I need help with this problem:
Prove that if a particule moves in constant speed, then its acceleration is orthogonal to its velocity.
I tried to prove it like this:
If the speed of $f$ is constant, that means that $\Vert f'(t)\Vert=\Vert(x'(t), y'(t))\Vert=c$. If the acceleration is orthogonal to the velocity, that means that $f'(t)\cdot f''(t)=0$, thus $x'(t)x''(t)+y'(t)y''(t)=0$ and that means that $x'(t)x''(t)= -y'(t)y''(t)$.
I really think I'm wrong. Please can you help me prove this correctly?
Best Answer
If the speed $\Vert f'(t)\Vert$ is constant then its derivative is zero. The calculation becomes even simpler if we consider the square of the speed (which is constant as well): $$ 0 = \frac{d}{dt} \Vert f'(t)\Vert^2 = \frac{d}{dt} \left( x'(t)^2 + y'(t)^2 \right) = 2 x'(t) x''(t) + 2 y'(t) y''(t) = 2 f'(t) \cdot f''(t) $$ so that $f'(t) \cdot f''(t) = 0$, i.e. the speed and the acceleration are orthogonal at each time.