Chapter 3.1, Problem 43E

To determine

To find : the equation of line tangent to the curve at the point $t=-\pi /4$ .

Answer to Problem 43E

The equation of line tangent to the curve at point $t=-\pi /4$ is $y+1=-\frac{1}{2}\left(x-1\right)$ .

Explanation of Solution

Given information :

  $x={\sec }^{2}t-1,y=\tan t$ .

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 $t$ .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left({\sec }^{2}t-1\right)\\ =2\sec t\left(\sec t\tan t\right)\\ \frac{dy}{dt}=\frac{d}{dt}\left(\tan t\right)\\ ={\sec }^{2}t\end{array}$

So,

  $\begin{array}{c}\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\\ =\frac{{\sec }^{2}t}{2{\sec }^{2}t\tan t}\\ =\frac{1}{2\tan t}\end{array}$

Slope is given by $m={\left.\frac{dy}{dx}\right|}_{t=-\pi /4}$ .

  $\begin{array}{c}m={\left.\frac{dy}{dx}\right|}_{t=-\pi /4}\\ =\frac{1}{2\tan \left(-\pi /4\right)}\\ =\frac{1}{-2}\end{array}$

The value of $x$ and $y$ at $t=-\pi /4$ is given by $x_1={\sec }^2\left(-\frac{\pi }{4}\right)-1,y_1=\tan \left(-\frac{\pi }{4}\right)$ .

  $\begin{array}{c}{x}_1={\sec }^2\left(-\frac{\pi }{4}\right)-1\\ ={\left(\sqrt{2}\right)}^2-1\\ =2-1\\ =1\end{array}$

And

  $\begin{array}{c}{y}_1=\tan \left(-\frac{\pi }{4}\right)\\ =-1\end{array}$

Thus, the equation of line tangent to the curve at point $t=-\pi /4$ is $y+1=-\frac{1}{2}\left(x-1\right)$ .