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}
$