Question

I just need help with question F

An object is moving around the unit circle with parametric equations x(t)=cos(t), y(t)=sin(t), so it's location at time t is P(t)=(cos(t), sin(t)). Assume 0 < t < pi/2. At a given time t, the tangent line to the unit circle at the position P(t) will determine a right triangle in the first quadrant. (Connect the origin with the y-intercept and x-intercept of the tangent line.) The identity sin(2t)=2sin(t)cost(t) might be useful in some parts of this question. The area of the right triangle is a(t)= 1/sin (2t). With our restriction on t, the smallest t so that a(t)=2 is pi/12 With our restriction on t, the largest t so that a(t)=2 is pi/6

Answer 1

$\underline{Solution}:$

From problem (a)

$Area\, \, of\, \, a\, \, triangle\,\, a(t)=\frac{1}{sin2t}$

$Given\,\, a(t)=2$

$\Rightarrow \frac{1}{sin2t}=2$

$\Rightarrow sin2t=\frac{1}{2}$

$\Rightarrow sin2t=\frac{1}{2}=sin\frac{\pi }{6}................................(1)$

$\Rightarrow Princple\, value\, (\alpha )=\frac{\pi }{6}$

$General\, solution\, of\, (1)\, is\,\, 2t=n\pi +(-1)^{n}\alpha$

$\Rightarrow 2t=n\pi +(-1)^{n}\frac{\pi }{6}$

$\Rightarrow t=\frac{n\pi}{2} +(-1)^{n}\frac{\pi }{12};\, \, \, n\epsilon Z$

$If\, n=0\Rightarrow t=0 +\frac{\pi }{12}\Rightarrow t=\frac{\pi }{12}=15^{0}$

$If\, n=1\Rightarrow t=\frac{\pi}{2}-\frac{\pi }{12}\Rightarrow t=\frac{5\pi }{12}=75^{0}$

$If\, n=2\Rightarrow t=\pi+\frac{\pi }{12}\Rightarrow t=\frac{13\pi }{12}=195^{0}$

$If\, n=-1\Rightarrow t=-\frac{\pi }{2}-\frac{\pi }{12}\Rightarrow t=-\frac{7\pi }{12}=-105^{0}$

$Since\, \, 0< t< \frac{\pi }{2}\, \, \left [ i.e.,0< t< 90^{0} \right ]$

$\Rightarrow t=\frac{\pi}{12},\, \, t=\frac{5\pi }{12}$

$\therefore Smalest \, \, t=\frac{\pi}{12}\, \, and\, \, Largest\, \, t=\frac{5\pi }{12}$

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