Chapter 3.3, Problem 37E

To determine

The equation of the line perpendicular to the tangent to the curve at the point $\left(2,3\right)$ .

Answer to Problem 37E

The required equation is $y=-\frac{1}{9}x+\frac{29}{9}$ .

Explanation of Solution

Given information:

The curve: $y=x^3-3x+1$

Formula used:

Power rule:

  $\frac{d}{dx}{x}^n=n{x}^{n-1}$

Point-slope form of a line:

  $y-y_1=m\left(x-x_1\right)$

Calculation:

The given curve is $y=x^3-3x+1$ and the point is $\left(2,3\right)$ .

To find the slope of the perpendicular to the tangent line, use the power rule of differentiation $\frac{d}{dx}{x}^n=n{x}^{n-1}$ .

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(x^3-3x+1\right)\\ \frac{dy}{dx}=\frac{d}{dx}\left(x^3\right)-\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(1\right)\\ \frac{dy}{dx}=3x^{\left(3-1\right)}-3x^{\left(1-1\right)}+0\\ \frac{dy}{dx}=3x^2-3\end{array}$

Substitute 2 for x in the above derivative and simplify.

  $\begin{array}{c}\frac{dy}{dx}=3{\left(2\right)}^2-3\\ =3\left(4\right)-3\\ =12-3\\ =9\end{array}$

Since, the required line is perpendicular to the tangent line; so the slope of the negative reciprocal to the slope of the normal that is, $-\frac{1}{9}$ .

To find the equation of the perpendicular line to the tangent line, use the point-slope form of the equation of a line $y-y_1=m\left(x-x_1\right)$ .

Substitute 2 for $x_1$ , 3 for $y_1$ and $-\frac{1}{9}$ for m in the above formula and simplify.

  $\begin{array}{c}y-3=-\frac{1}{9}\left(x-2\right)\\ y-3+3=-\frac{1}{9}x+\frac{2}{9}+3\\ y=-\frac{1}{9}x+\frac{29}{9}\end{array}$

Hence, the required equation is $y=-\frac{1}{9}x+\frac{29}{9}$ .