Question

At noon a ship S1 is 20 miles north of a ship S2. If S1 is sailing south at a rate of 6 miles per hour and S2 is sailing east at a rate of 8 miles per hour, find the time when they are nearest together.

Answer 1

At noon a ship S1 is 20 miles north of a ship S2. If S1 is sailing south at a rate of 6 miles per hour and S2 is sailing east at a rate of 8 miles per hour, find the time when they are nearest together.

Solution:

Let t - the travel time in hours of both ships.

L = 20 - the distance between the ships at the beginning of the movement.

$V_{1} = 6\frac{mile}{hour}$ - rate of the sheep S1

$V_{2} = 6\frac{mile}{hour}$ - rate of the sheep S1

Then:

$6t =$ travel dist of the S1 (sailing north toward the ref point C)

$8t =$ travel dist of the S2 (sailing west from the ref point C)

The two ships course form a right triangle from the ref points A, B and C. $(\angle ACB = 90{}^{\circ})$

The distance between the two ships, is the hypotenuse (S):

$$S ^ {2} = (2 0 - 6 t) ^ {2} + (8 t) ^ {2}$$ $$S ^ {2} = 4 0 0 - 2 4 0 t + 3 6 t ^ {2} + 6 4 t ^ {2}$$ $$S ^ {2} = 1 0 0 t ^ {2} - 2 4 0 t + 4 0 0$$

distance between ships is minimum when the derivative $\frac{ds}{dt}$ is zero (local minimum of the function):

$$\frac {d S}{d t} = \frac {S = \sqrt {1 0 0 t ^ {2} - 2 4 0 t + 4 0 0}}{2 \sqrt {1 0 0 t ^ {2} - 2 4 0 t + 4 0 0}} \cdot (2 0 0 t - 2 4 0) = 0$$ $$\begin{array}{l} 200t - 240 = 0 \\ t = \frac{240}{200} = 1.2 \text{ hour} \end{array}$$

After the time $t = 1.2$ hour ships will be nearest together at a minimum distance:

$$S_{\min} = S(1.2) = \sqrt{100 \cdot (1.2)^2 - 240 \cdot 1.2 + 400} = 16 \text{ miles}$$

Answer: after the time $t = 1.2$ hour ships will be nearest together at a minimum distance $S_{\min} = 16$ miles.