Chapter 3.5, Problem 30E

To determine

To find: The points on the curve $y=\tan x$ where the tangent is parallel to the line $y=2x$

Answer to Problem 30E

Points are $(\frac{\pi }{4},1)$ and $(\frac{-\pi }{4},-1)$

Explanation of Solution

Given information: We are given equation of curve, $y=\tan x$ and equation of $y=2x$

Explanation: The equation of curve is, $y=\tan x$

  $\frac{dy}{dx}=\frac{d}{dx}(\tan x)={\sec }^{2}x$

Slope of tangent = ${\sec }^{2}x$

Equation of line $y=2x$

Slope of line = 2

Because tangent is parallel to the line. Their slopes are equal

  $\begin{array}{l}{\sec }^{2}x=2\\ \sec x=\pm \sqrt{2}\end{array}$

It is given $\frac{-\pi }{2}<x<\frac{\pi }{2}$

  $x$ lies in IV and Ist quotient

  $x=\frac{\pi }{4},\frac{-\pi }{4}$

When $x=\frac{\pi }{4}$ then $y=\tan \frac{\pi }{4}=1$

Point is $(\frac{\pi }{4},1)$

When $x=\frac{-\pi }{4}$ then $y=\tan (\frac{\pi }{4})=-1$

Point is $(\frac{-\pi }{4},-1)$

Thus point on given curve are $(\frac{\pi }{4},1)$ and $(\frac{-\pi }{4},-1)$