Chapter 2, Problem 46RE

(a).

To determine

To find the equation for the tangent.

Answer to Problem 46RE

The equation of tangent:

  $y_{\text{tangent}}=-x+2+\frac{\pi }{2}$

Explanation of Solution

Given:

  $y=4+\cot \left(x\right)-2\csc \left(x\right)$ , $x=\frac{\pi }{2}$

Concept Used:

tangent $=\frac{dy}{dx}$

Equation of tangent:

  $\begin{array}{l}y=mx+c,\\ m=\text{slope}.\end{array}$

Calculation:

  $\begin{array}{c}y\text{at}x=\frac{\pi }{2}\text{is:}\\ \text{y=4+cot}\left(\frac{\pi }{2}\right)-2\csc \left(\frac{\pi }{2}\right)\\ =4+0-2\times 1\\ =2\end{array}$

Slope of tangent is as follows:

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left[4+\cot \left(x\right)-2\csc \left(x\right)\right]\\ =\frac{d}{dx}\left[4\right]+\frac{d}{dx}\left[\cot \left(x\right)\right]-2\frac{d}{dx}\left[\csc \left(x\right)\right]\\ =0-{\csc }^{2}x-2\cdot \left[-\cot (x)\cdot \text{csc}(x)\right]\\ =-{\csc }^{2}x+2\cot \left(x\right)\cdot \csc \left(x\right)\end{array}$

  $\begin{array}{c}{\left[\frac{dy}{dx}\right]}_{x=\frac{\pi }{2}}=\left[-{\csc }^2\left(\frac{\pi }{2}\right)+2\cot \left(\frac{\pi }{2}\right)\cdot \csc \left(\frac{\pi }{2}\right)\right]\\ =-1+0\\ =-1\end{array}$

Equation of tangent:

  $\begin{array}{l}\text{At}\left(\frac{\pi }{2},2\right):\\ 2=-\frac{\pi }{2}+c\\ c=2+\frac{\pi }{2}\\ y_{\text{tangent}}=-x+2+\frac{\pi }{2}\end{array}$

Conclusion:

The equation of tangent:

  $y_{\text{tangent}}=-x+2+\frac{\pi }{2}$

(b).

To determine

To find the equation of normal.

Answer to Problem 46RE

The equation of normal:

  $y_{\text{tangent}}=-x+2+\frac{\pi }{2}$

Explanation of Solution

Given:

  $y=4+\cot \left(x\right)-2\csc \left(x\right)$

Concept Used:

Slope of normal:

  $=\frac{-1}{m}$

Calculation:

Slope of normal:

  $m_{normal}=1$

Equation of normal:

  $\begin{array}{l}{y}_{normal}=x+c\\ c=2-\frac{\pi }{2}\\ y_{normal}=x+2-\frac{\pi }{2}\end{array}$

Conclusion:

The equation of normal:

  $y_{\text{tangent}}=-x+2+\frac{\pi }{2}$