Question

parametric calculus lecture participation day 1: problem 2
(3 points)
consider the parametric curve given by
$x = t+ln(t)$, $y = 4t - 4ln(t)$
(a) find $dy/dx$ and $d^{2}y/dx^{2}$ in terms of $t$.
$dy/dx=$
$d^{2}y/dx^{2}=$
(b) using \less than\ and \greater than\ notation, list the $t$-interval where the curve is concave upward.
use upper - case \inf\ for positive infinity and upper - case
inf\ for negative infinity. if the curve is never concave upward, type an upper - case
\ in the answer field.
$t$-interval: $note: you can earn partial credit on this problem.

Explanation:

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

For \(x = t+\ln(t)\), \(\frac{dx}{dt}=1 + \frac{1}{t}=\frac{t + 1}{t}\) (using the sum rule and the derivative of \(\ln(t)\)).
For \(y=4t-4\ln(t)\), \(\frac{dy}{dt}=4-\frac{4}{t}=\frac{4t - 4}{t}\) (using the sum rule and the derivative of \(\ln(t)\)).

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

By the formula \(\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\), we have \(\frac{dy}{dx}=\frac{\frac{4t - 4}{t}}{\frac{t + 1}{t}}=\frac{4(t - 1)}{t+1}\) (canceling out the \(t\) terms).

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

First, find \(\frac{d}{dt}(\frac{dy}{dx})\). Using the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\) where \(u = 4(t - 1)\), \(u^\prime=4\), \(v=t + 1\), \(v^\prime=1\). So \(\frac{d}{dt}(\frac{dy}{dx})=\frac{4(t + 1)-4(t - 1)}{(t + 1)^{2}}=\frac{4t+4-4t + 4}{(t + 1)^{2}}=\frac{8}{(t + 1)^{2}}\).
Then, \(\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}=\frac{\frac{8}{(t + 1)^{2}}}{\frac{t + 1}{t}}=\frac{8t}{(t + 1)^{3}}\).

Step4: Find where the curve is concave upward

The curve is concave upward when \(\frac{d^{2}y}{dx^{2}}>0\). Since \(\frac{8t}{(t + 1)^{3}}>0\).
The denominator \((t + 1)^{3}>0\) when \(t>- 1\) and \((t + 1)^{3}<0\) when \(t<-1\). The numerator \(8t>0\) when \(t>0\) and \(8t<0\) when \(t<0\).
We consider the domain of the original parametric equations. For \(x=t+\ln(t)\), the domain is \(t>0\).
When \(t>0\), \(\frac{8t}{(t + 1)^{3}}>0\) (because \(t>0\) and \(t + 1>0\)).

Answer:

\(\frac{dy}{dx}=\frac{4(t - 1)}{t + 1}\)
\(\frac{d^{2}y}{dx^{2}}=\frac{8t}{(t + 1)^{3}}\)
\(t-\text{interval}:0