Answer
$\frac{dy}{dx} = \frac{t+1}{t-1}$
$\frac{d^2y}{dx^2} =\frac{-2t}{(t-1)^3}$
The curve is concave upward when $~~0 \lt t \lt 1$
Work Step by Step
$x = t - ln~t$
$y = t+ln~t$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1+1/t}{1-1/t} = \frac{\frac{t+1}{t}}{\frac{t-1}{t}} = \frac{t+1}{t-1}$
We can find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{\frac{(t-1)-(t+1)}{(t-1)^2}}{(t-1)/t} = \frac{-2t}{(t-1)^3}$
The curve is concave upward when $\frac{d^2y}{dx^2} \gt 0$
$-2t \gt 0~~$ when $~~t \lt 0$
$(t-1)^3 \gt 0~~$ when $~~t \gt 1$
There are no values of $t$ such that both the numerator and the denominator are positive.
$-2t \lt 0~~$ when $~~t \gt 0$
$(t-1)^3 \lt 0~~$ when $~~t \lt 1$
The curve is concave upward when $~~0 \lt t \lt 1$
