Basically, I’m trying to solve the following differential equation, but for some reason I’m getting the wrong result and I don’t fully understand why.
$$ \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-1}{\frac{\mathrm{d} ax^2}{\mathrm{d} x}} $$
By $ \frac{\mathrm{d} y}{\mathrm{d} x} $ I'm referring to the slope of the curve I'm trying to find at every point it has (x,y). By $ \frac{\mathrm{d} ax^2}{\mathrm{d} x} $ I'm referring to the slope of the parabola with the peak at x=0. Since the tangents must be perpendicular one slope must be -1 over the other.
To me it seems like it describes what I mean, however when I try to solve it, I get:
$$ \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-1}{2ax} $$
$$ \mathrm{d} y = \frac{-1}{2ax}\mathrm{d} x $$
$$ y = \int \frac{-1}{2ax}dx $$
$$ y = \frac{-1}{2a}lnx + c $$
Which is not what I'm looking for:
(graphed x^2 and it's corresponding curve)
Sure, they ARE perpendicular, but they aren't perpendicular regardless of my choice of parabola, as in it isn't also perpendicular to, say, 2x^2, at their point of intersection.
I'm sorry if I'm not being concise with what I want to end up with, but hopefully I can clear things up with this picture. As you can see, $ y^2 + \frac{1}{2}x^2 = 1 $ is perpendicular at all points of intersection with any parabola of the form $ ax^2 $
Can anybody please tell me what I did wrong? Considering that my "wrong" result is still perpendicular to a parabola, I don't think that I made a mistake while solving the equation itself, but that rather I'm solving the wrong equation. If that is the case, I wish to know what is the corresponding (or "correct") equation, why, and what is the meaning of the one I solved here.
Best Answer
You obtained the curve which is perpendicular to the parabola $y=ax^2$, but not to all parabolas. If you want to have the curve which is perpendicular to all parabolas, you have to make $a$ dependent of $x$ and $y$ - at every point, $y(x)$ is perpendicular to $ax^2$ for different $a$.
Indeed you have $$ \frac{dy}{dx} = \frac{-1}{2ax}, $$ but at a point $x$, $y(x)$ crosses the parabola with $a=\frac{y}{x^2}$. So if we plug this $a$ into an equation, we have $$ \frac{dy}{dx} = \frac{-x}{2y} $$ and after solving it we get an ellipse.