If vector rt describes a path and vector rk a constant then

If vector r(t) describes a path and ||vector r||=k, a constant, then vector r. vector r\'=-

Solution

TRUE:

||r(t)||² =constant

<r(t),r(t)>=constant

Differentiation gives

<r(t),r\'(t)>+<r\'(t),r(t)>=0

implying

r(t) is orthogonal to r\'(t)

(note|| r(t) ||= constant describes a circle and the tangent is always perpendicular to the radius vector, as required to be proved)