Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 54

Find the equation of the tangent line and normal line of the curve $y = (1 + 2x)^2 = 1 + 4x + 4x^2$ at the point $(1,9)$

Required:

Equation of the tangent line and the normal line at $P(1,9)$

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{d}{dx} (1) + 4 \frac{d}{dx} (x) + 4 \frac{d}{dx} (x^2)
&& \text{}
\\
\qquad y' = m_T =& 0 + 4(1) + 4(2x)
&& \text{}
\\
\qquad y' = m_T =& 4 + 8x
&& \text{}
\\
\\
\qquad m_T =& 4 + 8x
&& \text{Substitute value of $x$ which is 1}
\\
\\
\qquad m_T =& 4 + 8 (1)
&& \text{Simplify the equation}
\\
\\
\qquad m_T =& 12
&& \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 - 9 =& 12(x - 1)
&& \text{Multiply $12 $ the equation}
\\
\\
\qquad y - 9 =& 12x - 12
&& \text{Add $9$ to each sides}
\\
\\
\qquad y =& 12x - 12 + 9
&& \text{Combine like terms}
\\
\\
\qquad y =& 12x - 3
&& \text{Equation of the tangent line to the curve at $P (1,9)$}


\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}{12}
&&
\\
\\
y- y_1 =& m_N(x - x_1)
&& \text{Substitute the value of the slope $(m_N)$ and the given point}
\\
\\
y - 9 =& \frac{-1}{12} (x - 1)
&& \text{Multiply $\large \frac{-1}{12}$ in the equation}
\\
\\
y - 9 =& \frac{-x + 1}{12}
&& \text{Add $9$ to each sides}
\\
\\
y =& \frac{-x + 1}{12} + 9
&& \text{Get the LCD}
\\
\\
y =& \frac{-x + 1 + 108}{12}
&& \text{Combine like terms}
\\
\\
y =& \frac{-x + 109}{12}
&& \text{Equation of the normal line at $P(1,9)$}

\end{aligned}
\end{equation}
$