Question

consider the curve parametrized by $x = t^{3}+1$ and $y = e^{2t}-1$.
a) find $\frac{dy}{dx}$ in terms of $t$.
b) find $\frac{d^{2}y}{dx^{2}}$ in terms of $t$.
c) for what intervals of $t$ is the curve concave up? enter your answer as intervals separated by a comma.

Explanation:

Step1: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

Differentiate $x = t^{3}+1$ with respect to $t$: $\frac{dx}{dt}=3t^{2}$. Differentiate $y = e^{2t}-1$ with respect to $t$: $\frac{dy}{dt}=2e^{2t}$.

Step2: Find $\frac{dy}{dx}$

By the chain - rule $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. So $\frac{dy}{dx}=\frac{2e^{2t}}{3t^{2}}$.

Step3: Find $\frac{d^{2}y}{dx^{2}}$

Use the quotient rule $\frac{d}{dt}(\frac{u}{v})=\frac{v\frac{du}{dt}-u\frac{dv}{dt}}{v^{2}}$ and the chain - rule $\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$.
First, find $\frac{d}{dt}(\frac{2e^{2t}}{3t^{2}})$. Let $u = 2e^{2t}$ and $v = 3t^{2}$. Then $\frac{du}{dt}=4e^{2t}$ and $\frac{dv}{dt}=6t$.
$\frac{d}{dt}(\frac{2e^{2t}}{3t^{2}})=\frac{3t^{2}\times4e^{2t}-2e^{2t}\times6t}{(3t^{2})^{2}}=\frac{12t^{2}e^{2t}-12te^{2t}}{9t^{4}}=\frac{12te^{2t}(t - 1)}{9t^{4}}=\frac{4e^{2t}(t - 1)}{3t^{3}}$.
Since $\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$, and $\frac{dx}{dt}=3t^{2}$, then $\frac{d^{2}y}{dx^{2}}=\frac{\frac{4e^{2t}(t - 1)}{3t^{3}}}{3t^{2}}=\frac{4e^{2t}(t - 1)}{9t^{5}}$.

Step4: Find intervals of concavity

The curve is concave up when $\frac{d^{2}y}{dx^{2}}>0$. Since $e^{2t}>0$ for all real $t$, we consider the sign of $\frac{4(t - 1)}{9t^{5}}$.
Set $\frac{4(t - 1)}{9t^{5}}>0$. The critical points are $t = 0$ and $t = 1$.
Using a sign - chart, we find that the inequality is satisfied for $t<0$ or $t > 1$.

Answer:

a) $\frac{dy}{dx}=\frac{2e^{2t}}{3t^{2}}$
b) $\frac{d^{2}y}{dx^{2}}=\frac{4e^{2t}(t - 1)}{9t^{5}}$
c) $(-\infty,0),(1,\infty)$