Question

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 center the origin.

Short answer

If a curve has the property that the position vector r(t) is always perpendicular to the tangent vector r’(t) then, the curve lies on sphere which is centered at origin. The required proof is explained inside.

Step-by-step solution

Step 1: Given and differentiation

It is given that \(r(t) \cdot r'(t) = 0 \to t \in \mathbb{R}\)

Now let us consider

\(\begin{array}{l}\frac{{d(r(t) \cdot r'(t))}}{{dt}} = r'(t) \cdot r(t) + r(t) \cdot r'(t)\\ = 2r(t)r'(t) = 0\end{array}\)

This implies that \(\frac{{d({{\left\| {r(t)} \right\|}^2})}}{{dt}} = 0\)

\({\left\| {r(t)} \right\|^2}\)is constant and therefore \({\left\| {r(t)} \right\|^2}\) should also be constant. Let K be the constant such that \(\left\| {r(t)} \right\| = K\). This implies that \({\left\| {r(t)} \right\|^2} = {K^2}\).

Let us say x, y ,z are the components of vector function r(t), this implies that we have,

\({x^2} + {y^2} + {z^2} = {K^2}\)

Which is centred at (0,0,0) and with radius K.

Step 2: The conclusion

Thus we have

\(\begin{array}{l}r(t) = \left\langle {x(t),y(t),z(t)} \right\rangle \\\left\| {r(t)} \right\| = \sqrt {{x^2} + {y^2} + {z^2}} \\{\left\| {r(t)} \right\|^2} = {x^2} + {y^2} + {z^2}\end{array}\)

Hence the curve r(t) must be a sphere centered at origin.

If a curve has the property that the position vector r(t) is always perpendicular to the tangent vector r’(t) then, the curve lies on sphere which is centered at origin. The required proof is explained inside.