Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 687: 10

$y=-3x+\dfrac{13}{4}$ and $6$

Here, we have $\dfrac{dy}{dx}=(\dfrac{-3}{2})t$ Now, $ \dfrac{dy}{dx}(t=2)=(\dfrac{-3}{2})(2)=-3$ Now, $y+\dfrac{1}{2}=(-3)(x-\dfrac{5}{4})$ or, $y=-3x+\dfrac{13}{4}$ Now, $\dfrac{d^2y}{dx^2}=\dfrac{dy'/dt}{dx/dt}=(\dfrac{3}{4})t^3$ and $\dfrac{d^2y}{dx^2}(t=2)=\dfrac{3}{4} (2)^3=(\dfrac{3}{4}) (8)=6$