# I Hi, how to find the magnitude of a rotating vector?

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1. Aug 19, 2016

### rashida564

i get stuck in how to find the magnitude of rotating vector . why say that |dA/dt|=A(dθ/dt) but who we can derive it or interpret this fact

2. Aug 20, 2016

### Simon Bridge

Start from the definitions.

3. Aug 20, 2016

### rashida564

the rotating vector have a constant magnitude but it change it's angle by the time

4. Aug 21, 2016

### Simon Bridge

OK. So write that out mathematically.

5. Aug 23, 2016

### rashida564

write the definition mathematically

6. Aug 24, 2016

### Simon Bridge

Do you not know how to write down the equation for a rotating vector?
A vector in rectangular coordinates is written: $\vec v = v_x\hat\imath + v_y\hat\jmath$
The magnitude of that vector is $|\vec v| = v = \sqrt{v_x^2+v_y^2}$

If the vector makes angle $\theta$ to the x-axis, then:
$v_x=v\cos\theta,\; v_y=v\sin\theta$
If the vector rotates, then $\theta$ is a function of time.

From there, and the definition of the derivative, you can derive and interpret the relation yourself.
So here is your task: prove $\big|\frac{d}{dt} \vec v(t) \big| = v\frac{d}{dt}\theta(t)$

Last edited: Aug 24, 2016
7. Aug 25, 2016

### rashida564

let v=r
dr/dt = lim Δt→0 Δr/Δt
Δr=2rsin(θ/2)
Δr≈rθ
dr/dt=lim Δt→0 Δrθ/Δt
|dt/dt| = r(dθ/dt)
is it correct

8. Aug 26, 2016

### Simon Bridge

No. dt/dt = 1, your delta-r is a non sequitur... basically, what you wrote is nonsense.
It looks like you need to revise your notes about writing vectors.

9. Aug 26, 2016

### rashida564

i saw the explanation from Kleppner and Kolenkow book

10. Aug 29, 2016

### Simon Bridge

What explanation?
Reread post #6. If you will not take advise I cannot help you.
Do you know how to write a vector?

Last edited: Aug 29, 2016
11. Aug 29, 2016

### rashida564

a vector in rectangular coordinates system is written by it's x and y component
then we should write the vector as vcos(θ)i+vsin(θ) j

12. Aug 30, 2016

### Simon Bridge

Well done... so write that out, A= or whatever letter you wsnt to use for a vector.
Then differentiate both sides with respect to time ... you will need the chain rule because $\theta$ is a function of time.

13. Sep 1, 2016

### rashida564

dA/dt=-sin(θ)dθ/dt i +vcos(θ)dθ/dt j

14. Sep 1, 2016

### Svein

You forgot the v in front of sin(θ). Now figure out what angle there is between A and dA/dt .

15. Sep 2, 2016

### Simon Bridge

Don't worry about the angle... it's a shortcut but you don't need to figure that out.
Your next step is to find the magnatude. Don't forget that |A| = v
Find the expression for |dA/dt| and simplify.

16. Sep 2, 2016

### rashida564

now i get it
|V|= √((-vsin(θ)dθ/dt)^2+(vcos(θ)dθ/dt)^2)
|V|=vdθ/dt*√(sin^2(θ)+cos^2(θ))
|V|=vdθ/dt

17. Sep 3, 2016

### Simon Bridge

Well done.
Sometimes you have to just go through the steps without knowing for sure where you will end up.
Of course I could just have told you ... but then you'd have missed out on that "ahah" moment.

18. Sep 4, 2016

### rashida564

thank you

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