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