Chapter 4.3, Problem 27E

a.

To determine

To calculate: The equation of the tangent for the given function at the given point.

Answer to Problem 27E

The equation of tangent is $2y-4x-2+\pi =0$ .

Explanation of Solution

Given information: The function is $y=\tan x$ and the point is $\left(x_1,y_1\right)=\left(\frac{\pi }{4},1\right)$ .

Formula used:

Slope-point form of line is:

  $\left(y-y_1\right)=m\left(x-x_1\right),\text{where}\left(x_1,y_1\right)\text{is}\text{the}\text{intercept}\text{and}m\text{is}\text{the}\text{slope}\text{of}\text{the}\text{line}$ .

  $\frac{d\tan x}{dx}={\sec }^{2}x$

Also product, quotient and chain rule of derivatives are used.

Calculation:

The point is $\left(x_1,y_1\right)=\left(\frac{\pi }{4},1\right)$ , in order to find the equation of the tangent the value of m is requiredat given point to get the slope point form of the tangent.

The function is $y=\tan x$ ............(1)

By differentiating equation (1), we get

  $y=\tan x$

  $\begin{array}{l}\frac{dy}{dx}=\frac{d\tan x}{dx}\\ \frac{dy}{dx}=m={\sec }^{2}x\mathrm{.........}\left(2\right)\end{array}$

Now put the value $\left(x_1,y_1\right)=\left(\frac{\pi }{4},1\right)$ in equation (2)

  $\begin{array}{l}m={\sec }^2\left(\frac{\pi }{4}\right)\\ m={\left(\sqrt{2}\right)}^2\\ m=2\end{array}$

Here $m=2$ is the slope of the tangent.

The slope point form of the tangent line is:

  $\left(y-y_1\right)=m\left(x-x_1\right)\text{,where}$$m=2$ and $\left(x_1,y_1\right)=\left(\frac{\pi }{4},1\right)$ .

  $\begin{array}{l}\left(y-1\right)=2\left(x-\frac{\pi }{4}\right)\\ \left(y-1\right)=2\left(\frac{4x-\pi }{4}\right)\\ \left(y-1\right)=\frac{1}{2}\left(4x-\pi \right)\\ 2y-2=4x-\pi \\ 2y-4x-2+\pi =0\end{array}$

Hence the equation of tangent is $2y-4x-2+\pi =0$ .

b.

To determine

To calculate: The equation of the tangent for the given function at the given point.

Answer to Problem 27E

The equation of tangent is $4y-2x-\pi +2=0$ .

Explanation of Solution

Given information: The function is $y={\tan }^{-1}x$ and the point is $\left(x_1,y_1\right)=\left(1,\frac{\pi }{4}\right)$ .

Formula used:

Slope-point form of line is:

  $\left(y-y_1\right)=m\left(x-x_1\right),\text{where}\left(x_1,y_1\right)\text{is}\text{the}\text{intercept}\text{and}m\text{is}\text{the}\text{slope}\text{of}\text{the}\text{line}$ .

  $\frac{d{\tan }^{-1}x}{dx}=\frac{1}{1+x^2}$

Also product, quotient and chain rule of derivatives are used.

Calculation:

The point is $\left(x_1,y_1\right)=\left(1,\frac{\pi }{4}\right)$ , in order to find the equation of the tangent the value of m is requiredat given point to get the slope point form of the tangent.

The function is $y={\tan }^{-1}x$ ................(1)

By differentiating equation (1),

  $y={\tan }^{-1}x$

  $\begin{array}{l}\frac{dy}{dx}=\frac{d\tan x}{dx}\\ \frac{dy}{dx}=m=\frac{1}{1+x^2}\mathrm{.........}\left(2\right)\end{array}$

Now put the value $\left(x_1,y_1\right)=\left(1,\frac{\pi }{4}\right)$ in equation (2)

  $\begin{array}{l}m=\frac{1}{1+{\left(1\right)}^2}\\ m=\frac{1}{2}\end{array}$

Here $m=\frac{1}{2}$ is the slope of the tangent.

The slope point form of the tangent line is:

  $\left(y-y_1\right)=m\left(x-x_1\right)\text{,where}$$m=\frac{1}{2}$ and $\left(x_1,y_1\right)=\left(1,\frac{\pi }{4}\right)$ .

  $\begin{array}{l}\left(y-\frac{\pi }{4}\right)=\frac{1}{2}\left(x-1\right)\\ 2\left(\frac{4y-\pi }{4}\right)=\left(x-1\right)\\ \frac{1}{2}\left(4y-\pi \right)=\left(x-1\right)\\ 4y-\pi =2x-2\\ 4y-2x-\pi +2=0\end{array}$

Hence the equation of tangent is $4y-2x-\pi +2=0$ .