Chapter 3.1, Problem 44E

To determine

To find : the equation of line tangent to the curve at the point $t=\pi /6$ .

Answer to Problem 44E

The equation of line tangent to the curve at point $t=\pi /6$ is $y-\frac{1}{\sqrt{3}}=2\left(x-\frac{2}{\sqrt{3}}\right)$ .

Explanation of Solution

Given information :

  $x=\sec t,y=\tan t$ .

Formula needed :

The equation of line tangent to the curve at point $t=a$ is $y-y_1=m\left(x-x_1\right)$ where $x_1,y_1$ are the value of $x$ and $y$ at $t=a$ respectively. Slope of line is $m={\left.\frac{dy}{dx}\right|}_{t=a}$ .

  $\frac{dy}{dx}$ partially expressed as $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ .

Partially differentiate $x$ and $y$ with respect to $t$ .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(\sec t\right)\\ =\sec t\tan t\\ \frac{dy}{dt}=\frac{d}{dt}\left(\tan t\right)\\ ={\sec }^{2}t\end{array}$

So,

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\\ =\frac{{\sec }^{2}t}{\sec t\tan t}\\ =\frac{\sec t}{\tan t}\\ =\csc t\end{array}$

Slope is given by $m={\left.\frac{dy}{dx}\right|}_{t=\pi /6}$ .

  $\begin{array}{c}m={\left.\frac{dy}{dx}\right|}_{t=\pi /6}\\ =\csc \left(\frac{\pi }{6}\right)\\ =2\end{array}$

The value of $x$ and $y$ at $t=\pi /6$ is given by $x_1=\sec \left(\frac{\pi }{6}\right),y_1=\tan \left(\frac{\pi }{6}\right)$ .

  $\begin{array}{c}{x}_1=\sec \left(\frac{\pi }{6}\right)\\ =\frac{2}{\sqrt{3}}\end{array}$

And

  $\begin{array}{c}{y}_1=\tan \left(\frac{\pi }{6}\right)\\ =\frac{1}{\sqrt{3}}\end{array}$

Thus, the equation of line tangent to the curve at point $t=\pi /6$ is $y-\frac{1}{\sqrt{3}}=2\left(x-\frac{2}{\sqrt{3}}\right)$ .