# Prove that v(t) is any vector that depends on time

1. Sep 18, 2007

### Oblio

Prove that v(t) is any vector that depends on time (for example the velocity of a moving particle) but which has constant magnitude, then $$\dot{v}$$(t) is orthogonal to v(t). Prove the converse that if $$\dot{v}$$(t) is orthogonal to v(t), then lv(t)l is constant. Hint: consder the derivative of v^2.

This is a very hand result. It explains why, in 2-D polars, d$$\hat{r}$$/dt has to be in the direction of $$\hat{\phi}$$ and vice versa. It also shows that the speed of a charged particle in a magnetic field is constant, since teh acceleration is perpendicular to the velocity.

How to I approach proving this? I'm also unclear on Orthogonal, which I'll look up now but if someone could explain this to compliment what I'm off to figure out, I'd appreciate it a great deal.
Thanks!

2. Sep 18, 2007

### learningphysics

The trick is to use this equation $$\vec{v}\cdot\vec{v} = |\vec{v}|^2$$

Try taking the derivative of both sides... see what happens.

3. Sep 18, 2007

### Oblio

How did you know that?

4. Sep 18, 2007

### learningphysics

Just kind of guessed... They want orthogonality... Orthogonality means perpendicular... so I thought it probably had to do with dot products. I looked up the derivative of the dot product, and checked that this idea worked.

When you have a problem that has to do with orthogonality... or angles between vectors... and the magnitudes of the vectors... you're probably dealing with dot product or cross product...

5. Sep 18, 2007

### Oblio

haha. You's brilliant eh?

I'll let you know how it turns out

6. Sep 18, 2007

### learningphysics

lol Nope. I wish. It's just experience. Just seen these types of problems so many times. So I kind of know what to expect.

cool.

7. Sep 18, 2007

### Oblio

I forget, do I treat v as a variable or does a vector differentiate differently?

8. Sep 18, 2007

### learningphysics

look up the rule for taking the derivative of a dot product.

http://mathworld.wolfram.com/DotProduct.html

The $$|\vec{v}|$$ is just a scalar... so just differentiate that normally, like an ordinary variable.

9. Sep 19, 2007

### Oblio

$$\vec{v}$$.$$\frac{d\vec{v}}{dt}$$ + $$\vec{v}$$.$$\frac{d\vec{v}}{dt}$$ = 2l$$\vec{v}$$l

Hows that look so far?

10. Sep 19, 2007

### Oblio

That's not actually equal is it?

11. Sep 19, 2007

### learningphysics

Left side looks good. Right side you need to multiply that by $$\frac{d|\vec{v}|}{dt}$$, using the chain rule. because you're taking the derviative wrt t.

12. Sep 19, 2007

### Oblio

Right... oops. (embarrasing)

that gives me
2$$\vec{v}$$$$\vec{a}$$ = 2l$$\vec{v}$$l $$\vec{a}$$

and v and a are perpendicular?

13. Sep 19, 2007

### learningphysics

careful. left side looks good, but not the right side. $$\vec{a} = \frac{d\vec{v}}{dt}$$. But $$\frac{d|\vec{v}|}{dt}$$ isn't the same as $$\frac{d\vec{v}}{dt}$$ (ie think of the velocity keeping the same magnitude but changing directions.)

You know the magnitude of the velocity is constant for this question, hence $$\frac{d|\vec{v}|}{dt}$$ is 0.

14. Sep 19, 2007

### Oblio

Right, the slope of a scalar.
So, 2va = 0?

15. Sep 19, 2007

### learningphysics

yeah, $$2\vec{v}\cdot\vec{a} = 0$$...hence $$\vec{v}\cdot\vec{a} = 0$$ if the dot product of two vectors is 0, then they are orthogonal/perpendicular.

16. Sep 19, 2007

### Oblio

To prove the converse that if $$\dot{v}$$(t) is orthogonal to v(t), (that v(t) is constant), do i start in the same manner?

17. Sep 19, 2007

### Oblio

a dv/dt + v da/dt = 0

this leaves me with the derivative of acceleration though...

18. Sep 19, 2007

### learningphysics

Do the same thing as the first part... except backwards...

basically it's just the reverse of the steps starting with the end...

start with $$\vec{v}\cdot\vec{a} = 0$$

19. Sep 19, 2007