## Problem Statement

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

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

$t_{0}=1$
## Solution

Since $t_{0}=1$, then

$x_{0}=2 \cdot 1-1^{2}=1$

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

Let's find the derivatives:

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

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

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

Then:

$y_{0}^{\prime}=\frac{3}{2} \cdot(1+1)=3$

Equation of the tangent line:

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

$y-2=3 \cdot(x-1)$

$y=3 x-1$

Equation of the normal line:

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

$y-2=-\frac{1}{3}(x-1)$

$y=-\frac{x}{3}+\frac{1}{3}+2$

$y=-\frac{x}{3}+2 \frac{1}{3}$

## Problem Kuznetsov Differentiation 17-4