Answer
$\frac{dy}{dx} = \frac{e^{t}}{2t}$
$\frac{d²y}{dx²} = \frac {e^{t}(t – 1)}{4t³}$
The curve is concave up for $t$ in $(–\infty, 0)$ and $(1, \infty)$.
Work Step by Step
We differentiate both equations with respect to $t$:
$x = t² + 1$
$\frac{dx}{dt} = 2t$
$y = e^{t} – 1$
$\frac{dy}{dt} = e^{t}$
Now, we know that $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, so:
$\frac{dy}{dx} = \frac{e^{t}}{2t}$
To find $\frac{d²y}{dx²}$, we differentiate $\frac{dy}{dx}$ using the chain rule:
$\frac{dy}{dx} = \frac{e^{t}}{2t}$
$\frac{d²y}{dx²} = \frac {\frac{d}{dt} (\frac{e^{t}}{2t})}{\frac{dx}{dt}}$
$= \frac {\frac{2t(e^{t}) – 2(e^{t})}{4t²}}{2t}$
$= \frac {e^{t}(2t – 2)}{8t³}$
$= \frac {e^{t}(t – 1)}{4t³}$
The curve will be concave up whenever $\frac{d²y}{dx²} \gt 0$, which will happen if $t – 1$ and $t³$ have the same sign (we don’t need to worry about $e^{t}$ because it is always positive). Since $t – 1 \gt 0$ when $t \gt 1$ and $t³ \gt 0$ when $t\gt 0$:
For $t$ in ($–\infty$ , 0), both expressions are negative and therefore $\frac{d²y}{dx²}$ is positive.
For $t$ in (0 , 1), the expressions have opposite signs and therefore $\frac{d²y}{dx²}$ is negative.
For $t$ in (1 , $\infty$), both expressions are positive and therefore $\frac{d²y}{dx²}$ is positive.
Thus the curve is concave up for $t$ in $(–\infty, 0)$ and $(1, \infty)$.
