Physics, 13.11.2019 06:31 tagerryawilson6

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)?

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Answer from: Alisewilcox

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

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.

Answer from: Quest

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Answer from: Quest

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The coordinates of a particle in the metric xy-plane are differentiable functions of time t with sta...