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Prove that the angle between F(t) and F'(t) is constant for a given vector function

by RoRi

Consider the vector-valued function

    \[ F(t) = \frac{2t}{1+t^2} \mathbf{i} + \frac{1-t^2}{1+t^2} \mathbf{j} + \mathbf{k}. \]

Prove that the angle between F(t) and F'(t) is a constant.

Proof. First, we have

    \[ F(t) = \left( \frac{2t}{1+t^2}, \frac{1-t^2}{1+t^2}, 1 \right) \quad \implies \quad F'(t) = \left( \frac{2-2t^2}{(1+t^2)^2}, \frac{-4t}{(1+t^2)^2}, 0 \right). \]

Thus,

    \begin{align*} F(t) \cdot F'(t) &= \left( \frac{2t}{1+t^2} \right) \left( \frac{2-2t^2}{(1+t^2)^2} \right) + \left( \frac{1-t^2}{1+t^2} \right) \left( \frac{-4t}{(1+t^2)^2} \right) \\[9pt] &= \frac{4t-4t^3 - 4t + 4t^3}{(1+t^2)^3} \\[9pt] &= 0. \end{align*}

Hence, F(t) and F'(t) are orthogonal, so the angle between them is constant, \theta = \frac{\pi}{2}. \qquad \blacksquare

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