# Consider the curve defined by the equation y+cosy=x+1 for 0≤y≤2pi, how do you find dy/dx in terms of y and write an equation for each vertical tangent to the curve?

Jun 30, 2016

$y ' = \frac{1}{1 - \sin y}$

with $y \in \left[0 , 2 \pi\right]$ specified, the only vertical tangent is $x = \frac{\pi}{2} - 1$

#### Explanation:

starting with $y + \cos y = x + 1$, you would differentiate

So $\frac{d}{\mathrm{dx}} \left(y + \cos y = x + 1\right)$

$\setminus \implies y ' - \sin y \setminus y ' = 1$

$\setminus \implies y ' = \frac{1}{1 - \sin y}$ in terms of y!!

vertical tangents have slope $\infty$ which means tyically looking fo a demoninator of 0 in the fraction

so we are interested in $\sin y = 1$

$\setminus \implies y = \frac{\pi}{2} , \frac{5 \pi}{2} , \ldots , \left(2 k + \frac{1}{2}\right) \pi q \quad q \quad k \in m a t h b f \left\{Z\right\}$

fortunately $\cos y = \cos \left(2 k \pi + \frac{\pi}{2}\right) = \cos 2 k \pi \textcolor{red}{\cos \left(\frac{\pi}{2}\right)} - \textcolor{red}{\sin 2 k \pi} \sin \frac{\pi}{2}$

and the terms in red are zero

$y + \cos y = x + 1$ becomes

$\left(2 k + \frac{1}{2}\right) \pi = x + 1$

$x = \left(2 k + \frac{1}{2}\right) \pi - 1$

that is generalised but $y \in \left[0 , 2 \pi\right]$ is specified. so we are limited to $k = 0$

that is to say $y = \frac{\pi}{2}$ when the vertical tangent is

$x = \frac{\pi}{2} - 1$