Question

consider the parametric curve given by

$x = 2 + t^{2}$ and $y=-6t^{2}-6t^{3}$, where $t$ is any real number.

a. find $\frac{dy}{dx}$ and $\frac{d^{2}y}{dx^{2}}$ in terms of $t$.

$\frac{dy}{dx}=$

$\frac{d^{2}y}{dx^{2}}=$

b. determine the open interval(s) of $t$-values where the curve is concave upward. enter dne if no such intervals exist. include the union symbol when entering multiple intervals.

concave up on:

Explanation:

Step1: Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)

Given \(x = 2+t^{2}\), then \(\frac{dx}{dt}=2t\).
Given \(y=-6t^{2}-6t^{3}\), then \(\frac{dy}{dt}=-12t - 18t^{2}\).

Step2: Calculate \(\frac{dy}{dx}\)

By the chain - rule \(\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), so \(\frac{dy}{dx}=\frac{-12t - 18t^{2}}{2t}=\frac{-6t(2 + 3t)}{2t}=-6 - 9t\) (\(t
eq0\)). When \(t = 0\), we can also use the limit definition of the derivative \(\lim_{\Delta t
ightarrow0}\frac{\Delta y}{\Delta x}=\lim_{\Delta t
ightarrow0}\frac{-6(\Delta t)^{2}-6(\Delta t)^{3}}{(\Delta t)^{2}}=- 6\), so \(\frac{dy}{dx}=-6 - 9t\).

Step3: Calculate \(\frac{d^{2}y}{dx^{2}}\)

We know that \(\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\).
Since \(\frac{dy}{dx}=-6 - 9t\), then \(\frac{d}{dt}(\frac{dy}{dx})=-9\).
So \(\frac{d^{2}y}{dx^{2}}=\frac{-9}{2t}\).

Step4: Determine where the curve is concave upward

The curve is concave upward when \(\frac{d^{2}y}{dx^{2}}>0\).
Set \(\frac{-9}{2t}>0\).
Multiply both sides by \(2t^{2}\) (since \(t^{2}>0\) for \(t
eq0\)), we get \(-9t>0\), which implies \(t < 0\).

Answer:

\(\frac{dy}{dx}=-6 - 9t\)
\(\frac{d^{2}y}{dx^{2}}=\frac{-9}{2t}\)
Concave up on: \((-\infty,0)\)