Chapter 2, Problem 45RE

(a).

To determine

To find the equation for the tangent.

Answer to Problem 45RE

The equation of tangent:

  $y_{\tan gent}=-2x+6$

Explanation of Solution

Given:

  $y=8x^{-2}$

Concept Used:

tangent $=\frac{dy}{dx}$

Equation of tangent:

  $\begin{array}{l}y=mx+c,\\ m=\text{slope, c is constant}.\end{array}$

Calculation:

  $\begin{array}{c}y\text{at}x=2\text{is:}\\ \text{y=8}\cdot {\left(2\right)}^{-2}\\ =\frac{8}{4}\\ =2\end{array}$

Slope of tangent is as follows:

  $\begin{array}{c}\frac{dy}{dx}=\frac{d}{dx}\left[8x^{-2}\right]\\ =8\cdot \frac{d}{dx}\left[x^{-2}\right]\\ =\frac{-16}{x^3}\end{array}$

  $\begin{array}{c}{\left[\frac{dy}{dx}\right]}_{x=2}={\left[\frac{-16}{{(2)}^3}\right]}_{x=2}\\ =\frac{-16}{8}\\ =-2\end{array}$

Equation of tangent:

Conclusion:

The equation of tangent:

  $\begin{array}{l}{y}_{\text{tangent}}=2\\ -2x+c=2\\ \text{At}x=2:\\ -4+c=2\\ c=6\\ \text{Put}c=6\text{in}{y}_{\text{tangent}}\\ y_{\text{tangent}}=-2x+6\end{array}$

(b).

To determine

To find the equation of normal.

Answer to Problem 45RE

The equation of normal:

  $y_{\text{normal}}=\frac{1}{2}x+1$

Explanation of Solution

Given:

  $y=8x^{-2}$

Concept Used:

The normal is perpendicular to the tangent.

Slope of normal:

  $=\frac{-1}{m}$

Calculation:

Slope of normal:

  $\begin{array}{l}=\frac{-1}{m}\\ =\frac{-1}{-2}\\ =\frac{1}{2}\end{array}$

  $m_{\text{normal}}=\frac{1}{2}$

Equation of normal:

  $\begin{array}{l}{y}_{normal}=\frac{1}{2}x+c\\ 2=\frac{1}{2}\times 2+c\\ c=1\\ y_{normal}=\frac{1}{2}x+1\end{array}$

Conclusion:

The equation of normal:

  $y_{\text{normal}}=\frac{1}{2}x+1$