Questions
17
With Worked SolutionClick Here -
Video Tutorials
1
Click Here -
HSC Questions
2
With Worked SolutionClick Here
Direction Fields Theory
![\begin{center} \resizebox{0.3\textwidth}{!}{ \begin{tikzpicture} \def\dx{0.5}; % x-spacing for ticks \def\dy{0.5}; % y-spacing for ticks \def\sx{-5}; % lower bound for x values \def\sy{-5}; % lower bound for y values \def\ex{5}; % upper bound for x values \def\ey{5}; % upper bound for y values \def\l{0.2}; % length of HALF of a segment % draw grid \draw[gray, thin] (-5,-5) grid (5,5); \draw[ultra thick,latex-latex,red] (-5.5,0) -- (5.5,0); \draw[ultra thick,latex-latex,red] (0,-5.5) -- (0,5.5); % draw slope ticks: \foreach \x in {-5,-4.5,...,5} { \foreach \y in {-5,-4.5,...,-0.5,0.5,1,...,5} { \pgfmathsetmacro{\m}{\x/\y}; \pgfmathsetmacro{\k}{\l/sqrt(1+\m*\m)}; \pgfmathsetmacro{\h}{\k*\m}; \draw[ thick,darkgray!90] (\x,\y) -- (\x+\k,\y+\h); \draw[ thick,darkgray!90] (\x,\y) -- (\x-\k,\y-\h); } } \foreach \i in {-5,...,-1,1,2,...,5} { \draw[thick] (\i,0)--(\i,-0.1); \node[below] at (\i,-0.1) {\large \i}; \draw[thick] (0,\i)--(-0.1,\i); \node[left] at (-0.1,\i) {\large \i}; } \end{tikzpicture} } \end{center} The direction field represents the first order differential equation \(y^{\prime}=\dfrac{x}{y}\). Consider the point \((2,2)\). \\ \(y^{\prime}=\dfrac{2}{2}=1\). On the direction field, the gradient \(\approx\) 1 \\ Consider the point \((-2,2)\)\\ \(y^{\prime}=\dfrac{-2}{2}=-1\). On the direction field the gradient \(\approx-1\).\\ $\begin{aligned} & \frac{d y}{d x}=\frac{x}{y} \\ & \displaystyle \int y \,d y=\displaystyle \int x \,d x \\ & \frac{y^2}{2}=\frac{x^2}{2}+c \\ & \therefore y^2=x^2+k \quad (k=2 c) \\ & \text{At }(2,2), \quad 4=4+k \quad \rightarrow k=0\\ & \therefore y=\pm x \text { is a particular solution } \\ & y^2-x^2=k \text { is a general solution } \\ & y^2-x^2=k \text { are a family of hyperbolas. } \end{aligned}$\\ \begin{multicols}{2} \textbf{Example}\\ %question 123253 Given the differential equation \(y^{\prime}=y \cos x\) and that the solution curve passes through the point \(\left(\dfrac{\pi}{3}, 1\right)\), then the gradient of the solution curve at this point is? \\ \columnbreak \textbf{Solution}\\ $\begin{aligned} \text { At }\left(\dfrac{\pi}{3}, 1\right) \qquad y^{\prime}&=1 \times \cos \frac{\pi}{3} \\ y^{\prime}&=\frac{1}{2} \end{aligned}$ \\ \end{multicols}](/media/l3mpbgat/direction-fields.png)