Chapter 3, Problem 50RE

To determine

To find:An equation for the tangent and normal to the curve at the indicated point.

Answer to Problem 50RE

Coordinates of all the points on the curve at which the tangent has slope 6 are $\left(3,\frac{9}{2}\right)$ and $\left(-2,-\frac{14}{3}\right)$ .

Explanation of Solution

Given information: $y=\frac{2x^3-3x^2}{6}\text{and with the tangent of slope }6$ .

Formula used:

• Derivative rule: $\left(x^n\right)\prime =n\left(x^{n-1}\right)$
• Slope $m_{\text{tangent}}=\frac{dy}{dx}$

Calculation:

The slope of the tangent $\frac{dy}{dx}=6$ ,

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left(\frac{2x^3-3x^2}{6}\right)\\ \frac{dy}{dx}=\frac{1}{6}\frac{d}{dx}\left(2x^3-3x^2\right)\\ \frac{dy}{dx}=\frac{1}{6}\left(6x^2-6x\right)\\ \frac{dy}{dx}=x^2-x\end{array}$

We know that $\frac{dy}{dx}=6$ ,

  $\begin{array}{c}{x}^2-x=6\\ x^2-x-6=0\\ \left(x-3\right)\left(x+2\right)=0\\ x=3\text{and}x=-2\end{array}$

Finding the coordinates when $x=3$ ,

  $\begin{array}{l}y=\frac{2{\left(3\right)}^3-3{\left(3\right)}^2}{6}\\ y=\frac{54-27}{6}\\ y=\frac{27}{6}=\frac{9}{2}\end{array}$

Finding the coordinates when $x=-2$ ,

  $\begin{array}{l}y=\frac{2{\left(-2\right)}^3-3{\left(-2\right)}^2}{6}\\ y=\frac{-16-12}{6}\\ y=\frac{-28}{6}=-\frac{14}{3}\end{array}$

Therefore we can conclude that the coordinates of all the points on the curve at which the tangent has slope 6 are $\left(3,\frac{9}{2}\right)$ and $\left(-2,-\frac{14}{3}\right)$ .