Clarification over Instantaneous Velocity and the Shape of a Curve Assume a ball sliding down a curve with no friction. Conventional thinking would suggest that the steeper the curve, the faster the ball will roll down. But when I derive the formula for velocity, I get:
$$v=\sqrt{2gy}$$
Which suggests that velocity only depends on the vertical height it falls down, upon which $GPE$ is converted into $KE$ which translates into velocity.
I am however interested in the point of maximum instantaneous velocity of different curves. So my question is: would these be a certain point on the curve (e.g. the point with maximum gradient), or is it just the lowest point on the curve (which my equation suggests).
 A: As the ball is subjected to gravitational (external) force throughout its fall, it is accelerating continuously. As a result, the velocity will be maximum at the lowest point for any given curve.
A: 
Conventional thinking would suggest that the steeper the curve, the faster the ball will roll down

Depends on velocity projection. The one along vertical (aligned with Earth gravity field) does not depend on path taken (steepness of ramp, number of curves, etc.) and will be the same at the end of path as per $$mg\Delta h = \frac {m{v_{_{||}}}^2}{2}.$$
But the one along horizontal,- perpendicular to gravity field indeed depends on exact path taken, because in this direction object gains speed due to superposition of gravity and normal force $\vec F_{accelerating}=\vec W + \vec N$. So, the longer the ramp or bigger length curve (even with same total height),- the greater speed object will achieve in $x$ direction (perpendicular to gravity field), because normal force will have more time to speed-up the object.
