Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 55

Find the equation of the tangent line and normal line of the curve $y = \displaystyle \frac{3x + 1}{x^2 + 1}$ at the point $(1,2)$


Required:

Equation of the tangent line and the normal line at $P(1,2)$

Solution:


$\begin{equation} \begin{aligned} \qquad y' = m_T =& \text{Slope of the tangent line}\\ m_N =& \text{Slope of the normal line} && \\ \\ \qquad y' = m_T =& \frac{(x^2 + 1) \displaystyle \frac{d}{dx} (3x + 1) - \left[ (3x + 1) \frac{d}{dx} (x^2 + 1)\right]}{(x^2 + 1)^2} && \text{} \\ \qquad y' = m_T =& \frac{(x^2 + 1) (3) - (3x + 1)(2x)}{(x^2 + 1)^2} && \text{} \\ \qquad y' = m_T =& \frac{3x^2 + 3 - 6x^2 - 2x}{(x^2 + 1)^2} = \frac{-3x^2 - 2x + 3}{ (x^2 + 1)^2} && \text{} \\ \\ \qquad m_T =& \frac{-3x^2 - 2x + 3}{(x^2 + 1)^2} && \text{Substitute the value of $x$ which is 1} \\ \\ \qquad m_T =& \frac{-3(1)^2 - 2 (1) + 3}{[(1)^2 + 1]^2} && \text{Simplify the equation} \\ \\ \qquad m_T =& \frac{-2}{4} && \text{Reduce it to lowest term} \\ \\ \qquad m_T =& \frac{-1}{2} && \text{} \\ \\ \end{aligned} \end{equation}$



Solving for the equation of the tangent line:


$\begin{equation} \begin{aligned} \qquad y - y_1 =& m_T(x - x_1) && \text{Substitute the value of the slope $(m_T)$ and the given point} \\ \\ \qquad y -2 =& \frac{-1}{2} (x - 1) && \text{Multiply $\large \frac{-1}{2}$ the equation} \\ \\ \qquad y - 2 =& \frac{-x + 1}{2} && \text{Add $2$ to each sides} \\ \\ \qquad y =& \frac{-x + 1}{2} + 2 && \text{Simplify the equation} \\ \\ \qquad y =& \frac{-x + 1 + 4}{2} && \text{Combine like terms} \\ \\ \qquad y =& \frac{-x + 5}{2} && \text{Equation of the tangent line to the curve at $P (1,2)$} \end{aligned} \end{equation}$


Solving for the equation of the normal line:


$\begin{equation} \begin{aligned} m_N =& \frac{-1}{m_T}\\ \\ m_N =& \frac{-1}{\displaystyle \frac{-1}{2}} \\ \\ m_N =& 2 && \\ \\ y - y_1 =& m_N (x - x_1) && \text{Substitute the value of slope $(m_N)$ and the given point} \\ \\ y - 2 =& 2 (x - 1) && \text{Multiply $2$ to the equation} \\ \\ y - 2 =& 2x - 2 && \text{Add 2 to each sides} \\ \\ y =& 2x - 2 + 2 && \text{Combine like terms} \\ \\ y =& 2x && \text{Equation of the normal line at $P(1,2)$} \end{aligned} \end{equation}$