Answer
$\frac{dy}{dx} = 6\sqrt t$
$\frac{d^2y}{dx^2}= 6$
The slope is equal to 6:
The graph is concave upward
Work Step by Step
1. Find dx/dt and dy/dt
$x=\sqrt t$
$dx =(\frac{1}{2\sqrt t})dt$
$dx/dt= (\frac{1}{2\sqrt t})$
$y=3t-1$
$dy=(3)dt$
$dy/dt=3$
2. Calculate dy/dx:
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3}{\frac{1}{2\sqrt t}} = (3) (2\sqrt t) = 6\sqrt t$
3. Find $d^2y/dx^2$
$\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}[dy/dx]}{dx/dt}=\frac{\frac{d}{dt}[6\sqrt t]}{\frac{1}{2\sqrt t}} = \frac{6\frac{1}{2\sqrt t}}{\frac{1}{2\sqrt t}} = 6$
4. Find the slope and the concavity:
The slope is dy/dx in t = 1, which is equal to:
$\frac{dy}{dx}(t=1) = 6\sqrt 1 = 6*1 = 6$
Since the second derivative ($d^2y/dx^2$) is equal to 6 (positive value), the graph is concave upward.
