Answer
Tangent line: $\quad y=2$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\pi/2}=-1$
Work Step by Step
$\displaystyle \frac{dx}{dt}=-\sin t,\qquad \displaystyle \frac{dy}{dt}=\cos t$
$\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\cos t}{-\sin t}=-\cot t$
The slope of the tangent line at $t=\pi/2$ is
$\left. \displaystyle \frac{dy}{dx} \right|_{t=\pi/2}=0$
(the tangent line is horizontal there)
The point $P$ on the curve corresponding to $t=\pi/2$
$x=\displaystyle \cos\frac{\pi}{2}=0,\quad,y=1+\sin\frac{\pi}{2}=2,$
is $P(0,2)$
Since the tangent line has slope 0, it is horizontal (and passes through P):
$y=2$
With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\pi/2}$,
$\displaystyle \frac{dy'}{dt}=\frac{d}{dt}[-\cot t]=\csc^{2}t$
$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{\csc^{2}t}{-\sin t}=-\csc^{3}t$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\pi/2}=-1$
