University Calculus: Early Transcendentals (3rd Edition)

$5~ft/sec$
Here, $\dfrac{dy}{dt}=(-3e^{-x/3})(\dfrac{dx}{dt})$ As we are given that $\dfrac{dy}{dt}=\dfrac{-1}{4}\sqrt {9-y}~ft/sec$ Then, we have $\dfrac{dx}{dt}=\dfrac{e^{x/3}}{12}\sqrt {9-y}$ Now, at $(9,\dfrac{9}{e^3})$ $\dfrac{dx}{dt}=\dfrac{e^{9/3}}{12}\sqrt {9-(\dfrac{9}{e^3})} \approx~5 ft/sec$