I really have no idea how to attack this one.
If a curve has the property that the position vector r(t) is always perpendicular to the tangent vector r'(t), show that the curve lies on a sphere with the center at the origin.
Thanks
I really have no idea how to attack this one.
If a curve has the property that the position vector r(t) is always perpendicular to the tangent vector r'(t), show that the curve lies on a sphere with the center at the origin.
Thanks
so we have $\displaystyle r(t) \cdot r'(t)=0,$ which gives us: $\displaystyle (r(t) \cdot r(t))'=2r(t) \cdot r'(t)=0.$ thus: $\displaystyle ||r(t)||^2=r(t) \cdot r(t)=c,$ for some constant $\displaystyle c.$ so the curve lies on: $\displaystyle x^2 + y^2 + z^2 = c. \ \ \square$