# Math Help - Parametric curves (derivative)

1. ## Parametric curves (derivative)

$x = e - e^t$
$y = t e^{-t}$

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

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

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

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.

2. Originally Posted by Open that Hampster!
$x = e - e^t$
$y = t e^{-t}$

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

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

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

$\frac{d^2y}{dx^2} = \frac{e^{2t} - (t-1) \cdot 2e^{2t}}{e^{6t}} = \frac{3-2t}{e^{4t}}$

3. Er, sorry. Something got lost when I was formatting. Big time.

The equations should be:

$x = t - e^t$
$y = t + e^{-t}$