Chapter 2.5, Problem 22E

To determine

To calculate: The equations of tangent and normal lines to the given curve.

Answer to Problem 22E

The equation of tangent line is $y=\sqrt{2}\left(1+x-\frac{\pi }{4}\right)$ and equation of normal line is $y=\sqrt{2}-\frac{1}{\sqrt{2}}\left(x-\frac{\pi }{4}\right)$ .

Explanation of Solution

Given information:

The equation of the curve is $y=\sec x$ , $x=\frac{\pi }{4}$

Concept used:

The formula used is $\frac{d}{dx}\left(\sec x\right)=\sec x\tan x$ , standard equation of line is $y-y_1=m\left(x-x_1\right)$ .

Calculation:

Slope of tangent line.

  $\begin{array}{c}m=\frac{dy}{dx}\\ =\frac{d}{dx}\left(\sec x\right)\\ =\sec x\tan x\end{array}$

Solve further by substituting $\pi$ for $x$ in $m=\sec x\tan x$

  $\begin{array}{c}m=\sec \frac{\pi }{4}\tan \frac{\pi }{4}\\ =\sqrt{2}\times 1\\ =\sqrt{2}\end{array}$

Substitute $\frac{\pi }{4}$ for $x$ in $y=\sec x$ for the value of $y_1$ .

  $\begin{array}{c}y=\sec \left(\frac{\pi }{4}\right)\\ =\sqrt{2}\end{array}$

So, the value of $\left(x_1,y_1\right)=\left(\frac{\pi }{4},\sqrt{2}\right)$

Substitute the values in the standard equation of line $y-y_1=m\left(x-x_1\right)$

  $\begin{array}{c}y-\sqrt{2}=\sqrt{2}\left(x-\frac{\pi }{4}\right)\\ y=\sqrt{2}+\sqrt{2}\left(x-\frac{\pi }{4}\right)\\ y=\sqrt{2}\left(1+x-\frac{\pi }{4}\right)\end{array}$

The equation of tangent line is $y=\sqrt{2}\left(1+x-\frac{\pi }{4}\right)$ .

Slope of normal line.

  $\begin{array}{c}{m}_1.m_2=-1\\ \sqrt{2}.m_2=-1\\ m_2=-\frac{1}{\sqrt{2}}\end{array}$

The slope of normal line is $-\frac{1}{\sqrt{2}}$

Substitute $\frac{\pi }{4}$ for $x$ in $y=\sec x$ for the value of $y_1$ .

  $\begin{array}{c}y=\sec \left(\frac{\pi }{4}\right)\\ =\sqrt{2}\end{array}$

So, the value of $\left(x_1,y_1\right)=\left(\frac{\pi }{4},\sqrt{2}\right)$

Substitute the values in the standard equation of line $y-y_1=m\left(x-x_1\right)$

  $\begin{array}{c}y-\sqrt{2}=-\frac{1}{\sqrt{2}}\left(x-\frac{\pi }{4}\right)\\ y=\sqrt{2}-\frac{1}{\sqrt{2}}\left(x-\frac{\pi }{4}\right)\end{array}$

The equation of normal line is$y=\sqrt{2}-\frac{1}{\sqrt{2}}\left(x-\frac{\pi }{4}\right)$.

Conclusion: The equation of tangent line is $y=\sqrt{2}\left(1+x-\frac{\pi }{4}\right)$ and equation of normal line is $y=\sqrt{2}-\frac{1}{\sqrt{2}}\left(x-\frac{\pi }{4}\right)$ .