Question

The coordinates of a particle in the metric​ xy-plane are differentiable functions of time t with StartFraction dx Over dt EndFraction equalsnegative 6 StartFraction m Over sec EndFraction and StartFraction dy Over dt EndFraction equalsnegative 4 StartFraction m Over sec EndFraction . How fast is the​ particle's distance from the origin changing as it passes through the point ​(12​,5​)?

Answer 1

Answer:

$\dfrac{dD}{dt}=-7.076\ m/s$

Explanation:

It is given that,

The coordinates of a particle in the metric​ xy-plane are differentiable functions of time t are given by :

$\dfrac{dx}{dt}=-6\ m/s$

$\dfrac{dy}{dt}=-4\ m/s$

Let D is the distance from the origin. It is given by :

$D^2=x^2+y^2$

Differentiate above equation wrt t as:

$2D\dfrac{dD}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$

$D\dfrac{dD}{dt}=x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$.............(1)

The points are given as, (12,5). Calculating D from these points as :

$D=\sqrt{12^2+5^2} =13\ m$

Put all values in equation (1) as :

$13\times \dfrac{dD}{dt}=12\times (-6)+5\times (-4)$

$\dfrac{dD}{dt}=-7.076\ m/s$

So, the particle is moving away from the origin at the rate of 7.076 m/s. Hence, this is the required solution.