Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 56

Find the equation of the tangent line and normal line of the curve $\displaystyle y = \frac{\sqrt{x}}{x + 1}$ at the point $(4,0.4)$


Required:

Equation of the tangent line and the normal line at $P(4,0.4)$

Solution:


$
\begin{equation}
\begin{aligned}

\qquad y' = m_T =& \frac{(x + 1) \displaystyle \frac{d}{dx} (x^{\frac{1}{2}}) - \left[ (x^{\frac{1}{2}}) \frac{d}{dx} (x + 1) \right] }{(x + 1)^2}
&& \text{Using Quotient Rule}
\\
\\
\\
\qquad y' = m_T =& \frac{(x + 1) \displaystyle \left[ \frac{1}{2 (x^{\frac{1}{2}})} \right] - (x^{\frac{1}{2}})(1)}{(x + 1)^2}
&& \text{Simplify the equation}
\\
\\
\\
\qquad y' = m_T =& \frac{\displaystyle \frac{x + 1}{2 (x^{\frac{1}{2}})} - (x^{\frac{1}{2}})}{(x + 1)^2}
&& \text{Get the LCD}
\\
\\
\\
\qquad y' = m_T =& \frac{\displaystyle \frac{x + 1 - 2x}{2(x^{\frac{1}{2}})}}{(x + 1)^2}
&& \text{Combine like terms}
\\
\\
\\
\qquad y' = m_T =& \frac{-x + 1}{2 \sqrt{x} (x + 1)^2}
&& \text{}
\\
\\
\\
\qquad m_T =& \frac{-x + 1}{2 \sqrt{x} (x + 1)^2}
&& \text{Substitute the value of $x$}
\\
\\
\\
\qquad m_T =& \frac{-4 + 1}{2 \sqrt{4} (4 + 1)^2}
&& \text{Simplify the equation}
\\
\\
\\
\qquad m_T =& \frac{-3}{100}
&& \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 - 0.4=& \frac{-3}{100} (x - 4)
&& \text{Multiply $\large \frac{-3}{100}$ in the equation}
\\
\\
\qquad y - 0.4 =& \frac{-3x + 12}{100}
&& \text{Add $0.4$ to each sides}
\\
\\
\qquad y =& \frac{-3x + 12}{100} + 0.4
&& \text{Simplify the equation}
\\
\\
\qquad y =& \frac{-3x + 12 + 40}{100}
&& \text{Combine like terms}
\\
\\
\qquad y =& \frac{-3x + 52}{100}
&& \text{Equation of the tangent line to the curve at $P (4,0.4)$}


\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}{100}}
&&
\\
\\
m_N =& \frac{100}{3}
&&
\\
\\
y - y_1 =& m_N (x - x_1)
&& \text{Substitute the value of slope $(m_N)$ and the given point}
\\
\\
y - 0.4 =& \frac{100}{3} (x - 4)
&& \text{Multiply $\large \frac{100}{3}$ to the equation}
\\
\\
y - 0.4 =& \frac{100 x - 400}{3}
&& \text{Add 0.4 to each sides}
\\
\\
y =& \frac{100x - 400}{3} +0.4
&& \text{Simplify the equation}
\\
\\
y =& \frac{100 - 400 + 1.2}{3}
&& \text{Combine like terms}
\\
\\
y =& \frac{100x - 398.8}{3}
&& \text{Equation of the normal line at $P(4,0.4)$}

\end{aligned}
\end{equation}
$