Chapter 3.5, Problem 47E

To determine

Which of the following is an equation of the normal line to $y=\sin x+\cos x$ at $x=\pi$ .

Answer to Problem 47E

Correct option is $B$ .

Explanation of Solution

Given information:Equation of line is $y=\sin x+\cos x$ .

Calculation:To find the slope of normal line we must find the derivative of the line.

  $\begin{array}{l}y=\sin x+\cos x\\ \frac{dy}{dx}=\cos x-\sin x\end{array}$

Put $x=\pi \mathrm{in}\frac{dy}{dx}$ we get:

  $\begin{array}{l}\frac{dy}{dx}=\cos x-\sin x\\ \frac{dy}{dx}=\cos \pi -\sin \pi \\ \frac{dy}{dx}=-1-0=-1\end{array}$

  $\begin{array}{l}{m}_1.m_2=-1\\ (-1).m_2=-1\\ m_2=\frac{-1}{-1}=1\end{array}$

Thereforeslope of normal line is $m=1$ .

Put $x=\pi$ in $y=\sin x+\cos x$ we get:

  $\begin{array}{l}y=\sin x+\cos x\\ y=\sin \pi +\cos \pi \\ y=0+(-1)\\ y=-1\end{array}$

So the line should be passing through $(\pi ,-1)$ .

Equation of line is $y=mx+b$ .

Put $x=\pi ,y=-1\mathrm{and}m=1$ in equation $y=mx+b$ we get:

  $\begin{array}{l}y=mx+b\\ -1=(1)\pi +b\\ -1=\pi +b\\ -\pi -1=b\end{array}$

Put $m=1\mathrm{and}b=\pi -1$ in equation $y=mx+b$ we get:

  $\begin{array}{l}y=1x-\pi -1\\ y=x-\pi -1\end{array}$

Therefore correct option is $B$ .