Answer
Tangent line: $\quad y=\displaystyle \sqrt{3}x-\frac{\pi\sqrt{3}}{3}+2$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\pi/3} =-4$
Work Step by Step
$\displaystyle \frac{dx}{dt}=1-\cos t,\qquad \displaystyle \frac{dy}{dt}=\sin t$
$\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\sin t}{1-\cos t}$
The slope of the tangent line at $t=\pi/3$ is
$\left. \displaystyle \frac{dy}{dx} \right|_{t=\pi/3}=\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}=\sqrt{3}$
With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\pi/3}$,
$\displaystyle \frac{dy'}{dt}=\frac{d}{dt}[\frac{\sin t}{1-\cos t}]=\frac{(1-\cos t)(\cos t)-(\sin t)(\sin t)}{(1-\cos t)^{2}}$
$=\displaystyle \frac{\cos t-(\cos^{2}t+\sin^{2}t)}{(1-\cos t)^{2}}$
$=\displaystyle \frac{\cos t-1}{(1-\cos t)^{2}}$
$=-\displaystyle \frac{(1-\cos t)}{(1-\cos t)^{2}}$
$=-\displaystyle \frac{1}{1-\cos t}$
$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{-\frac{1}{1-\cos t}}{1-\cos t}=-\frac{1}{(1-\cos t)^{2}}$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\pi/3} = -\frac{1}{(1-\frac{1}{2})^{2}}=-4$
The point $P$ on the curve corresponding to $t=\pi/3$
$x=\displaystyle \frac{\pi}{3}-\frac{\sqrt{3}}{2},\quad,y=1-\frac{1}{2}=\frac{1}{2},$
is $P(\displaystyle \frac{\pi}{3}-\frac{\sqrt{3}}{2},\frac{1}{2})$
Use the point-slope equation for the tangent line.
$y-\displaystyle \frac{1}{2}=\sqrt{3}\cdot(x-\frac{\pi}{3}+\frac{\sqrt{3}}{2})$
$y=\displaystyle \sqrt{3}x-\frac{\pi\sqrt{3}}{3}+\frac{3}{2}+\frac{1}{2}$
$y=\displaystyle \sqrt{3}x-\frac{\pi\sqrt{3}}{3}+2$
