## Calculus: Early Transcendentals 8th Edition

The distance from the particle to the origin is increasing at a rate of $9.16~cm/s$
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 $9.16~cm/s$