- 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
- Let $F$ be the vector-valued function given by: \begin{align*} F(t) = \frac{2t}{1+t^2}\mathbf{i} + \frac{1-t^2}{1+t^2}\mathbf{j} + \mathbf{k} \end{align*} Prove that the angle between $F(t)$ and $F'(t)$ is constant, that is, independent of $t.$
- Recall from section 12.9 that the cosine of the angle between two vectors can be expressed in terms of their dot product and their norms: $$\text{cos}\ \theta = \frac{A \cdot B}{\|A\|\|B\|}$$Then, if the cosine of the angle between the two vectors is constant, this imples that the angle itself is constant.
- First, we calculate $F'(t):$$$F'(t) = \left(\frac{2 - 2t^2}{(1 + t^2)^2}, \frac{-4t}{(1 + t^2)^2}, 0\right)$$Then, we calculate the dot product $F(t) \cdot F'(t):$$$F(t) \cdot F'(t) = \frac{4t(1 - t^2)}{(1 + t^2)^3} - \frac{4t(1 - t^2)}{(1 + t^2)^3} = 0$$But this means that the cosine of the angle between $F(t)$ and $F'(t)$ is constant. Thus, the angle between the two vectors must be constant, that is, independent of $t. \ \blacksquare$