Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 657: 18

$-4$

Here, we have $x (\cos t)+(\sin t) \dfrac{dx}{dt}+\dfrac{dx}{dt}(2)=1$ and $\dfrac{dy}{dt}=t(\cos t)+\sin t-2$ $x=\pi/2$ when $t=\pi$ $\implies \dfrac{dx}{dt}=2+\pi/4$ Also, $ \dfrac{dx}{dt}(t=\pi)=-\pi-2$ Thus, Slope: $\dfrac{dy}{dx}=\dfrac{-\pi-2}{2+(\dfrac{\pi}{4})}=-4$