Chapter 24, Problem 1RE

To determine

The slope of the line normal to the curve $y=3x^2-5$ at $\left(1,2\right)$ is $\frac{1}{6}$.

Answer to Problem 1RE

The given statement is $\underset{\_}{\text{False}}$.

Explanation of Solution

Differentiate $y$ with respect to the $x$.

$\begin{array}{c}y=\frac{d}{dx}\left(3x^2-5\right)\\ \frac{dy}{dx}=3\left(2x\right)-5\\ =6x-5\end{array}$

Slope of tangent of the curve at $x=1$.

$\begin{array}{c}\frac{dy}{dx}|\begin{array}{l}\\ x=1\end{array}=6x-5\\ =6\left(1\right)-5\\ =1\end{array}$

Note that, the slopes of perpendicular lines are negative reciprocals.

Hence, slope of normal is negative reciprocal of tangent of the curve.

$\begin{array}{c}\text{slope of normal}=\frac{-1}{\text{tangent of the curve}}\\ =\frac{-1}{1}\\ =-1\end{array}$

Hence, the given statement is $\underset{\_}{\text{False}}$.