Take the differentiable vector function $\vec{v}(t)$ (a velocity vector). If its speed, $|\vec{v}(t)|=constant$, then prove that at any point which $\frac{d\vec{v}}{dt}$ is non-zero, $\frac{d\vec{v}}{dt}$ is perpendicular to $\vec{v}(t)$.
In other words, if a velocity vector has constant speed, show that whenever its acceleration vector is non-zero, it is perpendicular to that velocity vector.
Intuitively this seems clear, since whenever the speed is remains constant, the acceleration is 0. Why though are the velocity and acceleration vectors perpendicular though? Because they share no intersection when the acceleration is non-zero since the speed is constant?
I'm assuming we should somehow show that $\vec{v} \cdot \vec{a} = 0$. The thing is, I can't figure out what knowing that the speed is constant tells us about the velocity. It could still be anything (like some trig functions squaring up and becoming 1 through an identity)?