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Prove that if vector A has constant magnitude then its derivative is perpendicular to vector A?
Wiki User
∙ 11y ago
Suppose A is a vector with real components.
A can be written as <f(t), g(t), h(t)>.
Since the magnitude of A is constant we have f(t)*f(t) + g(t)*g(t) + h(t)*h(t) = c, where c is a non-negative real number.
Take derivative of both sides of equation we get 2*f(t)*df(t)/dt + 2*g(t)*dg(t)/dt + 2*h(t)*dh(t)/dt = 0.
Divide both sides by 2, we get f(t)*df(t)/dt + g(t)*dg(t)/dt + h(t)*dh(t)/dt = 0.
Thus the dot product of A and its derivative is 0.
This implies the angle between A and its derivative is Pi/2. Hence they are perpendicular.
Wiki User
∙ 11y ago
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