Parametric curves (derivative)

**Open that Hampster!** · May 10th 2010, 04:25 PM

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

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.

**skeeter** · May 10th 2010, 04:43 PM

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 $\frac{e^{-t}-1+e^t-1}{(1-e^t)^3}$

$\frac{dy}{dx} = \frac{(1-t)e^{-t}}{-e^t} = \frac{t-1}{e^{2t}}$

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

**Open that Hampster!** · May 10th 2010, 06:02 PM

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

The equations should be:

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

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

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