Chapter 8.3, Problem 57E

Solution

The ship’s bearing and distance from point $O$ .

It has been determined that the cruise ship Princess Ann is about $886.7$ mi east and $1045.8$ mi north of point $O$ on a bearing of roughly $40°$ and is approximately at a distance of $1371$ mi from the point $O$ .

Given:

Three LORAN radio transmitters are positioned as shown in the figure, with $R$ due north of $O$ and $Q$ due east of $O$ . The cruise ship Princess Ann receives simultaneous signals from the three transmitters. The signal from $O$ arrives $323.27\mu \sec$ after the signal from $R$ and $646.53\mu \sec$ after the signal from $Q$ .

  

Concept used:

The speed of radio signals is $980ft/\mu \sec$ .

The difference between distances from the foci to any point on the hyperbola, is constant.

Calculation:

It can be seen from the figure that $R$ is $80$ mi due north of $O$ and $Q$ is $200$ mi due east of $O$ .

Now, the speed of radio signals is $980ft/\mu \sec$ .

Then, the speed of radio signals is $\frac{980}{5280}mi/\mu \sec$ .

Denote the position of the cruise ship as $P\left(x,y\right)$ .

It is given that the time difference between the reception of signals from $O$ and $R$ is $323.27\mu \sec$ .

Then, it follows that,

  $OP-PR=323.27\times \frac{980}{5280}\approx 60$ mi

Now, the set of all possible points such that $OP-PR\approx 60$ , is a hyperbola with focus at $O$ and $R$ , where $2a=60$ and so $a=30$ .

It is given that $OR=80$ .

This is nothing but the distance between two foci of the hyperbola.

So, $2c=80$ , which implies $c=40$ .

Now,

  $\begin{array}{c}{b}^2=c^2-a^2\\ ={\left(40\right)}^2-{\left(30\right)}^2\\ =1600-900\\ =700\end{array}$

So, $b^2=700$ .

Now, the center of the hyperbola is at the midpoint of $OR$ , which gives the center as $\left(0,40\right)$ .

Then, the equation of the hyperbola is,

  $\frac{{\left(y-40\right)}^2}{a^2}-\frac{x^2}{b^2}=1$

Put $a=30$ and $b^2=700$ in the above equation to get,

  $\frac{{\left(y-40\right)}^2}{900}-\frac{x^2}{700}=1$

Similarly, it is given that the time difference between the reception of signals from $O$ and $Q$ is $646.53\mu \sec$ .

Then, it follows that,

  $OP-PQ=646.53\times \frac{980}{5280}\approx 120$ mi

Now, the set of all possible points such that $OP-PQ\approx 120$ , is a hyperbola with focus at $O$ and $Q$ , where $2a=120$ and so $a=60$ .

It is given that $OQ=200$ .

This is nothing but the distance between two foci of the hyperbola.

So, $2c=200$ , which implies $c=100$ .

Now,

  $\begin{array}{c}{b}^2=c^2-a^2\\ ={\left(100\right)}^2-{\left(60\right)}^2\\ =10000-3600\\ =6400\end{array}$

So, $b^2=6400$ .

Now, the center of the hyperbola is at the midpoint of $OQ$ , which gives the center as $\left(100,0\right)$ .

Then, the equation of the hyperbola is,

  $\frac{{\left(x-100\right)}^2}{a^2}-\frac{y^2}{b^2}=1$

Put $a=60$ and $b^2=6400$ in the above equation to get,

  $\frac{{\left(x-100\right)}^2}{3600}-\frac{y^2}{6400}=1$

So, $P\left(x,y\right)$ lies on the point of intersection of the hyperbolas $\frac{{\left(y-40\right)}^2}{900}-\frac{x^2}{700}=1$ and $\frac{{\left(x-100\right)}^2}{3600}-\frac{y^2}{6400}=1$ .

Graphing the hyperbolas, the first quadrant point of intersection of these hyperbolas, such that the point of intersection is out in the sea, is $\left(886.664,1045.83\right)$ .

Now, the bearing of $P$ from the point $O$ is,

  $\theta \approx 90°-{\tan }^{-1}\left(\frac{1045.83}{886.664}\right)\approx 40.29°$

The distance of $P$ from $O$ is $\sqrt{{\left(886.664\right)}^2+{\left(1045.83\right)}^2}\approx 1371.11$ mi.

Conclusion:

It has been determined that the cruise ship Princess Ann is about $886.7$ mi east and $1045.8$ mi north of point $O$ on a bearing of roughly $40°$ and is approximately at a distance of $1371$ mi from the point $O$ .