Chapter 3.1, Problem 59E

To determine

To find : the equation of tangent line and normal line at $x=1$ .

Answer to Problem 59E

The tangent line for curve at $\left(1,2\right)$ is given by $y-2=\pi \left(x-1\right)$ .

The normal line for curve at $\left(1,2\right)$ is given by $y-2=\frac{-1}{\pi }\left(x-1\right)$ .

Explanation of Solution

Given information :

Curve $y=2\tan \left(\frac{\pi x}{4}\right)$ .

Formula needed :

The tangent line for curve is given by $y-y_1=m\left(x-x_1\right)$ and the normal line for curve is given by $y-y_1=\frac{-1}{m}\left(x-x_1\right)$ where $\left(x_1,y_1\right)$ is the point and $m$ is slope at point $\left(x_1,y_1\right)$ .

Differentiate $y=2\tan \left(\frac{\pi x}{4}\right)$ with respect to $x$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(2\tan \left(\frac{\pi x}{4}\right)\right)\\ =2\frac{d}{dx}\left(\tan \left(\frac{\pi x}{4}\right)\right)\\ =2{\sec }^2\left(\frac{\pi x}{4}\right)\cdot \frac{d}{dx}\left(\frac{\pi x}{4}\right)\\ =2{\sec }^2\left(\frac{\pi x}{4}\right)\cdot \left(\frac{\pi }{4}\right)\\ =\frac{\pi }{2}{\sec }^2\left(\frac{\pi x}{4}\right)\end{array}$

So, the slope at $x=1$ is $m={\left.\frac{dy}{dx}\right|}_{x=1}$ .

  $\begin{array}{c}m={\left.\frac{dy}{dx}\right|}_{x=1}\\ =\frac{\pi }{2}{\sec }^2\left(\frac{\pi }{4}\right)\\ =\frac{\pi }{2}{\left(\sqrt{2}\right)}^2\\ =\pi \end{array}$

The point of contact between curve and line at $x=1$ is $\left(1,2\tan \left(\frac{\pi }{4}\right)\right)$ .

So, the point of contact is $\left(1,2\right)$ .

The tangent line for curve at $\left(1,2\right)$ is given by $y-2=\pi \left(x-1\right)$ .

The normal line for curve at $\left(1,2\right)$ is given by $y-2=\frac{-1}{\pi }\left(x-1\right)$ .

The graph of curve in green, tangent line in blue and normal line in black is