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

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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 [tex]\dot{v}[/tex](t) is orthogonal to v(t). Prove the converse that if [tex]\dot{v}[/tex](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[tex]\hat{r}[/tex]/dt has to be in the direction of [tex]\hat{\phi}[/tex] 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

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The trick is to use this equation [tex]\vec{v}\cdot\vec{v} = |\vec{v}|^2[/tex]

Try taking the derivative of both sides... see what happens.
 
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How did you know that?
 

learningphysics

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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...
 
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haha. You's brilliant eh?

I'll let you know how it turns out
 

learningphysics

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

learningphysics

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[tex]\vec{v}[/tex].[tex]\frac{d\vec{v}}{dt}[/tex] + [tex]\vec{v}[/tex].[tex]\frac{d\vec{v}}{dt}[/tex] = 2l[tex]\vec{v}[/tex]l

Hows that look so far?
 
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That's not actually equal is it?
 

learningphysics

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[tex]\vec{v}[/tex].[tex]\frac{d\vec{v}}{dt}[/tex] + [tex]\vec{v}[/tex].[tex]\frac{d\vec{v}}{dt}[/tex] = 2l[tex]\vec{v}[/tex]l

Hows that look so far?
Left side looks good. Right side you need to multiply that by [tex]\frac{d|\vec{v}|}{dt}[/tex], using the chain rule. because you're taking the derviative wrt t.
 
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Right... oops. (embarrasing)

that gives me
2[tex]\vec{v}[/tex][tex]\vec{a}[/tex] = 2l[tex]\vec{v}[/tex]l [tex]\vec{a}[/tex]

and v and a are perpendicular?
 

learningphysics

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Right... oops. (embarrasing)

that gives me
2[tex]\vec{v}[/tex][tex]\vec{a}[/tex] = 2l[tex]\vec{v}[/tex]l [tex]\vec{a}[/tex]

and v and a are perpendicular?
careful. left side looks good, but not the right side. [tex]\vec{a} = \frac{d\vec{v}}{dt}[/tex]. But [tex]\frac{d|\vec{v}|}{dt}[/tex] isn't the same as [tex]\frac{d\vec{v}}{dt}[/tex] (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 [tex]\frac{d|\vec{v}|}{dt}[/tex] is 0.
 
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Right, the slope of a scalar.
So, 2va = 0?
 

learningphysics

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Right, the slope of a scalar.
So, 2va = 0?
yeah, [tex]2\vec{v}\cdot\vec{a} = 0[/tex]...hence [tex]\vec{v}\cdot\vec{a} = 0[/tex] if the dot product of two vectors is 0, then they are orthogonal/perpendicular.
 
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To prove the converse that if [tex]\dot{v}[/tex](t) is orthogonal to v(t), (that v(t) is constant), do i start in the same manner?
 
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a dv/dt + v da/dt = 0

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

learningphysics

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To prove the converse that if [tex]\dot{v}[/tex](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 [tex]\vec{v}\cdot\vec{a} = 0[/tex]
 
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integrate instead of derive?
 
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which just leaves you with a dot product between position and v?

v.x = 0
 

learningphysics

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

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