Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 53

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

Required:

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

Solution:

Let $y' = m$ (slope)


$
\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} (x) + \frac{d}{dx} (x^{\frac{1}{2}})
&&
\\
\\
\qquad m_T =& 1 + \frac{1}{2(x)^{\frac{1}{2}}}
&& \text{Substitute the value of $x$ which is 1}
\\
\\
\qquad m_T =& 1 + \frac{1}{2(1)^{\frac{1}{2}}}
&& \text{Simplify the equation}
\\
\\
\qquad m_T =& \frac{3}{2}
&&
\\
\\


\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{3}{2} (x - 1)
&& \text{Multiply $\large \frac{3}{2}$ the equation}
\\
\\
\qquad y - 2 =& \frac{3x - 3}{2}
&& \text{Add $2$ to each sides}
\\
\\
\qquad y =& \frac{3x - 3}{2} + 2
&& \text{Get the LCD}
\\
\\
\qquad y =& \frac{3x - 3 + 4}{2}
&& \text{Combine like terms}
\\
\\
\qquad y =& \frac{3x + 1}{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{3}{2}}
&&
\\
\\
m_N =& \frac{-2}{3}
&&
\\
\\
y - y_1 =& m_N (x - x_1)
&& \text{Substitute the value of slope $(m_N)$ and the given point}
\\
\\
y - 2 =& \frac{-2}{3} (x - 1)
&& \text{Multiply $\large \frac{-2}{3}$ to the equation}
\\
\\
y - 2 =& \frac{-2x + 2}{3}
&& \text{Add 2 to each sides}
\\
\\
y =& \frac{-2x + 2}{3} + 2
&& \text{Get the LCD}
\\
\\
y =& \frac{-2x + 2 + 6}{3}
&& \text{Combine like terms}
\\
\\
y =& \frac{-2x + 8}{3}
&& \text{Equation of the normal line at $P(1,2)$}

\end{aligned}
\end{equation}
$