Friday, December 2, 2011

Application of Hyperbolas

Two LORAN stations are positioned 252 miles apart along a straight shore. A ship records a time difference of 0.00129 seconds between the LORAN signals. (The radio signals travel at 186,000 miles per second).
(a) Where will the ship reach shore if it were to follow the hyperbola corresponding to this time difference?
(b) If the ship is 50 miles offshore, what is the position of the ship?

r = 186,000 miles
t = 0.00129 seconds
d = ?

d = r*t = 186,000*0.00129 = 239.94 miles (difference in distance)

(a) The ship will reach shore at the distance between the two stations minus the difference in distance between the two, divided by the two. So we have:
(252-239.94)/2 = 12.06/2 = 6.03 miles away from the station.

(b) To find the position of the ship 50 miles offshore, we have to find the equation of the hyperbola:
- the value of a will be difference in distance divided by two, so 239.94/2=119.97;
a = 119.97, a^2 = 14392.8009
- the value of c corresponds to the focus of the hyperbola, in this case distance between stations divided by two, therefore c = 252/2 = 126;
c = 126, c^2 = 15876
- to find b^2, we use the equation c^2 = a^2 + b^2, so 15876 = 14392.8009 + b^2; b^2 = 15876 – 14392.8009 = 1483.1991

Input information into hyperbola equation:
x^2/a^2 – y^2/b^2 = 1
x^2/14392.8009 – y^2/1483.1991 = 1
When ship is 50 miles offshore, y = 50.
Input value of y into equation to find the value of x:
x^2/14392.8009 – 50^2/1483.1991 = 1
x^2/14392.8009 – 2500/1483.1991 = 1
x^2/14392.8009 = (2500/1483.1991) + 1
x^2/14392.8009 = 2.68554579085168
x^2 = 14392.8009*2.68554579085168
x^2 = 38652. 5258755613
x = √38652. 5258755613
x = 196.6024 = ≈197
The position of the ship when 50 miles offshore is at (197, 50).

Thursday, October 27, 2011

Find the magnitude and the direction angle

Find the magnitude and the direction angle (to the nearest tenth) for the vector
\f[\left<3, 3 \right>\f]
1. Magnitude:
Formula: 
\f[\left|\left|v \right| \right|=\sqrt{x^{2}+y^{2}}\f]
\f[\left|\left|3, 3 \right| \right| = \sqrt{3^{2}+3^{2}} = \sqrt{9+9}=\sqrt{18}= 3\sqrt{2}\f]

2. Direction Angle:
Formula: 
\f[\theta = \tan^{-1} \frac{y}{x}\f]
Graph the vector:
Find the direction angle:\f[\tan \theta =\frac{3}{3} = 1\f]
\f[\tan^{-1} (1) = 45\deg \f]
Final results:
Magnitude 3√2
Direction angle 45 °

Images for my math classes