Chapter 8.3, Problem 58E

Solution

The bearing and distance of the cannon from point $B$ .

It has been determined that the cannon is about $8578.9$ feet east and $5664.1$ feet north of point $B$ on a bearing of roughly $56.57°$ and is approximately at a distance of $10280$ feet from the point $B$ .

Given:

Observers are located at positions $A$ , $B$ and $C$ with $A$ due north of $B$ and $C$ due east of $B$ . A cannon is located somewhere in the first quadrant, as illustrated in the figure. $A$ hears the sound of the cannon $2\sec$ before $B$ and $C$ hears the sound $4\sec$ before $B$ .

  

Concept used:

The speed of sound is $1100ft/\sec$ .

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

Calculation:

It can be seen from the figure that $A$ is $4000ft$ due north of $B$ and $C$ is $7000ft$ due east of $B$ .

Now, the speed of sound is $1100ft/\sec$ .

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

It is given that the time difference between the reception of sound from $A$ and $B$ is $2\sec$ .

Then, it follows that,

  $PB-PA=2\times 1100=2200ft$

Now, the set of all possible points such that $PB-PA=2200$ , is a hyperbola with foci at $A$ and $B$ , where $2a=2200$ and so $a=1100$ .

It is given that $AB=4000$ .

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

So, $2c=4000$ , which implies $c=2000$ .

Now,

  $\begin{array}{c}{b}^2=c^2-a^2\\ ={\left(2000\right)}^2-{\left(1100\right)}^2\\ =4000000-1210000\\ =2790000\end{array}$

So, $b^2=2790000$ .

Now, the center of the hyperbola is at the midpoint of $AB$ , which gives the center as $\left(0,2000\right)$ .

Then, the equation of the hyperbola is,

  $\frac{{\left(y-2000\right)}^2}{a^2}-\frac{x^2}{b^2}=1$

Put $a=1100$ and $b^2=2790000$ in the above equation to get,

  $\frac{{\left(y-2000\right)}^2}{1210000}-\frac{x^2}{2790000}=1$

Similarly, it is given that the time difference between the reception of signals from $B$ and $C$ is $4\sec$ .

Then, it follows that,

  $PB-PC=4\times 1100=4400ft$

Now, the set of all possible points such that $PB-PC=4400$ , is a hyperbola with foci at $B$ and $C$ , where $2a=4400$ and so $a=2200$ .

It is given that $BC=7000$ .

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

So, $2c=7000$ , which implies $c=3500$ .

Now,

  $\begin{array}{c}{b}^2=c^2-a^2\\ ={\left(3500\right)}^2-{\left(2200\right)}^2\\ =12250000-4840000\\ =7410000\end{array}$

So, $b^2=7410000$ .

Now, the center of the hyperbola is at the midpoint of $BC$ , which gives the center as $\left(3500,0\right)$ .

Then, the equation of the hyperbola is,

  $\frac{{\left(x-3500\right)}^2}{a^2}-\frac{y^2}{b^2}=1$

Put $a=2200$ and $b^2=7410000$ in the above equation to get,

  $\frac{{\left(x-3500\right)}^2}{4840000}-\frac{y^2}{7410000}=1$

So, $P\left(x,y\right)$ lies on the point of intersection of the hyperbolas $\frac{{\left(y-2000\right)}^2}{1210000}-\frac{x^2}{2790000}=1$ and $\frac{{\left(x-3500\right)}^2}{4840000}-\frac{y^2}{7410000}=1$ .

Graphing the hyperbolas, the first quadrant point of intersection of these hyperbolas such that the point of intersection farther from $B$ than the points $A$ or $C$ , is $\left(8578.898,5664.119\right)$ .

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

  $\theta \approx 90°-{\tan }^{-1}\left(\frac{5664.119}{8578.898}\right)\approx 56.57°$

The distance of $P$ from $O$ is $\sqrt{{\left(8578.898\right)}^2+{\left(5664.119\right)}^2}\approx 10280.06$ feet.

Conclusion:

It has been determined that the cannon is about $8578.9$ feet east and $5664.1$ feet north of point $B$ on a bearing of roughly $56.57°$ and is approximately at a distance of $10280$ feet from the point $B$ .