## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=t^{3}+1 \\
y=t^{2}
\end{array}\right. \\
& t_{0}=-2
\end{aligned}
$$
## Solution

Since $t_{0}=-2$, then

$x_{0}=(-2)^{3}+1=9$

$y_{0}=(-2)^{2}=4$

Find the derivatives:

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

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

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

Then:

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

Equation of the tangent line:

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

$y-4=-\frac{1}{3} \cdot(x-9)$

$y=-\frac{x}{3}+\frac{9}{3}+4$

$y=-\frac{x}{3}+3+4$

$y=-\frac{x}{3}+7$

Equation of the normal line:

$$
\begin{aligned}
& y-y_{0}=-\frac{1}{y_{0}^{\prime}}\left(x-x_{0}\right) \\
& y-4=-\frac{1}{\left(-\frac{1}{3}\right)} \cdot(x-9) \\
& y=3 \cdot(x-9)+4 \\
& y=3 x-27+4 \\
& y=3 x-23
\end{aligned}
$$

## Problem Kuznetsov Differentiation 17-28