Answer
$\frac{dy}{dx} = -\frac{2~cos~2t}{sin~t}$
$\frac{d^2y}{dx^2} = -\frac{2~cos~t~(2~sin^2~t+1)}{sin^3~t}$
The curve is concave upward when $~~\frac{\pi}{2} \lt t \lt \pi$
Work Step by Step
$x = cos~t$
$y = sin~2t$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2~cos~2t}{-sin~t} = -\frac{2~cos~2t}{sin~t}$
We can find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{\frac{4~sin~2t~sin~t+2~cos~2t~cos~t}{sin^2~t}}{-sin~t} = -\frac{4~(2~sin~t~cos~t)~sin~t+2~(1-2sin^2~t)~cos~t}{sin^3~t}= -\frac{8~sin^2~t~cos~t-4~sin^2~t~cos~t+2~cos~t}{sin^3~t} = -\frac{4~sin^2~t~cos~t+2~cos~t}{sin^3~t} = -\frac{2~cos~t~(2~sin^2~t+1)}{sin^3~t}$
The curve is concave upward when $\frac{d^2y}{dx^2} \gt 0$
On the interval $0 \lt t \lt \pi~~$, note that $~~sin~t \gt 0$
$\frac{d^2y}{dx^2} \gt 0~~$ when $~~-[2~cos~t~(2~sin^2~t+1)] \gt 0$
$2~cos~t~(2~sin^2~t+1) \lt 0$
$cos~t \lt 0$
$\frac{\pi}{2} \lt t \lt \pi$
The curve is concave upward when $~~\frac{\pi}{2} \lt t \lt \pi$
