Parametric curves (derivative)

Oct 2009
77
5
\(\displaystyle x = e - e^t\)
\(\displaystyle y = t e^{-t}\)

Find \(\displaystyle \frac{dy}{dx}\) and \(\displaystyle \frac{d^2y}{dx^2}\) and state where the graph is concave upward and downward.

I found dy/dx already: \(\displaystyle \frac{1-e^{-t}}{1-e^t}\)

But when I used the quotient rule to find \(\displaystyle \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\) I got \(\displaystyle \frac{e^{-t}-1+e^t-1}{(1-e^t)^3}\)

I think this is right, but I might have entered it wrong. Is it?

I know the concavity is based on when the second derivative is positive and negative.
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
\(\displaystyle x = e - e^t\)
\(\displaystyle y = t e^{-t}\)

Find \(\displaystyle \frac{dy}{dx}\) and \(\displaystyle \frac{d^2y}{dx^2}\) and state where the graph is concave upward and downward.

I found dy/dx already: \(\displaystyle \frac{1-e^{-t}}{1-e^t}\)

But when I used the quotient rule to find \(\displaystyle \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\) I got \(\displaystyle \frac{e^{-t}-1+e^t-1}{(1-e^t)^3}\)
\(\displaystyle \frac{dy}{dx} = \frac{(1-t)e^{-t}}{-e^t} = \frac{t-1}{e^{2t}}\)

\(\displaystyle \frac{d^2y}{dx^2} = \frac{e^{2t} - (t-1) \cdot 2e^{2t}}{e^{6t}} = \frac{3-2t}{e^{4t}}\)
 
Oct 2009
77
5
Er, sorry. Something got lost when I was formatting. Big time.

The equations should be:

\(\displaystyle x = t - e^t\)
\(\displaystyle y = t + e^{-t}\)

My bad. I meant that I had found dy/dx and it had been marked as correct.