Prove that a non-zero acceleration is perpendicular to a constant speed 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)?
 A: If $v(t) = \bigl(x_1(t),x_2(t),\ldots,x_n(t)\bigr)$ (I don't know what dimension you are working on), then the speed is given by
$$||v(t)|| = \sqrt{ x_1(t)^2  + \cdots + x_n(t)^2}.$$
Since this is constant, say equal to $k$, then you have
$$k^2 = x_1(t)^2 + \cdots + x_n(t)^2.$$
Now take derivatives with respect to $t$. Compare that with $v\cdot a$ (remembering that $a(t) = v'(t)$). 
(Of course, if $\frac{dv}{dt}$ is zero, then it is also perpendicular to $v(t)$, since $\mathbf{0}$ is perpendicular to every vector...)
A: You can apply the product rule when differentiating $\vec v\cdot\vec v =$constant.
The intuitive idea is that $\vec v(t)$ traces out a curve on a sphere centered at the origin (i.e., picture $\vec v(t)$ as a moving radius vector), while $\vec v'(t)=\vec a(t)$ is tangent to the sphere, and hence perpendicular to the radius at the point of tangency, namely $\vec v(t)$.
There is also a geometric description of what this says in terms of the original curve, say $\vec x(t)$, of which $\vec v(t)$ gives the velocity.  Since $\vec v(t)$ gives a vector tangent to the trajectory of $\vec x(t)$, and in your case $\vec a(t)$ is perpendicular to $\vec v(t)$, $\vec a(t)$ is perpendicular to the trajectory of the original curve $\vec x(t)$.  
Acceleration can occur for 2 reasons.  In general, the component of $\vec a(t)$ in the direction of the trajectory (or opposite this direction) tells you how the speed is changing, while the component of $\vec a(t)$ perpendicular to the trajectory tells you how the direction is changing.  So, again, in case the speed is constant, $\vec a(t)$ is perpendicular to the curve.  Its direction tells you which direction $\vec x(t)$ is turning, while its magnitude tells you how quickly $\vec x(t)$ is changing direction.  This last quantity is a constant multiple of the curvature of $\vec x(t)$.
A: The shortest way is to notice that:
$$v^2=\vec{v}\cdot\vec{v}$$  
After differentiating: $$2v\frac{dv}{dt}=2\vec{v}\cdot\frac{d\vec{v}}{dt}$$ Done.
A: Lemma 1: Velocity vector is tangent to the velocity curve.
Then consider the acceleration as a limit of velocity difference. Since the magnitude of the velocity vector is constant, the velocity vector traverse the circle of a constant radius. The difference of the two infinitely close velocity vectors is perpendicular to the radius, hence is perpendicular to velocity vector. Here the solution follows immediately.
A: The dot product of the vector with itself is the square of its magnitude, which is given to be constant
$$\vec{v} \cdot \vec{v} = c.$$
Differentiate both sides with respect to $t$, use the vector product rule for the left-side:
$$2\vec{v} \cdot \frac{d\vec{v}}{dt} = 0.$$
Notice that $\frac{d\vec{v}}{dt}$ is the acceleration vector. This proves that non-zero acceleration must be orthogonal to velocity if the speed is constant. Run it backwards to prove the converse.
