prove that

d/dt(rsub1(t) dot rsub2(t) ) = r'sub1(t) dot r'sub2(t) + rsub1(t) dot r'sub2(t)

Thank you very much

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- February 22nd 2007, 09:42 AMJenny20a calculus proof
prove that

d/dt(rsub1(t) dot rsub2(t) ) = r'sub1(t) dot r'sub2(t) + rsub1(t) dot r'sub2(t)

Thank you very much - February 22nd 2007, 10:13 AMJhevon
The dot here means dot product right? cause if it meant multiply it wouldn't be a prob, it would just be the product rule.

- February 22nd 2007, 10:52 AMJenny20
Hi Jhevon,

You are right. I got this part. It is actually very simple. Thank you very much.

what about the below part :

use

d/dt(rsub1(t) dot rsub2(t) ) = r'sub1(t) dot r'sub2(t) + rsub1(t) dot r'sub2(t)

to show that for a differentiable 3-space vector valued function r, the graph of r lies on a sphere centered at the orgin if and only if r(t) and r'(t) are orthogonal. - February 22nd 2007, 11:41 AMPlato
If R(t) is on the surface of such a sphere then ||R(t)||=C, is constant.

Taking the derivative we get R’(t)·R(t)/[R(t)·R(t)]=0 or R’(t)·R(t)=0.