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