Chapter 3, Problem 48RE

To determine

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

Answer to Problem 48RE

Equation for the tangent to the curve at the indicated point $y=3$ .

Equation for the normal to the curve at the indicated point $x=1$ .

Explanation of Solution

Given information: $y=2x^2+\frac{1}{x^4},x=1$ .

Formula used:

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

Calculation:

Finding $y-$ coordinated when $x=1$ ,

  $\begin{array}{l}y=2{\left(1\right)}^2+\frac{1}{{\left(1\right)}^4}\\ y=2+1\\ y=3\end{array}$

Finding slope of the tangent at $x=1$ ,

  $\begin{array}{l}\frac{dy}{dx}=\frac{d}{dx}\left(2x^2+\frac{1}{x^4}\right)\\ \frac{dy}{dx}=4x-\frac{4}{x^{-5}}\end{array}$

Slope $m_{\text{tangent}}=\frac{dy}{dx}$ at $x=1$ is,

  $\begin{array}{c}\frac{dy}{dx}=4x-\frac{4}{x^5}\\ =4\left(1\right)-\frac{4}{{\left(1\right)}^5}\\ =4-4\\ =0\end{array}$

The slope of the tangent $m_{\text{tangent}}=0$ .

Finding equation of the tangent at $\left(1,3\right)$ ,

  $\begin{array}{c}\frac{y-y_1}{x-x_1}=m_{\text{tangent}}\\ \frac{y-3}{x-1}=0\\ y-3=0\\ y=3\end{array}$

Finding slope of the normal,

  $\begin{array}{l}{m}_{\text{normal}}=-\frac{1}{m_{\text{tangent}}}\\ m_{\text{normal}}=-\frac{1}{0}\\ m_{\text{normal}}=\infty \end{array}$

Equation of normal at $\left(1,3\right)$ ,

  $\begin{array}{l}\frac{y-y_1}{x-x_1}=m_{\text{normal}}\\ \frac{y-3}{x-1}=\infty ,\text{by taking inverse on both sides we get}\\ \frac{x-1}{y-3}=\frac{1}{\infty }\\ \text{t}\\ \frac{x-1}{y-3}=0\\ \text{t}\\ x-1=0\\ x=1\end{array}$

Therefore we can conclude that the equation of the tangent is $y=3$ and the equation of the normal is $x=1$ .