## Problem Statement

Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.

\[
\left\{\begin{array}{l}
x=\frac{1}{2} \cdot t^{2}-\frac{1}{4} \cdot t^{4} \\
y=\frac{1}{2} \cdot t^{2}+\frac{1}{3} \cdot t^{3}
\end{array}\right.
\]

$t_{0}=0$
## Solution

Since $t_{0}=0$, then

$$
\begin{aligned}
& x_{0}=\frac{1}{2} \cdot 0^{2}-\frac{1}{4} \cdot 0^{4}=0 \\
& y_{0}=\frac{1}{2} \cdot 0^{2}+\frac{1}{3} \cdot 0^{3}=0
\end{aligned}
$$

Find the derivatives:
$x_{t}^{\prime}=\left(\frac{1}{2} \cdot t^{2}-\frac{1}{4} \cdot t^{4}\right)^{\prime}=t-t^{3}$

$y_{t}^{\prime}=\left(\frac{1}{2} \cdot t^{2}+\frac{1}{3} \cdot t^{3}\right)^{\prime}=t+t^{2}$

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{t+t^{2}}{t-t^{3}}=\frac{1+t}{1-t^{2}}$

Then:

$y_{0}^{\prime}=\frac{1+0}{1-0^{2}}=1$

Equation of the tangent line:

$y-y_{0}=y_{0}^{\prime}\left(x-x_{0}\right)$

$y-0=1 \cdot(x-0)$

$y=x$

Equation of the normal line:

$y-y_{0}=-\frac{1}{y_{0}^{\prime}}\left(x-x_{0}\right)$

$y-0=-\frac{1}{1} \cdot(x-0)$

$y=-x$

Kuznetsov Differentiation Problem 17-10