## Problem Statement

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

\[
\left\{\begin{array}{l}
x=a \cdot \sin ^{3} t \\
y=a \cdot \cos ^{3} t
\end{array}\right.
\]

$t_{0}=\frac{\pi}{6}$
## Solution

Since $t_{0}=\frac{\pi}{6}$, then

$x_{0}=a \sin ^{3} \frac{\pi}{6}=a \cdot\left(\frac{1}{2}\right)^{3}=\frac{a}{8}$

$y_{0}=a \cos ^{3} \frac{\pi}{6}=a \cdot\left(\frac{\sqrt{3}}{2}\right)=\frac{3 \sqrt{3} \cdot a}{8}$

Let's find the derivatives:

$x_{t}^{\prime}=\left(a \sin ^{3} t\right)^{\prime}=a \cdot 3 \cdot \sin ^{2} t \cdot \cos t$

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

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{-a \cdot 3 \cdot \cos ^{2} t \cdot \sin t}{a \cdot 3 \cdot \sin ^{2} t \cdot \cos t}=-\mathrm{ctg} t$

Then:

$y_{0}^{\prime}=-\operatorname{ctg} \frac{\pi}{6}=-\sqrt{3}$

Equation of the tangent line:

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

$y-\frac{3 \sqrt{3} \cdot a}{8}=-\sqrt{3} \cdot\left(x-\frac{a}{8}\right)$

$y=-\sqrt{3} \cdot x+\frac{\sqrt{3} \cdot a}{8}+\frac{3 \sqrt{3} \cdot a}{8}$

$y=-\sqrt{3} \cdot x+\frac{4 \sqrt{3} \cdot a}{8}$

Equation of the normal line:

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

$y-\frac{3 \sqrt{3} \cdot a}{8}=-\frac{1}{-\sqrt{3}}\left(x-\frac{a}{8}\right)$

$y=\frac{x}{\sqrt{3}}-\frac{a}{8 \sqrt{3}}+\frac{3 \sqrt{3} \cdot a}{8}$

$y=\frac{x}{\sqrt{3}}-\frac{a}{8 \sqrt{3}}+\frac{9 a}{8 \sqrt{3}}$

$y=\frac{x}{\sqrt{3}}+\frac{a}{\sqrt{3}}$

## Problem Kuznetsov Differentiation 17-16