Derivative of a vector with constant magnitude

Question

I was reading the definition of work done in terms of a kinetic energy.

It read that when a force is applied to a particle moving with constant velocity v, its kinetic energy changes as follows:- $$\frac{dK}{dt} = \frac{1}{2}\frac{d(m\vec v.\vec v)}{dt} = \frac{1}{2}\frac{d(mv^2)}{dt} = m\vec v\frac{d\vec v}{dt}$$ The speed of a particle does not change when the force applied on it is perpendicular to the velocity and as a result, the kinetic energy of the particle also doesn't change.

Hence the force acting on the particle must have a tangential component.
Therefore:- $$\frac{dK}{dt} = \vec F.\vec v$$

My query:

I know that the symbol $\vec v$ represents velocity(which is a vector) and $v^2$ or $\vec v.\vec v$ is a scalar. But then why is $\frac{d\vec v}{dt}$ equal to zero when the force applied on a particle is perpendicular since the velocity clearly is changing,at least its direction is.

Answer 1

The last expression should be a dot product:

$$\frac{dK}{dt} = m\vec v \cdot \frac{d\vec v}{dt}$$

So when a force is applied perpendicularly to the direction of motion, $\frac{d\vec v}{dt}$ is a vector that is perpendicular to $\vec v$, so their dot product is zero, giving you zero change in the kinetic energy as expected.

Answer 2

the magnitude of the vector $\vec v $ is constant, so is it's square $v^2= \vec v \cdot \vec v$ , therefore the derivative over time is zero (the derivative of a constant is zero) therefore from the expression: $$\frac{d(\vec v\cdot\vec v)}{dt} = \vec v \cdot \frac{d\vec v}{dt} +\vec v\cdot \frac{d\vec v}{dt} = 0$$ it implies that $\vec v $ is orthogonal to $\frac{d\vec v}{dt}$(from the definitions of orthogonal vector) if $\vec v$ is different to zero.

Answer 3

\begin{equation} \frac{d\tilde{v}}{dt} \neq 0. \end{equation}

It is not necessary that the vector change in velocity be zero to prove constant Kinetic energy. If the Force is perpendicular to the velocity direction, then $\tilde{F} \cdot \tilde{v} = 0$, but $\tilde{F} \neq 0,$ in general. The other way to see this is

\begin{equation} \frac{dK}{dt} = \tilde{F} \cdot \tilde{v} = \frac{1}{2} \frac{d}{dt} (v^2) = 0, \end{equation}

if the speed is constant. Thus it's a matter of force being perpendicular to the motion, not zero.

Answer 4

that first equation seems fishy; you're getting a 'vector kinetic energy' by differentiating the velocity? it is the 'speed' or simply the magnitude of velocity 'v' (without the vector sign) that needs to be differentiated.

the speed is constant so that turns out to be zero!