Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 657: 17

$-6$

Here, we have $ \dfrac{dx}{dt}=\dfrac{(2t+1)}{1+3x^{1/2}}$ and $\dfrac{y}{2\sqrt {t+1}}+[\sqrt{t+1}+(\dfrac{t}{\sqrt y})]\dfrac{dy}{dt}+2 (\sqrt y)=0$ Since, $x=0$ when $t=0$ $\implies \dfrac{dx}{dt}(t=0)=1$ Also, $y=4$ when $t=0$ $ \implies \dfrac{dx}{dt}(t=0)=-6$ Thus, Slope: $\dfrac{dy}{dx}=-\dfrac{6}{1}=-6$