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

[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!

 

[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.

 

[Oblio]

How did you know that?

 

[learningphysics]

How did you know that?

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...

 

[Oblio]

haha. You's brilliant eh?

I'll let you know how it turns out

 

[learningphysics]

haha. You's brilliant eh?

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.

I'll let you know how it turns out

cool.

 

[Oblio]

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

 

[learningphysics]

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

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.

 

[Oblio]

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

Hows that look so far?

 

[Oblio]

That's not actually equal is it?

 

[learningphysics]

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

Hows that look so far?

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.

 

[Oblio]

Right... oops. (embarrasing)

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

and v and a are perpendicular?

 

[learningphysics]

Right... oops. (embarrasing)

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

and v and a are perpendicular?

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.

 

[Oblio]

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

 

[learningphysics]

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

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.

 

[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?

 

[Oblio]

a dv/dt + v da/dt = 0

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

 

[learningphysics]

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?

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

 

[Oblio]

integrate instead of derive?

 

[learningphysics]

integrate instead of derive?

yes, exactly.

 

[Oblio]

which just leaves you with a dot product between position and v?

v.x = 0

 

[learningphysics]

which just leaves you with a dot product between position and v?

v.x = 0

No... do the exact opposite of what you did last time...

so start with the end:

v.a = 0

now you can go to the previous step by multiplying by 2:

2v.a = 0

now take the integral... from the last part, you know that the derivative of v.v is 2v.a , so what is the integral of 2v.a ?

 

[Oblio]

without doing part a, what would make you want to multiply by 2?

 

[Oblio]

just v.v

 

[learningphysics]

just v.v

Yes, and what happens to the right side?

 

[Oblio]

i know i shoul get the magnitude of v.. but how from 0?

 

[learningphysics]

i know i shoul get the magnitude of v.. but how from 0?

You're integrating both sides of:

2v.a = 0

what's the integral of the 0?

 

[Oblio]

the scalar of v..

 

[learningphysics]

the scalar of v..

yeah, it will turn out to be |v|^2... but when we integrate 0, we get C...

ie: integral of 0, is just a constant, so you can just call the right side C.

ie:

$$\vec{v}\cdot\vec{v} = C$$

 

[Oblio]

I need to work it to v^2?

Was the multiplying by two only by the knowledge of doing the first part?

 

[learningphysics]

I need to work it to v^2?

Was the multiplying by two only by the knowledge of doing the first part?

yeah sort of... for the second part you somehow need to figure out that $$\frac{d(\vec{v}\cdot\vec{v})}{dt} = 2\vec{\frac{dv}{dt}}\cdot\vec{v}$$

we already proved this in the first part.

so yes, v.v = |v|^2

|v|^2 = C, so |v|=sqrt(C), which is a constant.

 

[Oblio]

Alright, I wasn't sure if multiplying by two was also a reasonable 'step' without the preknowledge.