Answer to Question #173432 in Calculus for Jon jay Mendoza

Question

Real life applications of Hyperbole

B.   Read and analyze the problem below. Solve and show complete solution. (5 points)

Two stations, located at M(1.5, 0) and N(1.5, 0) (units are in km), simultaneously send sound signals to a ship, with the signal traveling at the speed of 0.33 km/s. If the signal from N was received by the ship four seconds before the signal it received from M, find the equation of the curve containing the possible location of the ship.  

Answer 1

M=(-1.5,0)

SOLUTION:

GIVEN:

Co-ordinate of M (-1.5 , 0)

Co-ordinate of N (1.5 , 0)

Let,

co-ordinate of ship as S (x,y)

$\therefore SM=\sqrt{(x+1.5)^2+y^2}\\ \therefore SN=\sqrt{(x-1.5)^2+y^2}\\$

now, time taken given by :$\boxed{t={d\over s}}$

Condition given: time taken by signal to reach N is 4 seconds less than time taken for M.

i.e,

${SN\over0.33}+4={SM\over 0.33}$

so we get,

$4={SM\over0.33}-{SN\over0.33}={SM-SN\over0.33}$

$\implies SM-SN={4\over3}$

$\sqrt{(x+1.5)^2+y^2}-\sqrt{(x-1.5)^2+y^2}={4\over3}$

now ,squaring both sides;

${(x+1.5)^2+y^2}+{(x-1.5)^2+y^2}-2 \sqrt{(x+1.5)^2+y^2}\sqrt{(x-1.5)^2+y^2}={16\over9}$

$2x^2+2y^2+4.5-2\sqrt{x^4+2x^2y^2-4.5x^2+y^4+4.5y^2+5.0625}=1.7$

$2x^2+2y^2+4.5-1.7=-2\sqrt{x^4+2x^2y^2-4.5x^2+y^4+4.5y^2+5.0625}$

$2x^2+2y^2+2.8=-2\sqrt{x^4+2x^2y^2-4.5x^2+y^4+4.5y^2+5.0625}$

solving further we get,

$(2x^2+2y^2+2.8)-(2\sqrt{x^4+2x^2y^2-4.5x^2+y^4+4.5y^2+5.0625})=0$

after solving above equation we get:

$29.2x^2-6.8y^2=12.41$

so equation of required curve (HYPERBOLA) is

$\implies$

$\boxed{{x^2\over0.4356}-{y^2\over1.8144}=1}$

this contains all possible locations of ship.