To be fair, there is the added condition that r'' is orthogonal to r'.
Okay.
r(t) = ((x(t), y(t), z(t))
r't=((x'(t),y'(t),z'(t))
r't·r't=(x'(t)^2)+(y'(t)^2)+((z't)^2)
d/dt (r'·r') = 2(x'(t))(x''(t))+2(y'(t))(y''(t))+(z'(t))(z''(t))
d/dt (r'·r') = 2(r'(t))·(r''(t))
r'·r' =...
Okay, the magnitude of the velocity is a constant.
Therefore the magnitude of the acceleration is zero, |r''|=0,
And certainly then r'.r'' =|r'||r''|cos(theta)=0 , because |r''|=0
Is this really all there is to it?
Also getting a bit tripped up because while in circular motion the magnitude...
Thanks for your quick reply.
The question is only given in general terms. And reads explicitly as:
Let r(t)=(x(t),y(t),z(t)) be the position vector along some curve for -infinity to +infinity.
Suppose t has been chosen so that 1=r'.r' for all t. Show that 0=r'.r''.
Homework Statement
r(t)=(x(t),y(t),z(t))
t has been chosen so that r'.r'=1
show that r'.r''=0
Homework Equations
v'.w'=|v||w|cos(theta)
The Attempt at a Solution
Clearly what is being described is circular motion about a unit circle. And using the equation for a unit circle its easy to...