Chapter 3, Problem 45RE

(a)

To determine

To calculate: To find an equation for the tangent to the curve at $x=2$

Answer to Problem 45RE

  $\text{The equation of the tangent line is }y-2=-2\left(x-2\right)$ .

Explanation of Solution

Given Information: The equation of the curve is $y=8x^{-2}$ .

Calculation:

The y- coordinate of the curve $y=8x^{-2}$ at $x=2$ is,

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

Therefore, a point on the curve is $\left(x,y\right)=\left(2,2\right)$

Find $\frac{dy}{dx}$ as follows:

  $\begin{array}{c}y=8x^{-2}\\ \frac{dy}{dx}=\frac{d}{dx}\left(8x^{-2}\right)\left(\text{Substitute the value }y\text{ in }\frac{dy}{dx}\right)\\ =\left(8\times -2\times x^{-2-1}\right)\left(\frac{d}{dx}{x}^m=m{x}^{m-1}\right)\\ =-16x^{-3}\end{array}$

The slope of the tangent line at the point $\left(2,2\right)$ is equal to the derivative of the equation of the curve at $\left(2,2\right)$ . Therefore, the slope of the tangent is,

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

• Here, $m_{\text{tangent}}$ is the slope of the tangent line.

The equation of the tangent line using point-slope form is,

  $\begin{array}{c}y-y_1=m_{\text{tangent}}\left(x-x_1\right)\left(\text{Point-slope form}\right)\\ y-2=-2\left(x-2\right)\left(x=2,y=2\right)\end{array}$

Conclusion:

  $\text{The equation of the tangent line is }y-2=-2\left(x-2\right)$ .

(b)

To determine

To calculate: To find an equation for the normal to the curve at $x=2$

Answer to Problem 45RE

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

Explanation of Solution

Given Information: The equation of the curve is $y=8x^{-2}$ .

Calculation:

The y- coordinate of the curve $y=8x^{-2}$ at $x=2$ is,

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

Therefore, a point on the curve is $\left(x,y\right)=\left(2,2\right)$

Find $\frac{dy}{dx}$ as follows:

  $\begin{array}{c}y=8x^{-2}\\ \frac{dy}{dx}=\frac{d}{dx}\left(8x^{-2}\right)\left(\text{Substitute the value }y\text{ in }\frac{dy}{dx}\right)\\ =\left(8\times -2\times x^{-2-1}\right)\left(\frac{d}{dx}{x}^m=m{x}^{m-1}\right)\\ =-16x^{-3}\end{array}$

The slope of the tangent line at the point $\left(2,2\right)$ is equal to the -2, as obtained in part (a).

The slope of the normal line is perpendicular to the slope of tangent line. Therefore, the products of both the slopes is equal to 1. That is,

  $\begin{array}{c}{m}_{\text{tangent}}\times m_{\text{normal}}=-1\\ -2\times m_{\text{normal}}=-1\\ m_{\text{normal}}=\frac{1}{2}\end{array}$

• Here, $m_{\text{normal}}$ is the slope of the normal line.

The equation of the normal line using point-slope form is,

  $\begin{array}{c}y-y_1=m_{normal}\left(x-x_1\right)\left(\text{Point-slope form}\right)\\ y-2=\frac{1}{2}\left(x-2\right)\left(x=2,y=2\right)\end{array}$

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