37-38 How fast a ship leaving from its starting point

Problem 37
A ship sails east 20 miles and then turns N 30° W. If the ship's speed is 10 mi/hr, find how fast it will be leaving the starting point 6 hr after the start.

Solution 37

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By cosine law,
$s^2 = 20^2 + (10t)^2 - 2(20)(10t) \cos 60^\circ$

$s = \sqrt{100t^2 - 200t + 400}$

$\dfrac{ds}{dt} = \dfrac{200t - 200}{2\sqrt{100t^2 - 200t + 400}}$

$\dfrac{ds}{dt} = \dfrac{100t - 100}{\sqrt{100t^2 - 200t + 400}}$

Ship sails east then turned to N 30 deg W

after 6 hrs from start, t = 6 - 2 = 4 hrs
$\dfrac{ds}{dt} = \dfrac{100(4) - 100}{\sqrt{100(4^2) - 200(4) + 400}}$

$\dfrac{ds}{dt} = \dfrac{100(4) - 100}{\sqrt{100(4^2) - 200(4) + 400}}$

$\dfrac{ds}{dt} = 8.66 \, \text{ mi/hr}$ ← answer

Problem 38
Solve Problem 37, if the ship turns N 30° E.

Solution 38

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By cosine law,
$s^2 = 20^2 + (10t)^2 - 2(20)(10t) \cos 120^\circ$

$s = \sqrt{100t^2 + 200t + 400}$

$\dfrac{ds}{dt} = \dfrac{200t + 200}{2\sqrt{100t^2 + 200t + 400}}$

$\dfrac{ds}{dt} = \dfrac{100t + 100}{\sqrt{100t^2 + 200t + 400}}$

after 6 hrs from start, t = 6 - 2 = 4 hrs
$\dfrac{ds}{dt} = \dfrac{100(4) + 100}{\sqrt{100(4^2) + 200(4) + 400}}$

$\dfrac{ds}{dt} = 9.45 \, \text{ mi/hr}$ ← answer

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Dear Sir,

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