Vector velocity and acceleration proof i) Show that for a particle moving with velocity $v(t), $if $ v(t)·v′(t) = 0$ for all $t$ then the speed $v$ is constant. I did $(v(t))^2=|v(t)|^2=(v(t)\bullet(v(t)))$. Therefore $\frac{d}{dt}(v(t))^2=2v(t)$ Also, $\frac{d}{dt}(v(t)\bullet(v(t))=2(v(t)·v′(t))$ I'm stuck here. ii) A particle of mass m with position vector r(t) at time t is acted on by a total force F (t) = λr(t) × v(t), where λ is a constant and v(t) is the velocity of the particle. Show that the speed v of the particle is con- stant. (Note that Newton’s second law of motion in its vector form is F = ma.) Therefore, ma(t)=λ(r(t) × v(t)) after which I don't know what to do.

Since $$\frac{d}{dt}|v(t)|^2 = \frac{d}{dt}[v(t)\cdot v(t)] = 2v(t)\cdot v'(t) = 0$$

we have that $|v(t)|^2$ is constant, hence $|v(t)|$ is constant.

For the second problem, note that we can write

$$mv'(t) = \lambda r(t)\times v(t).$$

So

$$\frac{d}{dt}|v(t)|^2 = 2v(t)\cdot v'(t) = 2v(t)\cdot \left(\frac{\lambda}{m}r(t)\times v(t)\right) = 0,$$

since $\frac{\lambda}{m}r(t)\times v(t)$ is orthogonal to $v(t)$. Therefore, $|v(t)|$ is constant.