How to prove?

Question

If length of a position vector for an object is constant, then the position vector must be perpendicular to the velcocity vector.

Can you prove this using the dot product?
Umm.... ???

Answer 1

Let the position vector, as a function of time, be given by P(t) = (x(t),y(t),z(t)). Then its velocity vector is given by V(t)=(x'(t),y'(t),z'(t)).

Then its length is L(t)=sqrt(x(t)^2 + y(t)^2 + z(t)^2).

Because its length is constant, the derivative of the length, as a function of t, must be zero.

That derivative is

1/2*(x(t)^2 + y(t)^2 + z(t)^2)^(-1/2) * 2(x(t)x'(t) + y(t)y'(t) + z(t)z'(t)).

However, this is just [P(t) dot V(t)] / L(t), and since this must be equal to zero, it follows that P(t) dot V(t) = 0, so P(t) and V(t) must be orthogonal.

Answer 2

If the position vector length is constant, then V(t)∙P(t) must be zero and that can only happen if the two are at right angles.


Doug

Answer 3

I tried, but I just keep going in circles !!