University Calculus: Early Transcendentals (3rd Edition)

$2 m/sec$
Here, $(2s)\dfrac{ds}{dt}=[(2x) \dfrac{dx}{dt}+(2y) \dfrac{dy}{dt}]$ $s=\sqrt {(3)^2+(-4)^2}=5$ Plug $\dfrac{dx}{dt}=10m/sec; \dfrac{dy}{dt}=5m/sec$ Then , we have $\dfrac{ds}{dt}=(\dfrac{1}{s})(x \dfrac{dx}{dt}+y\dfrac{dy}{dt})$ This implies that $\dfrac{ds}{dt}=(\dfrac{1}{5})[(10)(3)+(-4)(5)]=2 m/sec$