## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.1 Inverse Circular Functions - 6.1 Exercises - Page 266: 106b

#### Answer

As $v$ increases without bound, the equation of the asymptote is $\theta = 45^{\circ}$

#### Work Step by Step

Let $~~h=6$ $\theta = arcsin \left(\sqrt{\frac{v^2}{2v^2+64h}}\right)$ $\theta = arcsin \left(\sqrt{\frac{v^2}{2v^2+(64)(6)}}\right)$ $\theta = arcsin \left(\sqrt{\frac{v^2}{2v^2+384}}\right)$ As $v$ increases, the value of ${\frac{v^2}{2v^2+384}}$ approaches $\frac{1}{2}$. We can find the equation of the asymptote: $\theta = arcsin \left(\sqrt{\frac{1}{2}}\right)$ $\theta = arcsin \left(0.707\right)$ $\theta = 45^{\circ}$ As $v$ increases without bound, the equation of the asymptote is $\theta = 45^{\circ}$

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