Chapter 3.5, Problem 21E

To determine

To find:Equations for the lines that are tangent and normal to the graph of $y=\sin x+3$ at $x=\pi$ .

Answer to Problem 21E

Equation of tangent line is $y=-x+3+\pi$ .

Equation of normal line is $y=x+3-\pi$ .

Explanation of Solution

Given information:Given equation is $y=\sin x+3$ .

Calculation:Given equation is $y=\sin x+3$ .

  $y\text{'}=\cos x$

To find the slope of line and ordinate put $x=\pi \mathrm{in}y\text{'}\mathrm{and}y$

  $\begin{array}{l}y\text{'}=\cos \pi \\ y\text{'}=-1\\ \mathrm{so}m=-1\\ y=\sin \pi +3\\ y=0+3=3\end{array}$

Equation of tangent line is:

  $\begin{array}{l}y-3=-1(x-\pi )\\ y-3=-x+\pi \\ y=-x+3+\pi \end{array}$

For equation of normal line firstly we find slope:

  $\begin{array}{l}m.m_1=-1\\ -1.m_1=-1\\ m_1=\frac{-1}{-1}=1\end{array}$

Equation of normal line is:

  $\begin{array}{l}y-3=1(x-\pi )\\ y-3=x-\pi \\ y=x+3-\pi \end{array}$

Therefore equation of tangent line is $y=-x+3+\pi$ .

Equation of normal line is $y=x+3-\pi$ .