So I am given a problem stated as: a point moves in the plane at speed 1 along the curve $y = x^2$. Find the acceleration at the point (x,y).
I know that the velocity is y' = 2x, and that at a constant speed the acceleration is a vector perpendicular to the velocity.
What I would like to say is that the velocity vector is (x, 2x), so the perpendicular vector to that is (-2x, x) since two vectors x * y = 0 (dot product) are perpendicular. I'm asking this because it seems too damn easy to be true compared to all my other problems.
I think you are a little confused. If the curve is parametrized by the $x$ coordinate, we have that the velocity is given by $$\vec{v}(t)=\left(\frac{dx}{dt},\frac{dy}{dt}\right)=\left(\frac{dx}{dt},\frac{dy}{dx}\frac{dx}{dt}\right)=\frac{dx}{dt}\left(1,2x(t)\right)$$ Since we want $\|\vec{v}(t)\|=1$, we must have $\dfrac{dx}{dt}=\dfrac{1}{\sqrt{1+4x(t)^2}}$. For the acceleration, we have $$\vec{a}(t)=\vec{v}'(t)=\frac{d^2x}{dt^2}(1,2x(t))+\frac{dx}{dt}\left(0,2\frac{dx}{dt}\right)$$ But, $$\frac{d^2x}{dt^2}=\frac{d}{dx}\left(\frac{dx}{dt}\right)\cdot\frac{dx}{dt}=-\frac{4 x}{\left(1+4 x^2\right)^{3/2}}\dfrac{1}{\sqrt{1+4x^2}}=-\frac{4x}{(1+4x^2)^2}$$ Hence, $$\vec{a}(x)=\left(-\frac{4x}{(1+4x^2)^2},-\frac{8x^2}{(1+4x^2)^2}+\frac{2}{1+4x^2}\right)=\left(-\frac{4x}{(1+4x^2)^2},\frac{2}{(1+4x^2)^2}\right),$$ or $$\vec{a}(x,y)=\left(-\frac{4x}{(1+4y)^2},\frac{2}{(1+4y)^2}\right)$$