Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 52

a.) $\displaystyle y = \frac{x}{(1 + x^2)}$ is a curve called Serpentine. Find the equation of the tangent line to this curve at $P ( 3, 0.3)$

Required:

Equation of the tangent line to the curve at $P(3,0.3)$

Solution:

Let $y' = m$ (slope)


$
\begin{equation}
\begin{aligned}

\qquad y' = m =& \frac{\displaystyle (1 + x^2) \frac{d}{dx} (x) - \left[ (x) \frac{d}{dx} (1 + x^2) \right]}{(1 + x^2)^2}
&&
\\
\\
\qquad y' = m =& \frac{(1 + x^2) (1) - (x)(2x)}{(1 + x^2)^2}
&&
\\
\\
\qquad y' = m =& \frac{1 + x^2 - 2x^2}{(1 + x^2)^2}
&&
\\
\\
\qquad m =& \frac{1 - x^2}{(1 + x^2)^2}
&& \text{Substitute the value of $x$ which is $3$}
\\
\\
\qquad m =& \frac{1 - (3)^2}{[ 1 + (3)^2]^2}
&& \text{Simplify the equation}
\\
\\
\qquad m =& \frac{-8}{100}
&& \text{Reduce to lowest term}
\\
\\
\qquad m =& \frac{-2}{25}
&&


\end{aligned}
\end{equation}
$


Solving for the equation of the tangent line:


$
\begin{equation}
\begin{aligned}

\qquad y - y_1 =& m(x - x_1)
&& \text{Substitute the value of the slope $(m)$ and the given point}
\\
\\
\qquad y - 0.3 =& \frac{-2}{25} (x - 3)
&& \text{Multiply $\large \frac{-2}{25}$ the equation}
\\
\\
\qquad y - 0.3 =& \frac{-2x + 6}{25}
&& \text{Add $0.3$ to each sides}
\\
\\
\qquad y =& \frac{-2x + 6}{25} + 0.3
&& \text{Get the LCD}
\\
\\
\qquad y =& \frac{-2x + 6 + 7.5}{25}
&& \text{Combine like terms}
\\
\\
\qquad y =& \frac{-2x + 13.5}{25} \text{ or } -0.08x + 0.54
&& \text{Equation of the tangent line to the curve at $P (3, 0.3)$}


\end{aligned}
\end{equation}
$


b.) Graph the curve and the tangent line on part (a).