## Problem Statement

Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.

$$
\left\{\begin{array}{l}
x=\sin t - t \cdot \cos t \\
y=\cos t + t \cdot \sin t
\end{array}\right.
$$
## Solution

$x_{t}^{\prime}=(\sin t-t \cdot \cos t)^{\prime}=\cos t-\cos t+t \cdot \sin t=t \cdot \sin t$

$y_{t}^{\prime}=(\cos t+t \cdot \sin t)^{\prime}=-\sin t+\sin t+t \cdot \cos t=t \cdot \cos t$

We obtain:

$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{t \cdot \cos t}{t \cdot \sin t}=\frac{\cos t}{\sin t}=\operatorname{ctg} t \\
& \left(y_{x}^{\prime}\right)_{t}^{\prime}=(\operatorname{ctg} t)^{\prime}=-\frac{1}{\sin ^{2} t}
\end{aligned}
$$

Then:

$y_{x x}^{\prime \prime}=\frac{\left(y_{x}^{\prime}\right)_{t}^{\prime}}{x_{t}^{\prime}}=\left(-\frac{1}{\sin ^{2} t}\right) /(t \cdot \sin t)=-\frac{1}{t \cdot \sin ^{3} t}$

## Problem Kuznetsov Differentiation 20-28