# 44 - Angle of elevation of the rope tied to a rowboat on shore

**Problem 44**

A rowboat is pushed off from a beach at 8 ft/sec. A man on shore holds a rope, tied to the boat, at a height of 4 ft. Find how fast the angle of elevation of the rope is decreasing, after 1 sec.

**Solution 44**

$\theta = \arctan \dfrac{1}{2t}$

$\dfrac{d\theta}{dt} = \dfrac{\dfrac{-1(2)}{(2t)^2}}{1 + \left( \dfrac{1}{2t} \right)^2}$

$\dfrac{d\theta}{dt} = \dfrac{\dfrac{-1}{2t^2}}{1 + \dfrac{1}{4t^2}}$

$\dfrac{d\theta}{dt} = -\dfrac{1}{2t^2 \left( \dfrac{4t^2 + 1}{4t^2} \right)}$

$\dfrac{d\theta}{dt} = -\dfrac{1}{\dfrac{4t^2 + 1}{2}}$

$\dfrac{d\theta}{dt} = -\dfrac{2}{4t^2 + 1}$

after t = 1 sec

$\dfrac{d\theta}{dt} = -\dfrac{2}{4(1^2) + 1}$

$\dfrac{d\theta}{dt} = -\dfrac{2}{5} \, \text{ rad/sec}$ *answer*

*The negative sign indicates that the angle is decreasing.*

**Another Solution**

$\tan \theta = \dfrac{1}{2t}$

$\sec^2 \theta \, \dfrac{d\theta}{dt} = \dfrac{-1(2)}{(2t)^2}$

$\dfrac{1}{\cos^2 \theta} \, \dfrac{d\theta}{dt} = -\dfrac{1}{2t^2}$

$\dfrac{d\theta}{dt} = -\cos^2 \theta \, \dfrac{1}{2t^2}$

after t = 1 sec

$8t = 8(1) = 8$

$\text{hypotenuse} = \sqrt{8^3 + 4^2 \,} = 4\sqrt{5}$

$\cos \theta = \dfrac{8}{4\sqrt{5}}$

$\dfrac{d\theta}{dt} = -\left( \dfrac{8}{4\sqrt{5}} \right)^2 \, \dfrac{1}{2(1^2)}$

$\dfrac{d\theta}{dt} = -\left( \dfrac{64}{80} \right) \, \dfrac{1}{2}$

$\dfrac{d\theta}{dt} = -\left( \dfrac{64}{80} \right) \, \dfrac{1}{2}$

$\dfrac{d\theta}{dt} = -\dfrac{2}{5} \, \text{ rad/sec}$ (*okay!*)