Liner & Angular Speed Questions

x(t)   = -5 + 5 sin (2 pi * t / 13)

y(t)  = 3 + 5cos (2pi * t / 13)

circle is centered at (-5,3) and has radius of 5. A particle starts at point (-5,8) and travels clockwise on the circle, making one complete revolution every 13 seconds. {nl} Write the parametric equations for the motion of the particle, where tt represents time in seconds:

x(t)=

y(t)=

$\begin{array}{|rcll|} \hline \dbinom{x(t)}{y(t)} &=& \dbinom{-5}{3} + 5 \cdot \dbinom{ \cos(-\omega t + 90^{\circ}) } { \sin(-\omega t + 90^{\circ}) } \\\\ \dbinom{x(t)}{y(t)} &=& \dbinom{-5}{3} + 5 \cdot \dbinom{ \cos( 90^{\circ}-\omega t ) } { \sin( 90^{\circ}-\omega t ) } \quad | \quad \small{\cos(90^{\circ}-\varphi)=\sin(\varphi) \quad \sin(90^{\circ}-\varphi)=\cos(\varphi)} \\\\ \dbinom{x(t)}{y(t)} &=& \dbinom{-5}{3} + 5 \cdot \dbinom{ \sin( \omega t ) } { \cos( \omega t ) } \qquad | \qquad \omega=\dfrac{2\pi}{13\ s} \\\\ x(t) &=& -5 + 5\cdot \sin(\frac{2\pi}{13\ s} \cdot t) \\ y(t) &=& 3+5\cdot \cos( \frac{2\pi}{13\ s} \cdot t ) \\ \hline \end{array}$

Proof:

$\begin{array}{|lrcrl|} \hline \text{if } t = 0\quad \Rightarrow & x(t) &=& -5\\ & y(t) &=& 8 \\ \hline \end{array}$