- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
14.4 Exercises
- Given a nonzero vector $B$ and a vector-valued function $F$ such that $F(t) \cdot B = t$ for all $t,$ and such that the angle between $F'(t)$ and $B$ is constant (independent of $t$). Prove that $F''(t)$ is orthogonal to $F'(t).$
- Recall the dot product identity from Theorem 14.1:$$\left(F \cdot G \right)' = F' \cdot G + F \cdot G'$$If the angle between $F'$ and $B$ is constant, then the cosine of this angle is also constant. What does this imply about $F' \cdot B$?
- We know from the dot product identity in Theorem 14.1 that $$\left(F \cdot B \right)' = F' \cdot B + F \cdot B'$$But $B$ is a vector of constants, so $B' = O,$ and $F \cdot B = t,$ so $(F \cdot B)' = 1.$ Combining this with the above, we get $$\left(F \cdot B\right)' = F' \cdot B = 1$$Taking the derivative of $(F' \cdot B),$ we see that $$F'' \cdot B = 0$$But by definition, $B$ is nonzero, thus $F''(t)$ must be the zero vector, meaning that $F'' \cdot F = 0$ for any $t.$ Put another way, $F''(t)$ is orthogonal to $F'(t)$ for all $t. \quad \blacksquare$