Exercise 9.13

Question

Exercise 13: Suppose that $\v f$ is a differentiable mapping of $\R1$ into $\R3$ such $\big|\v f(t)\big|=1$ for every $t$. Prove that $\v f'(t)\cdot\v f(t)=0$. Interpret this result geometrically.

analambanomenos · Accepted Answer

Since $\big|\v f(t)|^2=1$ is constant, we have
    $$
        0=\frac{d}{dt}\big|\v f(t)\big|^2=\frac{d}{dt}\sum_{i=1}^3f_i^2(t)=2\sum_{i=1}^3f_i'(t)f_i(t)=2\v f'(t)\cdot\v f(t).
    $$
    So $\v f'(t)$ is perpendicular to $\v f(t)$. I suppose this hasn't been shown, but if $\v a\cdot\v b=0$, then $|\v a+\v b|^2=|\v a|^2+|\v b|^2$ and you can apply the law of cosines to the
    triangle formed by $\v a$, $\v b$, and $\v a+\v b$ to conclude that $\v a$ and $\v b$ form a right angle. Also, the book hasn't mentioned tangent vectors, but this says that tangent vectors of
    curves on a sphere are perpendicular to the radial vectors, or that the tangent plane at the point on the sphere is perpendicular to the radius.

Exercise 9.13

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Exercise 9.13

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