Proof that v'(t) is orthogonal to v(t)

[Nano-Passion]

Homework Statement

Prove that if v(t) is any vector that depends on time, then v'(t) is orthogonal to v(t).

Hint given: Consider the derivative of v^2.

The Attempt at a Solution

V'^2 = d/dt (v * v)
= v d/dt + v d/dt
= d/dt (v+v)
= 2v d/dt

??

 

[voko]

For ANY vector that's not true. E.g., the displacement, velocity and acceleration vectors of a body falling from rest are all parallel.

The statement is true for vectors whose magnitude is constant (but direction changes). Can you see the relevance of the hint given?

 

[Curious3141]

Homework Statement

Prove that if v(t) is any vector that depends on time, then v'(t) is orthogonal to v(t).

Hint given: Consider the derivative of v^2.

The Attempt at a Solution

V'^2 = d/dt (v * v)
= v d/dt + v d/dt
= d/dt (v+v)
= 2v d/dt

??

Your notation is atrocious, frankly. What you wrote, ${(v')}^2$ represents the square of the derivative, not the derivative of the square.

There are many other errors in the rest of the working as well. It's hard to tell if they represent typos or errors in thinking. Please use LaTex.

What you're supposed to be focussing on is the time derivative of $\overrightarrow{v}.\overrightarrow{v}$, i.e. $\frac{d}{dt}(\overrightarrow{v}.\overrightarrow{v})$. What happens when the velocity has a constant magnitude?