Bereken de coördinaten van de keerpunten
Gegeven voor t=0..2\pi:
x(t)= 4sin(t)+2sin(2t)
y(t)= 4cos(t)-2cos(2t)Bereken de coördinaten van de keerpunten.
Uitwerking
Stap 1: het vinden van de t-waarden van de keerpunten. Dat geeft een stelsel dat je op moet lossen:
\begin{array}{l} \left\{ \begin{array}{l} \frac{{dx}}{{dt}} = 0 \\ \frac{{dy}}{{dt}} = 0 \\ \end{array} \right. \\ \left\{ \begin{array}{l} 4\cos (2t) + 4\cos (t) = 0 \\ 4\sin (2t) - 4\sin (t) = 0 \\ \end{array} \right. \\ \left\{ \begin{array}{l} \cos (2t) + \cos (t) = 0 \\ \sin (2t) - \sin (t) = 0 \\ \end{array} \right. \\ \left\{ \begin{array}{l} 2\cos ^2 (t) - 1 + \cos (t) = 0 \\ 2\sin (t)\cos (t) - \sin (t) = 0 \\ \end{array} \right. \\ \left\{ \begin{array}{l} 2\cos ^2 (t) + \cos (t) - 1 = 0 \\ \sin (t)\left( {2\cos (t) - 1} \right) = 0 \\ \end{array} \right. \\ \left\{ \begin{array}{l} (\cos (t) + 1)(2\cos (t) - 1) = 0 \\ \sin (t) = 0 \vee 2\cos (t) - 1 = 0 \\ \end{array} \right. \\ \left\{ \begin{array}{l} \cos (t) = - 1 \vee \cos (t) = \frac{1}{2} \\ \sin (t) = 0 \vee \cos (t) = \frac{1}{2} \\ \end{array} \right. \\ \left\{ \begin{array}{l} t = \pi + k \cdot 2\pi \vee t = \frac{1}{3}\pi + k \cdot 2\pi \vee t = 1\frac{2}{3}\pi + k \cdot 2\pi \\ t = k \cdot \pi \vee t = \frac{1}{3}\pi + k \cdot 2\pi \vee t = 1\frac{2}{3}\pi + k \cdot 2\pi \\ \end{array} \right. \\ Voor\,\,[0,2\pi ]: \\ t = \pi \vee t = \frac{1}{3}\pi \vee t = 1\frac{2}{3}\pi \\ \end{array}
Stap 2: vul je de gevonden waarden voor t in de oorspronkelijke parametervoorstelling in om de coördinaten te bepalen.
\begin{array}{l} \left\{ \begin{array}{l} x(t) = {\rm{ }}4\sin (t) + 2\sin (2t) \\ y(t) = {\rm{ }}4\cos (t) - 2\cos (2t) \\ \end{array} \right. \\ t = \pi : \\ \left\{ \begin{array}{l} x(\pi ) = {\rm{ }}4\sin (\pi ) + 2\sin (2\pi ) = 0 \\ y(\pi ) = {\rm{ }}4\cos (\pi ) - 2\cos (2\pi ) = - 6 \\ \end{array} \right. \\ t = \frac{1}{3}\pi : \\ \left\{ \begin{array}{l} x(\frac{1}{3}\pi ) = {\rm{ }}4\sin (\frac{1}{3}\pi ) + 2\sin (2 \cdot \frac{1}{3}\pi ) = 3\sqrt 3 \\ y(\frac{1}{3}\pi ) = {\rm{ }}4\cos (\frac{1}{3}\pi ) - 2\cos (2 \cdot \frac{1}{3}\pi ) = 3 \\ \end{array} \right. \\ t = 1\frac{2}{3}\pi : \\ \left\{ \begin{array}{l} x(1\frac{2}{3}\pi ) = {\rm{ }}4\sin (1\frac{2}{3}\pi ) + 2\sin (2 \cdot 1\frac{2}{3}\pi ) = - 3\sqrt 3 \\ y(1\frac{2}{3}\pi ) = {\rm{ }}4\cos (1\frac{2}{3}\pi ) - 2\cos (2 \cdot 1\frac{2}{3}\pi ) = 3 \\ \end{array} \right. \\ (0, - 6),\,\,(3\sqrt 3 ,3)\,\,en\,\,( - 3\sqrt 3 ,3) \\ \end{array}
Je hebt nu de coördinaten van de keerpunten gevonden. Zo moet het kunnen...
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