Answer
$\frac{dy}{dx} = \frac 1 4 sec(t)$
$\frac{d^2y}{dx^2}= -\frac {1}{16} sec^3(t)$
The curve is concave upward for these values of t : $\frac{\pi}{2} \lt t \lt \pi$
Work Step by Step
1. Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$
$\frac{dy}{dt} = \frac{d(cos(t))}{dt} = -sin(t)$
$\frac{dx}{dt} = \frac{d(cos(2t))}{dt} = -2sin(2t)$
Therefore: $\frac{dy}{dx} = \frac{-sin(t)}{-2sin(2t)} = \frac{sin(t)}{2sin(2t)}$
** $sin(2t) = 2sin(t)cos(t)$
$\frac{dy}{dx} = \frac{sin(t)}{2(2sin(t)cos(t))} = \frac 1 {4cos(t)} = \frac 1 4 sec(t)$
2. Calculate $\frac{d^2y}{dx^2}$:
$\frac{d(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{\frac 1 4 (tan(t)sec(t))}{-2sin(2t)} = \frac{\frac 1 4 \frac{sin(t)}{cos^2(t)}}{-2(2sin(t)cos(t))} = -\frac{1}{16} \frac{1}{cos^3(t)} = -\frac {1}{16} sec^3(t)$
3. Determine for which values the curve is concave upward:
- That will happen when $\frac{d^2y}{dx^2} \lt 0$:
$-\frac {1}{16} sec^3(t) \lt 0$
$sec^3(t) \lt 0$
$sec(t) \lt 0$
$\frac{1}{cos(t)} \lt 0$
- Since $1$ is always going to be positive, then $cos(t)$ will determine if the result is positive or negative.
$cos(t) \lt 0$
Therefore: $\frac{\pi}{2} \lt t \lt \frac{3\pi}{2}$, but, since t is between $0$ and $\pi$:
$\frac{\pi}{2} \lt t \lt \pi$
