Chapter 3.5, Problem 22E

To determine

To find:Equations for the lines that are tangent and normal to the graph of $y=\sec x$ at $x=\frac{\pi }{4}$ .

Answer to Problem 22E

Equation of tangent line is $y=\sqrt{2}x-\frac{\sqrt{2}\pi }{4}+\sqrt{2}$ .

Equation of normal line is $y=\frac{-x}{\sqrt{2}}+\frac{\pi }{\sqrt{2}}+3$ .

Explanation of Solution

Given information:Given equation is $y=\sec x$ .

Calculation:Given equation is $y=\sec x$ .

  $y\text{'}=\sec x\tan x$

To find the slope of line and ordinate put $x=\frac{\pi }{4}\mathrm{in}y\text{'}\mathrm{and}y$

  $\begin{array}{l}y\text{'}=\sec \frac{\pi }{4}\tan \frac{\pi }{4}\\ y\text{'}=\sqrt{2}(1)=\sqrt{2}\\ \mathrm{so}m=\sqrt{2}\\ y=\sec \frac{\pi }{4}\\ y=\sqrt{2}\end{array}$

Equation of tangent line is:

  $\begin{array}{l}y-\sqrt{2}=\sqrt{2}(x-\frac{\pi }{4})\\ y=\sqrt{2}x-\frac{\sqrt{2}\pi }{4}+\sqrt{2}\end{array}$

For equation of normal line firstly we find slope:

  $\begin{array}{l}m.m_1=-1\\ \sqrt{2}.m_1=-1\\ m_1=\frac{-1}{\sqrt{2}}\end{array}$

Equation of normal line is:

  $\begin{array}{l}y-3=\frac{-1}{\sqrt{2}}(x-\pi )\\ y-3=\frac{-x}{\sqrt{2}}+\frac{\pi }{\sqrt{2}}\\ y=\frac{-x}{\sqrt{2}}+\frac{\pi }{\sqrt{2}}+3\end{array}$

Therefore equation of tangent line is $y=\sqrt{2}x-\frac{\sqrt{2}\pi }{4}+\sqrt{2}$ .

Equation of normal line is $y=\frac{-x}{\sqrt{2}}+\frac{\pi }{\sqrt{2}}+3$ .