How to find the critical angle

**Theressa** · October 14th 2009, 08:21 PM

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!

**redsoxfan325** · October 14th 2009, 08:37 PM

Originally Posted by Theressa

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!

$22500\cos^2(t)-(22500+22500\sin(t))\sin(t)=$ $22500(1-\sin^2(t))-22500\sin^2(t))-22500\sin(t)$ $=22500-22500\sin^2(t)-22500\sin^2(t)-22500\sin(t)$ $=22500(1-\sin(t)-2\sin^2(t))=22500(1-2\sin(t))(1+\sin(t))$

Solve $22500(1-2\sin(t))(1+\sin(t))=0$ for $t$ to get $t=\left\{\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{ 2}\right\}$

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