Answer:
[tex]\dfrac{dD}{dt}=-7.076\ m/s[/tex]
Explanation:
It is given that,
The coordinates of a particle in the metric xy-plane are differentiable functions of time t are given by :
[tex]\dfrac{dx}{dt}=-6\ m/s[/tex]
[tex]\dfrac{dy}{dt}=-4\ m/s[/tex]
Let D is the distance from the origin. It is given by :
[tex]D^2=x^2+y^2[/tex]
Differentiate above equation wrt t as:
[tex]2D\dfrac{dD}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}[/tex]
[tex]D\dfrac{dD}{dt}=x\dfrac{dx}{dt}+y\dfrac{dy}{dt}[/tex].............(1)
The points are given as, (12,5). Calculating D from these points as :
[tex]D=\sqrt{12^2+5^2} =13\ m[/tex]
Put all values in equation (1) as :
[tex]13\times \dfrac{dD}{dt}=12\times (-6)+5\times (-4)[/tex]
[tex]\dfrac{dD}{dt}=-7.076\ m/s[/tex]
So, the particle is moving away from the origin at the rate of 7.076 m/s. Hence, this is the required solution.