## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{10\sqrt 3}{9}$
Take the derivative of the given equations and isolate the variables. $\dfrac{dx}{dt}=(\dfrac{-1}{2 \sqrt t})\dfrac{1}{2 \sqrt {5-\sqrt t}}$ and $(\dfrac{\sqrt t}{t-1})+(t-1)\dfrac{dy}{dt}=\dfrac{1}{2\sqrt t}$ At $t=4$, we have $\dfrac{dx}{dt}=\dfrac{-1}{8\sqrt {3}}$ and $\dfrac{dx}{dt}=\dfrac{-5}{36}$ Now, Slope: $\dfrac{dy}{dx}=\dfrac{\dfrac{-5}{36}}{\dfrac{-1}{8\sqrt {3}}}=\dfrac{10\sqrt 3}{9}$