In a pursuit problem, the target moves along a given curve,
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In a pursuit problem, the target moves along a given curve, another object, the pursuer pursues the first, that is, at all times the pursuer moves in the direction of the target.
Suppose that the target is a ship that moves along a straight line and that the pursuer is a destroyer that moves so that the distance between it and the ship is constant, but it is $a> 0 $
a) If $(x, 0) $ represents the position of the ship at a given moment and $y (x) $ is the ordinate of the position of the destroyer at that same moment, deduce that $y$ satisfies the following initial value problem: $$left {begin{array}{c} y'(x) = - dfrac{y (x)}{sqrt{a^{2} -y^{ 2}(x)}} \ y(0) = a end{array}right. $$
b) Show that the previous problem has a unique solution,
I have problems with the first literal because, I do not know what it is that I am asking ... and for the second one, how do I demonstrate the uniqueness of the solution?
real-analysis calculus ordinary-differential-equations
$endgroup$
add a comment |
$begingroup$
In a pursuit problem, the target moves along a given curve, another object, the pursuer pursues the first, that is, at all times the pursuer moves in the direction of the target.
Suppose that the target is a ship that moves along a straight line and that the pursuer is a destroyer that moves so that the distance between it and the ship is constant, but it is $a> 0 $
a) If $(x, 0) $ represents the position of the ship at a given moment and $y (x) $ is the ordinate of the position of the destroyer at that same moment, deduce that $y$ satisfies the following initial value problem: $$left {begin{array}{c} y'(x) = - dfrac{y (x)}{sqrt{a^{2} -y^{ 2}(x)}} \ y(0) = a end{array}right. $$
b) Show that the previous problem has a unique solution,
I have problems with the first literal because, I do not know what it is that I am asking ... and for the second one, how do I demonstrate the uniqueness of the solution?
real-analysis calculus ordinary-differential-equations
$endgroup$
add a comment |
$begingroup$
In a pursuit problem, the target moves along a given curve, another object, the pursuer pursues the first, that is, at all times the pursuer moves in the direction of the target.
Suppose that the target is a ship that moves along a straight line and that the pursuer is a destroyer that moves so that the distance between it and the ship is constant, but it is $a> 0 $
a) If $(x, 0) $ represents the position of the ship at a given moment and $y (x) $ is the ordinate of the position of the destroyer at that same moment, deduce that $y$ satisfies the following initial value problem: $$left {begin{array}{c} y'(x) = - dfrac{y (x)}{sqrt{a^{2} -y^{ 2}(x)}} \ y(0) = a end{array}right. $$
b) Show that the previous problem has a unique solution,
I have problems with the first literal because, I do not know what it is that I am asking ... and for the second one, how do I demonstrate the uniqueness of the solution?
real-analysis calculus ordinary-differential-equations
$endgroup$
In a pursuit problem, the target moves along a given curve, another object, the pursuer pursues the first, that is, at all times the pursuer moves in the direction of the target.
Suppose that the target is a ship that moves along a straight line and that the pursuer is a destroyer that moves so that the distance between it and the ship is constant, but it is $a> 0 $
a) If $(x, 0) $ represents the position of the ship at a given moment and $y (x) $ is the ordinate of the position of the destroyer at that same moment, deduce that $y$ satisfies the following initial value problem: $$left {begin{array}{c} y'(x) = - dfrac{y (x)}{sqrt{a^{2} -y^{ 2}(x)}} \ y(0) = a end{array}right. $$
b) Show that the previous problem has a unique solution,
I have problems with the first literal because, I do not know what it is that I am asking ... and for the second one, how do I demonstrate the uniqueness of the solution?
real-analysis calculus ordinary-differential-equations
real-analysis calculus ordinary-differential-equations
asked Dec 5 '18 at 21:36
Santiago SeekerSantiago Seeker
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