Question

consider the parametric curve given by

$x = t+ln(t)$ and $y = 8t - 8ln(t)$ with $t>0$.

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 \(\frac{dy}{dt}\)

For \(x = t+\ln(t)\), by the sum - rule and the derivative of \(\ln(u)\) (\(u = t\), \(\frac{d}{dt}\ln(u)=\frac{1}{u}\frac{du}{dt}\)), \(\frac{dx}{dt}=1 + \frac{1}{t}=\frac{t + 1}{t}\).
For \(y=8t-8\ln(t)\), using the sum - rule and the derivative of \(\ln(u)\) (\(u = t\)), \(\frac{dy}{dt}=8-\frac{8}{t}=\frac{8t-8}{t}=\frac{8(t - 1)}{t}\).

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

By the formula \(\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), we substitute the values of \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\):
\(\frac{dy}{dx}=\frac{\frac{8(t - 1)}{t}}{\frac{t + 1}{t}}=\frac{8(t - 1)}{t+1}\).

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

First, find \(\frac{d}{dt}(\frac{dy}{dx})\). Using the quotient rule \(\frac{d}{dt}(\frac{u}{v})=\frac{u^{\prime}v - uv^{\prime}}{v^{2}}\), where \(u = 8(t - 1)\), \(u^{\prime}=8\), \(v=t + 1\), \(v^{\prime}=1\).
\(\frac{d}{dt}(\frac{dy}{dx})=\frac{8(t + 1)-8(t - 1)}{(t + 1)^{2}}=\frac{8t+8-8t + 8}{(t + 1)^{2}}=\frac{16}{(t + 1)^{2}}\).
Then, by the formula \(\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\), substitute \(\frac{d}{dt}(\frac{dy}{dx})\) and \(\frac{dx}{dt}\):
\(\frac{d^{2}y}{dx^{2}}=\frac{\frac{16}{(t + 1)^{2}}}{\frac{t + 1}{t}}=\frac{16t}{(t + 1)^{3}}\).

Step4: Determine the concavity

The curve is concave upward when \(\frac{d^{2}y}{dx^{2}}>0\).
Since \(\frac{d^{2}y}{dx^{2}}=\frac{16t}{(t + 1)^{3}}\) and \(t>0\), for \(t>0\), the numerator \(16t>0\) and the denominator \((t + 1)^{3}>0\) (because \(t>0\)).

Answer:

\(\frac{dy}{dx}=\frac{8(t - 1)}{t + 1}\)
\(\frac{d^{2}y}{dx^{2}}=\frac{16t}{(t + 1)^{3}}\)
Concave up on: \((0,\infty)\)