Chapter 7.2, Problem 43E

To determine

To rewrite: the given equation in terms of $\tan \theta$ only.

Answer to Problem 43E

  $y=\frac{-g{x}^2}{2v_0^2}\left(1+{\tan }^2\theta \right)+x\tan \theta$

Explanation of Solution

Given information:

  $y=\frac{-g{x}^2}{2v_0^2{\cos }^2\theta }+\frac{x\sin \theta }{\cos \theta }$

Calculation:

The given equation can be rewritten in terms of $\tan \theta$ only as shown below.

  $\begin{array}{c}y=\frac{-g{x}^2}{2v_0^2{\cos }^2\theta }+\frac{x\sin \theta }{\cos \theta }\\ =\frac{-g{x}^2}{2v_0^2}\left({\sec }^2\theta \right)+x\tan \theta \left[\begin{array}{l}\because \frac{\mathrm{sinx}}{\mathrm{cosx}}=\mathrm{tanx}\\ \because \frac{1}{\cos x}=\sec x\end{array}\right]\\ =\frac{-g{x}^2}{2v_0^2}\left(1+{\tan }^2\theta \right)+x\tan \theta \left[\because {\sec }^{2}x=1+{\tan }^{2}x\right]\end{array}$

Thus, the equation is $y=\frac{-g{x}^2}{2v_0^2}\left(1+{\tan }^2\theta \right)+x\tan \theta$ .