I'm having trouble with this problem I was wondering if someone could help me out.
If r(t) is not equal to zero, show that
(d/dt)|r(t)| = 1/|r(t)|r(t)*r'(t)
$\displaystyle r(t)=<x(t),y(t),z(t)>$
$\displaystyle r'(t)=<\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}>$
$\displaystyle \mid r(t) \mid =\sqrt{x^2+y^2+z^2}$
$\displaystyle \frac{d}{dt}\mid r(t) \mid =\frac{1}{2\sqrt{x^2+y^2+z^2}}\frac{d}{dt}(x^2+y^2 +z^2)$
Notice that the derivative of the inner function is just the dot product of the vector and its derivative.
$\displaystyle \frac{d}{dt}(x^2+y^2+z^2)=2x\frac{dx}{dt}+2y\frac{ dy}{dt}+2z\frac{dz}{dt}=2(r(t)\cdot r'(t))$
$\displaystyle \frac{d}{dt}\mid r(t) \mid =\frac{2(r(t)\cdot r'(t))}{2\sqrt{x^2+y^2+z^2}}$
$\displaystyle =\frac{r(t)\cdot r'(t)}{\mid r(t) \mid}$