Condition of the problem

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

$\left\{\begin{array}{l}x=t^{3}+1 \\ y=t^{2}+t+1\end{array}\right.$

$t_{0}=1$
## Solution

Since $t_{0}=1$, then

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

$y_{0}=1^{2}+1+1=3$

Let's find the derivatives:

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

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

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

Then:

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

Equation of the tangent line:

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

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

$y=x+1$

Equation of the normal line:

$$
\begin{aligned}
& y-y_{0}=-\frac{1}{y_{0}^{\prime}}\left(x-x_{0}\right) \\
& y-3=-\frac{1}{1}(x-2) \\
& y=-x+5
\end{aligned}
$$