Proving speed is constant if the velocity and acceleration vectors are perpendicular.

Question

The question is: Prove that if the velocity and acceleration vectors of a parameterized curve X(t) are always perpendicular, the speed equals a constant.

My attempt: I figured I could let the curve be given by a function that I know to always have perpendicular velocity and acceleration vectors, which is the equation of a circle. Then I made the parameters be x(t) = cos(t) and y(t) = sin(t), and since speed is the square root of the velocity vector dotted with itself, it is easy to see here that the speed would always be 1, a constant.

That is the only way I can figure out how to arrive at an answer, but it doesn't feel right to me. I feel like I'm not being general enough, or that I'm not being rigorous in my supposition that X(t) is necessarily given by a circle. Was this a valid way of doing this proof?

Answer 1

HINT

\begin{align*} \|v(t)\| = k & \Longleftrightarrow \|v(t)\|^{2} = \langle v(t),v(t)\rangle = k^{2}\\\\ & \Longleftrightarrow \langle v'(t),v(t)\rangle + \langle v(t),v'(t)\rangle = 0\\\\ & \Longleftrightarrow \langle a(t),v(t)\rangle = 0 \end{align*}

Answer 2

Let $(x(t),y(t)$ be the parametrisation of the curve.

the velocity is $$\vec{v}=(x'(t),y'(t))$$

the acceleration is $$ \vec{a}= (x''(t),y''(t))$$

if they are perpendicular, we will have $$\vec{v}•\vec{a} = 0$$ or

$$2 \Bigl(x'(t)x''(t) + y'(t)y''(t)\Bigr)=0$$

and this is the derivative of

$$x'^2(t) + y'^2(t)=||\vec{v}||^2$$

which is Constante.