# D/dt |r(t)| = (1/|r(t)|) r(t)dot r'(t) [this has been posted before.I need more help]

1. Sep 27, 2012

### bubbers

1. The problem statement, all variables and given/known data

If vector r(t) is not 0, show that d/dt |r(t)| = (1/|r(t)|) r(t)dot r'(t).

2. Relevant equations

The hint given was that |r(t)|^2 = r(t) dot r(t)

3. The attempt at a solution

the post I saw about this question said that you should take the derivatives of both sides of the hint equation using chain rule, so I got:

2|r(t)|(1)=r'(t) dot r(t) + r(t) dot r'(t)
which is equal to:
|r(t)|= r'(t) dot r(t)

aaaaand now I don't know what to do

Last edited: Sep 27, 2012
2. Sep 27, 2012

### jmcelve

Re: d/dt |r(t)| = (1/|r(t)|) r(t)dot r'(t) [this has been posted before.I need more h

This step is incorrect. You have the equivalent of f(g(x)) -- the chain rule says d/dx f(g(x)) = f'(g(x)) * g'(x).

3. Sep 27, 2012

### bubbers

Re: d/dt |r(t)| = (1/|r(t)|) r(t)dot r'(t) [this has been posted before.I need more h

Okay, now I'm confused... in this situation, wouldn't f'(g(x)) be equivalent to 2|r(t)| and g'(x) be equivalent to 0 because |r(t)| is a length and the derivative of a number is 0? So then it would be 2|r(t)|(0)=r'(t) dot r(t) + r(t) dot r'(t) --> 0=r'(t) dot r(t) + r(t) dot r'(t)?
If this is right, then I still don't know where to go from here.

4. Sep 27, 2012

### jmcelve

Re: d/dt |r(t)| = (1/|r(t)|) r(t)dot r'(t) [this has been posted before.I need more h

No. |r(t)| has t dependence. Accordingly, we can't assume the magnitude is constant.

5. Sep 27, 2012

### bubbers

Re: d/dt |r(t)| = (1/|r(t)|) r(t)dot r'(t) [this has been posted before.I need more h

Oh man...I totally get it now...i feel kinda dumb :P
the derivative would just be:
2|r(t)|d/dt|r(t)|= r'(t) dot r(t) + r(t) dot r'(t)
2|r(t)|d/dt|r(t)|= 2(r'(t) dot r(t))
|r(t)|d/dt|r(t)|= r'(t) dot r(t)
d/dt|r(t)|=(1/|r(t)|) ( r'(t) dot r(t))

Thanks for letting me waste your time :)

6. Sep 27, 2012

### jmcelve

Re: d/dt |r(t)| = (1/|r(t)|) r(t)dot r'(t) [this has been posted before.I need more h

No problem. Glad to have helped. :)