Chapter 2, Problem 47RE

(a).

To determine

To find the equation for the tangent.

Answer to Problem 47RE

The equation of tangent:

  $y_{\text{tangent}}=\sqrt{2}$

Explanation of Solution

Given:

  $y=\sin x+\cos x,\text{ at x=}\frac{\pi }{4}$

Concept Used:

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

Equation of tangent:

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

  $\begin{array}{c}y\text{at}x=\frac{\pi }{4}\text{is:}\\ \text{y=sin}\left(\frac{\pi }{4}\right)+\cos \left(\frac{\pi }{4}\right)\\ =\frac{2}{\sqrt{2}}\\ =\sqrt{2}\end{array}$

Calculation:

Slope of tangent is as follows:

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left[\sin x+\cos x\right]\\ =\frac{d}{dx}\left[\sin x\right]+\frac{d}{dx}\left[\cos x\right]\\ =\cos x-\sin x\end{array}$

  $\begin{array}{c}{\left[\frac{dy}{dx}\right]}_{x=\frac{\pi }{4}}=\cos \left(\frac{\pi }{4}\right)-\sin \left(\frac{\pi }{4}\right)\\ =0\end{array}$

Equation of tangent:

  $\begin{array}{l}{y}_{\text{tangent}}=0+c\\ \sqrt{2}=c\\ y_{\text{tangent}}=\sqrt{2}\end{array}$

Conclusion:

The equation of tangent:

  $y_{\text{tangent}}=\sqrt{2}$

(b).

To determine

To find the equation of normal.

Answer to Problem 47RE

The equation of normal:

  $x=\frac{\pi }{4}$

Explanation of Solution

Given:

  $y=\sin x+\cos x,\text{ at x=}\frac{\pi }{4}$

Concept Used:

Slope of normal:

  $=\frac{-1}{m}$

Calculation:

Slope of normal:

  $m_{normal}=\infty$

The equation of normal:

  $x=\frac{\pi }{4}$

Conclusion:

The equation of normal:

  $x=\frac{\pi }{4}$