Orthogonal trajectories

**SyNtHeSiS** · November 2nd 2010, 09:03 AM

Show that the orthogonal trajectories for the family of curves $y = ae^{-x}$ (a constant) are a family of parabolas.

Attempt:

$y = ae^{-x}$

$\frac{dy}{dx} = -ae^{-x}$

$\frac{dy}{dx} = \frac{1}{ae^{-x}}$

$\int dy = \int \frac{1}{ae^{-x}}dx$

$y = \frac{1}{a}\int e^{x}dx$

$y = \frac{1}{a}e^{x} + C$

but the correct answer was:

$\frac{1}{2}y^2 = x +C$

I cant seem to see where I went wrong though.

**Unknown008** · November 2nd 2010, 09:09 AM

Welll, maybe you could just change this part:

$\dfrac{dy}{dx} = \dfrac{1}{ae^{-x}}$

$\dfrac{dy}{dx} = \dfrac{1}{y}$

$\int y \ dy = \int \ dx$

$\frac12 y^2 = x + C$

From what you did though.. maybe you can do this:

$y = \frac{1}{a}e^{x} + C$

$y = \frac{1}{ae^{-x}} + C$

$y = \frac{1}{y} + C$

$y^2 = 1 + yC$

$y^2 = 1 + ae^{-x}C$

But then, it gets complicated for me... the constant becomes another constant...

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