How would I find the critical angle of
V'(t)=22500*(cos(t)^2)-(100*(225+225*sin(t)))*sin(t)
(just to let you know, t us used in place of )
thanks!
$\displaystyle 22500\cos^2(t)-(22500+22500\sin(t))\sin(t)=$ $\displaystyle 22500(1-\sin^2(t))-22500\sin^2(t))-22500\sin(t)$ $\displaystyle =22500-22500\sin^2(t)-22500\sin^2(t)-22500\sin(t)$ $\displaystyle =22500(1-\sin(t)-2\sin^2(t))=22500(1-2\sin(t))(1+\sin(t))$
Solve $\displaystyle 22500(1-2\sin(t))(1+\sin(t))=0$ for $\displaystyle t$ to get $\displaystyle t=\left\{\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{ 2}\right\}$