Question

Consider the curve defined by the equation $ y + \cos y = x + 1 $ for $ 0 \leqslant y \leqslant 2\pi $ , how do you find $ \dfrac{{dy}}{{dx}} $ in terms of y and write an equation for each vertical tangent to the curve.

Answer 1

Hint: In the given question, we are provided with an equation of curve $ y + \cos y = x + 1 $ for $ 0 \leqslant y \leqslant 2\pi $ and we are required to compute $ \dfrac{{dy}}{{dx}} $ in terms of y and write an equation for each vertical tangent or vertical asymptote to the curve $ y + \cos y = x + 1 $ . So, we differentiate both sides of the equation of the curve with respect to x so as to find $ \dfrac{{dy}}{{dx}} $ .

Complete step by step solution:
We have $ y + \cos y = x + 1 $ .
Differentiating both sides of the equation with respect to x, we get,
 $ \Rightarrow \dfrac{{dy}}{{dx}} + \left( { - \sin y} \right)\dfrac{{dy}}{{dx}} = 1 $
 $ \Rightarrow \dfrac{{dy}}{{dx}}\left( {1 - \sin y} \right) = 1 $
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\left( {1 - \sin y} \right)}} $
For vertical tangents, $ \dfrac{{dy}}{{dx}} $ should be equal to $ 0 $ .
But $ \dfrac{{dy}}{{dx}} = \dfrac{1}{{\left( {1 - \sin y} \right)}} $ cannot be equal to zero for any value of y belonging to $ 0 \leqslant y \leqslant 2\pi $ .
So, value of $ \dfrac{{dy}}{{dx}} $ in terms of y is $ \dfrac{1}{{\left( {1 - \sin y} \right)}} $ and there is no vertical tangent to the curve defined by the equation $ y + \cos y = x + 1 $ .
So, the correct answer is “ $ \dfrac{1}{{\left( {1 - \sin y} \right)}} $ ”.

Note: Such equations or functions that involve two variables and the two variables cannot be separated from each other in the given expression are known as implicit functions or expressions. In the given question, the expression given to us involves two variables but the variables can be easily separated from each other and hence it is an explicit expression. The equations of the vertical tangents can be verified by sketching the graph of the curve defined by the given equation.