# Parametric curves (derivative)

#### Open that Hampster!

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

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

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 $$\displaystyle \frac{e^{-t}-1+e^t-1}{(1-e^t)^3}$$
$$\displaystyle \frac{dy}{dx} = \frac{(1-t)e^{-t}}{-e^t} = \frac{t-1}{e^{2t}}$$

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

#### Open that Hampster!

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

The equations should be:

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