Answer
The distance from the particle to the origin is increasing at a rate of $9.16~cm/s$
Work Step by Step
To find $\frac{dy}{dt}$, we can differentiate both sides of the equation with respect to $t$:
$y = 2~sin(\frac{\pi~x}{2})$
$\frac{dy}{dt} = \pi~cos(\frac{\pi~x}{2})~\frac{dx}{dt}$
$\frac{dy}{dt} = \pi~cos(\frac{\pi~(\frac{1}{3})}{2})~(\sqrt{10})$
$\frac{dy}{dt} = \frac{\sqrt{30}~\pi}{2}$
Let $z$ be the distance from the origin:
$z^2 = x^2+y^2$
$z = \sqrt{x^2+y^2}$
$z = \sqrt{(\frac{1}{3})^2+(1)^2}$
$z = \frac{\sqrt{10}}{3}$
We can differentiate both sides of the equation with respect to $t$:
$z^2 = x^2+y^2$
$2z~\frac{dz}{dt} = 2x~\frac{dx}{dt} + 2y~\frac{dy}{dt}$
$\frac{dz}{dt} = \frac{1}{z}~(x~\frac{dx}{dt} + y~\frac{dy}{dt})$
$\frac{dz}{dt} = \frac{1}{(\frac{\sqrt{10}}{3})}~[(\frac{1}{3})(\sqrt{10}) + (1)(\frac{\sqrt{30}~\pi}{2})]$
$\frac{dz}{dt} = 9.16~cm/s$
The distance from the particle to the origin is increasing at a rate of $9.16~cm/s$
