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}$