Question

A curve has equation $\sin x + \cos y = \frac{1}{2}, 0 \leq x, y < 2\pi$. Then the number of the points on the curve, where the tangent to the curve is parallel to the y-axis, is

Answer 1

2

Answer 2

To find the points on the curve where the tangent is parallel to the y-axis, we need to find the points where $\frac{dx}{dy} = 0$.

The given equation of the curve is $\sin x + \cos y = \frac{1}{2}$.

Differentiate the equation implicitly with respect to $y$: $\frac{d}{dy}(\sin x) + \frac{d}{dy}(\cos y) = \frac{d}{dy}\left(\frac{1}{2}\right)$ $\cos x \frac{dx}{dy} - \sin y = 0$

Now, solve for $\frac{dx}{dy}$: $\cos x \frac{dx}{dy} = \sin y$ $\frac{dx}{dy} = \frac{\sin y}{\cos x}$

For the tangent to be parallel to the y-axis, $\frac{dx}{dy}$ must be 0. This implies: $\frac{\sin y}{\cos x} = 0$

This condition is satisfied when the numerator is zero and the denominator is non-zero. So, we need $\sin y = 0$ and $\cos x \neq 0$.

First, let's find the values of $y$ for which $\sin y = 0$ in the interval $0 \leq y < 2\pi$. $\sin y = 0 \implies y = 0 \text{ or } y = \pi$.

Now, we will consider each value of $y$ and substitute it back into the original equation to find the corresponding values of $x$.

Case 1: $y = 0$ Substitute $y=0$ into the curve equation: $\sin x + \cos(0) = \frac{1}{2}$ $\sin x + 1 = \frac{1}{2}$ $\sin x = \frac{1}{2} - 1$ $\sin x = -\frac{1}{2}$

For $0 \leq x < 2\pi$, the values of $x$ for which $\sin x = -\frac{1}{2}$ are: $x = \pi + \frac{\pi}{6} = \frac{7\pi}{6}$ (in the 3rd quadrant) $x = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$ (in the 4th quadrant)

Now, we must check the condition $\cos x \neq 0$ for these $x$ values: For $x = \frac{7\pi}{6}$, $\cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2} \neq 0$. This point is valid. For $x = \frac{11\pi}{6}$, $\cos\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2} \neq 0$. This point is valid.

So, we have two points: $\left(\frac{7\pi}{6}, 0\right)$ and $\left(\frac{11\pi}{6}, 0\right)$.

Case 2: $y = \pi$ Substitute $y=\pi$ into the curve equation: $\sin x + \cos(\pi) = \frac{1}{2}$ $\sin x - 1 = \frac{1}{2}$ $\sin x = \frac{1}{2} + 1$ $\sin x = \frac{3}{2}$

Since the range of $\sin x$ is $[-1, 1]$, there is no real value of $x$ for which $\sin x = \frac{3}{2}$. Therefore, there are no points on the curve corresponding to $y=\pi$.

Check for $\frac{0}{0}$ case: If both $\sin y = 0$ and $\cos x = 0$, then $\frac{dx}{dy}$ would be indeterminate. $\sin y = 0 \implies y = 0, \pi$. $\cos x = 0 \implies x = \frac{\pi}{2}, \frac{3\pi}{2}$. Let's check if any of these combinations of $(x,y)$ satisfy the original equation $\sin x + \cos y = \frac{1}{2}$:

1. $(\frac{\pi}{2}, 0)$: $\sin(\frac{\pi}{2}) + \cos(0) = 1 + 1 = 2 \neq \frac{1}{2}$. Not on the curve.
2. $(\frac{\pi}{2}, \pi)$: $\sin(\frac{\pi}{2}) + \cos(\pi) = 1 - 1 = 0 \neq \frac{1}{2}$. Not on the curve.
3. $(\frac{3\pi}{2}, 0)$: $\sin(\frac{3\pi}{2}) + \cos(0) = -1 + 1 = 0 \neq \frac{1}{2}$. Not on the curve.
4. $(\frac{3\pi}{2}, \pi)$: $\sin(\frac{3\pi}{2}) + \cos(\pi) = -1 - 1 = -2 \neq \frac{1}{2}$. Not on the curve. Since none of these points are on the curve, the condition $\cos x \neq 0$ is implicitly satisfied for the points we found.

In summary, the only points on the curve where the tangent is parallel to the y-axis are $\left(\frac{7\pi}{6}, 0\right)$ and $\left(\frac{11\pi}{6}, 0\right)$. The number of such points is 2.