Q 58. - Calculus

Question

Complete the proof of Theorem 9.14 by verifying that

${\left(f\text{'}\left(\theta \right)\cos \theta -f\left(\theta \right) \sin \theta \right)}^2+{\left(f\text{'}\left(\theta \right) \sin \theta +f\left(\theta \right) \cos \theta \right)}^2$ is equal to ${\left(f\text{'}\left(\theta \right)\right)}^2+{\left(f\left(\theta \right)\right)}^2$

Step-by-Step Solution

Answer

It is proved that ${\left(f\text{'}\left(\theta \right)\cos \theta -f\left(\theta \right) \sin \theta \right)}^2+{\left(f\text{'}\left(\theta \right) \sin \theta +f\left(\theta \right) \cos \theta \right)}^2={\left(f\text{'}\left(\theta \right)\right)}^2+{\left(f\left(\theta \right)\right)}^2$

Step 1. Given information

Left side of the equation is ${\left(f\text{'}\left(\theta \right)\cos \theta -f\left(\theta \right) \sin \theta \right)}^2+{\left(f\text{'}\left(\theta \right) \sin \theta +f\left(\theta \right) \cos \theta \right)}^2$

Step 2. Explanation

$\begin{array}{rcl}{\left(f\text{'}\left(\theta \right)\cos \theta -f\left(\theta \right) \sin \theta \right)}^2+{\left(f\text{'}\left(\theta \right) \sin \theta +f\left(\theta \right) \cos \theta \right)}^2& =& {\left(f\text{'}\left(\theta \right)\right)}^2\left({\cos }^2\theta +{\sin }^2\theta \right)-2f\text{'}\left(\theta \right)f\left(\theta \right)\cos \theta \sin \theta +{\left(f\left(\theta \right)\right)}^2\left({\sin }^2\theta +{\cos }^2\theta \right)\\ & & +2f\text{'}\left(\theta \right)f\left(\theta \right)\cos \theta \sin \theta \\ & =& {\left(f\text{'}\left(\theta \right)\right)}^2+{\left(f\left(\theta \right)\right)}^2\end{array}$