Chapter 2.3, Problem 37E

To determine

To calculate: The equation of line perpendicular to the tangent.

Answer to Problem 37E

The equation of the line is $y=3-\frac{1}{9}\left(x-2\right)$ .

Explanation of Solution

Given information:

The equation of the curve is $y=x^3-3x+1$ , $\left(x_1,y_1\right)=\left(2,3\right)$

Concept used:

The formulae used are $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

Calculation:

Slope of tangent line.

  $\begin{array}{c}m=\frac{dy}{dx}\\ =\frac{d}{dx}\left(x^3-3x+1\right)\\ =\frac{d}{dx}\left(x^3\right)-3\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(1\right)\\ =3x^2-3\end{array}$

Solve further by substituting $2$ for $x$ in $m=3x^2-3$

  $\begin{array}{c}m=3x^2-3\\ =3{\left(2\right)}^2-3\\ =3\times 4-3\\ =9\end{array}$

The slope of tangent line is $9$

Slope of normal line.

  $\begin{array}{c}{m}_1.m_2=-1\\ 9.m_2=-1\\ m_2=-\frac{1}{9}\end{array}$

The slope of normal line is $-\frac{1}{9}$

Substitute the values in the standard equation of line $y-y_1=m\left(x-x_1\right)$

  $\begin{array}{l}y-3=-\frac{1}{9}\left(x-2\right)\\ y=3-\frac{1}{9}\left(x-2\right)\end{array}$

The equation of normal line is $y=3-\frac{1}{9}\left(x-2\right)$.

Conclusion: The equation of normal line is $y=3-\frac{1}{9}\left(x-2\right)$ .