Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 31

a.) Find an equation of the tangent line to the curve $y^2 = 5x^4 - x^2$ which is also called Kampyle of Eudoxus, at the point $(1, 2)$.

Solving for the slope $(m)$


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} (y^2) =& 5 \frac{d}{dx} (x^4) - \frac{d}{dx} (x^2)
\\
\\
2y \frac{dy}{dx} =& (5)(4x^3) - 2x
\\
\\
2yy' =& 20x^3 - 2x
\\
\\
\frac{\cancel{2y} y'}{\cancel{2y}} =& \frac{20x^3 - 2x}{2y}
\\
\\
y' =& \frac{20x^3 - 2x}{2y}
\\
\\
y' =& \frac{\cancel{2} (10x^3 - x)}{\cancel{2}y}
\\
\\
y' =& \frac{10x^3 - x}{y}

\end{aligned}
\end{equation}
$


For $x = 1$ and $y = 2$, we obtain


$
\begin{equation}
\begin{aligned}

y' =& \frac{10(1)^3 - 1}{2}
\\
\\
y' =& \frac{10 - 1}{2}
\\
\\
y' =& \frac{9}{2} \text{ or } m = \frac{9}{2}

\end{aligned}
\end{equation}
$


Using Point Slope Form


$
\begin{equation}
\begin{aligned}

y - y_1 =& m(x - x_1)
\\
\\
y - 2 =& \frac{9}{2} (x - 1)
\\
\\
y =& \frac{9x - 9}{2} + 2
\\
\\
y =& \frac{9x - 9 + 4}{2}
\\
\\
y =& \frac{9x - 5}{2} \qquad \text{Equation of the tangent line at $(1,2)$}

\end{aligned}
\end{equation}
$


b.) Graph the curve and the tangent line on a common screen