## Problem Statement

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

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

$t_{0}=1$
## Solution

Since $t_{0}=1$, then

$x_{0}=\frac{2 \cdot 1+1^{2}}{1+1^{3}}=\frac{3}{2}$

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

Let's find the derivatives:

$$
\begin{aligned}
& x_{t}^{\prime}=\left(\frac{2 t+t^{2}}{1+t^{3}}\right)^{\prime}=\frac{(2+2 t) \cdot\left(1+t^{3}\right)-\left(2 t+t^{2}\right) \cdot 3 t^{2}}{\left(1+t^{3}\right)^{2}}= \\
& =\frac{2+2 t+2 t+2 t^{4}-6 t^{3}-3 t^{4}}{\left(1+t^{3}\right)^{2}}=\frac{2+4 t-6 t^{3}-t^{4}}{\left(1+t^{3}\right)^{2}} \\
& y_{t}^{\prime}=\left(\frac{2 t-t^{2}}{1+t^{3}}\right)^{\prime}=\frac{(2-2 t) \cdot\left(1+t^{3}\right)-\left(2 t-t^{2}\right) \cdot 3 t^{2}}{\left(1+t^{3}\right)^{2}}= \\
& =\frac{2+2 t^{2}-2 t-2 t^{4}-6 t^{3}+3 t^{4}}{\left(1+t^{3}\right)^{2}}=\frac{2-2 t+2 t^{2}-6 t^{3}+t^{4}}{\left(1+t^{3}\right)^{2}} \\
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\left(\frac{2-2 t+2 t^{2}-6 t^{3}+t^{4}}{\left(1+t^{3}\right)^{2}}\right) /\left(\frac{2+4 t-6 t^{3}-t^{4}}{\left(1+t^{3}\right)^{2}}\right)=
\end{aligned}
$$

$=\frac{2-2 t+2 t^{2}-6 t^{3}+t^{4}}{2+4 t-6 t^{3}-t^{4}}$

Then:

$$
y_{0}^{\prime}=\frac{2-2 \cdot 1+2 \cdot 1^{2}-6 \cdot 1^{3}+1^{4}}{2+4 \cdot 1-6 \cdot 1^{3}-1^{4}}=\frac{-3}{-1}=3
$$

Equation of the tangent line:

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

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

$y=3 x-\frac{9}{2}+\frac{1}{2}$

$y=3 x-4$

Equation of the normal line:

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

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

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

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

## Problem Kuznetsov Differentiation 17-5