## Problem Statement

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

$$
\left\{\begin{array}{l}
x=t-\sin t \\
y=2-\cos t
\end{array}\right.
$$
## Solution

$x_{t}^{\prime}=(t-\sin t)^{\prime}=1-\cos t$

$y_{t}^{\prime}=(2-\cos t)^{\prime}=\sin t$

## We obtain:

$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{\sin t}{1-\cos t}$

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

$=\frac{\cos t-\cos ^{2} t-\sin ^{2} t}{(1-\cos t)^{2}}=\frac{\cos t-1}{(1-\cos t)^{2}}=-\frac{1}{1-\cos t}$

Then:

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

Kuznetsov Differentiation Problem 20-20