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.

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kaluitagf · Accepted Answer

r(t)×r′(t)=0&amp;nbsp;for all&amp;nbsp;t∈Rd(r(t)×r′(t))&amp;nbsp;dt&amp;nbsp;=r(t)×r(t)+r(t)×r′(t)=2×r(t)×r′(t)=0d(∥r(t)∥2)dt=0⇒∥r(t)∥2&amp;nbsp;is constant,&amp;nbsp;∥r(t)∥&amp;nbsp;also a constant. Let K be the constant such that&amp;nbsp;∥r(t)∥=K.∥r(t)∥2=K2x, y, z are the components of vector function&amp;nbsp;r(t),x2+y2+z2=K2, a sphere centered at&amp;nbsp;(0,&amp;nbsp;0,&amp;nbsp;0)&amp;nbsp;and with radius Kr(t)=⟨x(t),&amp;nbsp;y(t),&amp;nbsp;z(t)⟩&amp;nbsp;such that,&amp;nbsp;∥r(t)∥=x2+y2+z2⇒∥r(t)∥2=x2+y2+z2The curve&amp;nbsp;r(t)&amp;nbsp;must be a sphere centered at origin.

zurilomk4 · Accepted Answer

Let&amp;nbsp;r(t)&amp;nbsp;be&amp;nbsp;r(t)=⟨x(t),&amp;nbsp;y(t),&amp;nbsp;z(t)⟩&amp;nbsp;then,|r(t)|=x2+y2+z2|r(t)|2=x2+y2+z2Since the position vector&amp;nbsp;r(t)&amp;nbsp;is always perpendicular to the tangent vector&amp;nbsp;r′(t)r(t)×r′(t)=02×r(t)×r′(t)=2×02×r(t)×r′(t)=0d&amp;nbsp;dt&amp;nbsp;(|r(t)|2)=0∫d&amp;nbsp;dt&amp;nbsp;(|r(t)|2)=∫0|r(t)|2=cSo&amp;nbsp;|r(t)|=c&amp;nbsp;then&amp;nbsp;x2+y2+z2=c, the position vector of the curve is a constant that is the distance is same in all directions.So, the curve lies on a sphere whose centre is origin.

nick1337 · Accepted Answer

r(t)×r′(t)=0(r(t)×r(t))′=2r(t)×r′(t)=0∥r(t)∥2=r(t)×r(t)=cfor some constant cThe curve lies on:x2+y2+z2=c

If a curve has the property that the position vector

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