## Problem Statement

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

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

$t_{0}=1$
## Solution

Since $t_{0}=1$, then

$x_{0}=1-1^{4}=0$
$y_{0}=1^{2}-1^{3}=0$

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

$y_{t}^{\prime}=\left(t^{2}-t^{3}\right)^{\prime}=2 t-3 t^{2}$

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2 t-3 t^{2}}{1-4 t^{3}}$

Then:

$y_{0}^{\prime}=\frac{2 \cdot 1-3 \cdot 1^{2}}{1-4 \cdot 1^{3}}=\frac{-1}{-3}=\frac{1}{3}$

Equation of the tangent line:

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

$y-0=\frac{1}{3} \cdot(x-0)$

$y=\frac{x}{3}$

Equation of the normal line:

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

$y-0=-\frac{1}{\left(\frac{1}{3}\right)}(x-0)$

$y=-3 x$

## Problem Kuznetsov Differentiation 17-24