## Problem Statement

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

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

$t_{0}=2$
## Solution

Since $t_{0}=1$, then

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

$y_{0}=2-2^{3}=-6$

Let's find the derivatives:

$x_{t}^{\prime}=\left(1-t^{2}\right)^{\prime}=-2 t$

$y_{t}^{\prime}=\left(t-t^{3}\right)^{\prime}=1-3 t^{2}$
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{1-3 t^{2}}{-2 t}=\frac{3 t^{2}-1}{2 t}$

Then

$y_{0}^{\prime}=\frac{3 \cdot 2^{2}-1}{2 \cdot 2}=\frac{11}{4}$

Equation of the tangent line:

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

$y-(-6)=\frac{11}{4} \cdot(x-(-3))$

$y=\frac{11 x}{4}+\frac{33}{4}-6$

$y=\frac{11 x}{4}+\frac{9}{4}$

Equation of the normal line:

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

$y-(-6)=-\frac{1}{\left(\frac{11}{4}\right)} \cdot(x-(-3))$

$y=-\frac{4}{11} \cdot(x-(-3))-6$

$y=-\frac{4 x}{11}-\frac{12}{11}-6$

$y=-\frac{4 x}{11}-\frac{78}{11}$

Kuznetsov Differentiation 17-19