46-48 Rate of rotation of the searchlight

$h = 8(t - 2) \sin 60^\circ$

$h = 8(t - 2)\left( \dfrac{\sqrt{3}}{2} \right)$

$h = 4\sqrt{3}(t - 2)$
 

$x = 8(t - 2) \cos 60^\circ$

$x = 8(t - 2)(1/2)$

$x = 4(t - 2)$
 

$b = 16 - x = 16 - 4(t - 2)$

$b = 16 - 4t + 8 = 24 - 4t$

$b = 4(6 - t)$
 

$\theta = \arctan \dfrac{h}{b} = \arctan \dfrac{4\sqrt{3}(t - 2)}{4(6 - t)}$

$\theta = \arctan \dfrac{\sqrt{3}(t - 2)}{6 - t}$

$\dfrac{d\theta}{dt} = \dfrac{\dfrac{(6 - t)\sqrt{3} - \sqrt{3}(t - 2)(-1)}{(6 - t)^2}}{1 + \left[ \dfrac{\sqrt{3}(t - 2)}{6 - t} \right]^2}$

$\dfrac{d\theta}{dt} = \dfrac{\dfrac{6\sqrt{3} - t\sqrt{3} + t\sqrt{3} - 2\sqrt{3}}{(6 - t)^2}}{1 + \dfrac{3(t - 2)^2}{(6 - t)^2}}$

$\dfrac{d\theta}{dt} = \dfrac{\dfrac{4\sqrt{3}}{(6 - t)^2}}{\dfrac{(6 - t)^2 + 3(t - 2)^2}{(6 - t)^2}}$

$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{(6 - t)^2 + 3(t - 2)^2}$
 

(a) 3 hours after the start, t = 3
$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{(6 - 3)^2 + 3(3 - 2)^2}$

$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{9 + 3}$

$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{12}$

$\dfrac{d\theta}{dt} = \frac{1}{3}\sqrt{3} \, \text{ rad/sec}$           answer
 

(b) Just after the turn, t = 2
$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{(6 - 2)^2 + 3(2 - 2)^2}$

$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{16 + 0}$

$\dfrac{d\theta}{dt} = \dfrac{4\sqrt{3}}{16}$

$\dfrac{d\theta}{dt} = \frac{1}{4}\sqrt{3} \, \text{ rad/sec}$           answer

The rotation of light is most rapid at the instant when the light is perpendicular to the path of the ship.

$\cos 60^\circ = \dfrac{8(t - 2)}{16}$

$16\cos 60^\circ = 8(t - 2)$

$8 = 8(t - 2)$

$1 = t - 2$

$t = 3 \, \text{ hrs}$           answer

Let r = rate or speed of the ship
$h = r(t - 2) \sin 60^\circ$

$h = \frac{1}{2}\sqrt{3}r(t - 2)$
 

$x = r(t - 2)\cos 60^\circ$

$x = \frac{1}{2}r(t - 2)$
 

$b = 2r - x = 2r - \frac{1}{2}r(t - 2)$

$b = 2r - \frac{1}{2}rt + r = 3r - \frac{1}{2}rt$

$b = \frac{1}{2}r(6 - t)$
 

$\theta = \arctan \dfrac{h}{b} = \arctan \dfrac{\frac{1}{2}\sqrt{3}r(t - 2)}{\frac{1}{2}r(6 - t)}$

$\theta = \arctan \dfrac{h}{b} = \arctan \dfrac{\sqrt{3}(t - 2)}{6 - t}$
 

The above value of angle θ did not include the speed of the ship r. The derivative of θ in terms of time t will of course be independent from r. Solutions 46 & 47 are both derivative of θ in terms of time t.