Question

For the curve $\sin x + \sin y = 1$ lying in the first quadrant there exists a constant $\alpha$ for which $\lim_{x \to 0} x^{\alpha} \frac{d^2y}{dx^2} = L$ (not zero)

The value of $L$:

• A. $\frac{1}{2}$
• B. 1
• C. $\frac{1}{2\sqrt{2}}$
• D. $\frac{1}{2\sqrt{3}}$

Answer 1

Answer: (C)

$\frac{1}{2\sqrt{2}}$

Answer 2

The given curve is $\sin x + \sin y = 1$, lying in the first quadrant. We are asked to find the value of $L$ for which $\lim_{x \to 0} x^{\alpha} \frac{d^2y}{dx^2} = L$ (not zero) for some constant $\alpha$.

First, let's find the first derivative $\frac{dy}{dx}$ by differentiating the equation with respect to $x$: $\frac{d}{dx}(\sin x + \sin y) = \frac{d}{dx}(1)$ $\cos x + \cos y \frac{dy}{dx} = 0$ $\frac{dy}{dx} = -\frac{\cos x}{\cos y}$

Next, let's find the second derivative $\frac{d^2y}{dx^2}$ by differentiating the first derivative with respect to $x$: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{\cos x}{\cos y}\right)$ $\frac{d^2y}{dx^2} = -\frac{\frac{d}{dx}(\cos x) \cos y - \cos x \frac{d}{dx}(\cos y)}{(\cos y)^2}$ $\frac{d^2y}{dx^2} = -\frac{(-\sin x) \cos y - \cos x (-\sin y \frac{dy}{dx})}{(\cos y)^2}$ $\frac{d^2y}{dx^2} = \frac{\sin x \cos y - \cos x \sin y \frac{dy}{dx}}{\cos^2 y}$ Substitute $\frac{dy}{dx} = -\frac{\cos x}{\cos y}$: $\frac{d^2y}{dx^2} = \frac{\sin x \cos y - \cos x \sin y (-\frac{\cos x}{\cos y})}{\cos^2 y}$ $\frac{d^2y}{dx^2} = \frac{\sin x \cos y + \frac{\cos^2 x \sin y}{\cos y}}{\cos^2 y}$ $\frac{d^2y}{dx^2} = \frac{\sin x \cos^2 y + \cos^2 x \sin y}{\cos^3 y}$

We need to evaluate the limit as $x \to 0$. As $x \to 0$ in the first quadrant, $\sin x \to 0$. Since $\sin x + \sin y = 1$, we must have $\sin y \to 1$. In the first quadrant, this implies $y \to \frac{\pi}{2}$. As $x \to 0$, $y \to \frac{\pi}{2}$, we have: $\sin x \to 0$ $\cos x \to \cos 0 = 1$ $\sin y \to \sin(\frac{\pi}{2}) = 1$ $\cos y \to \cos(\frac{\pi}{2}) = 0$

The expression for $\frac{d^2y}{dx^2}$ has $\cos^3 y$ in the denominator, which approaches 0. The numerator approaches $\sin(0)\cos^2(\pi/2) + \cos^2(0)\sin(\pi/2) = 0 \cdot 0 + 1^2 \cdot 1 = 1$. So $\frac{d^2y}{dx^2}$ approaches infinity as $x \to 0$.

To evaluate the limit $\lim_{x \to 0} x^{\alpha} \frac{d^2y}{dx^2}$, we need to understand how $\cos y$ behaves as $x \to 0$. From $\sin x + \sin y = 1$, we have $\sin y = 1 - \sin x$. Since $y \to \frac{\pi}{2}$, $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - (1 - \sin x)^2} = \sqrt{1 - (1 - 2\sin x + \sin^2 x)} = \sqrt{2\sin x - \sin^2 x}$. As $x \to 0$, $\sin x \approx x - \frac{x^3}{6}$. So, $\cos y \approx \sqrt{2(x - \frac{x^3}{6}) - (x - \frac{x^3}{6})^2} = \sqrt{2x - \frac{x^3}{3} - (x^2 - \frac{x^4}{3} + \frac{x^6}{36})} = \sqrt{2x - x^2 - \frac{x^3}{3} + \frac{x^4}{3} + \dots}$. For small $x$, the dominant term is $2x$. $\cos y \approx \sqrt{2x}$.

Now substitute this approximation into the expression for $\frac{d^2y}{dx^2}$ as $x \to 0$: Numerator: $\sin x \cos^2 y + \cos^2 x \sin y$ As $x \to 0$, $\sin x \to 0$, $\cos^2 y \approx 2x$, $\cos^2 x \to 1$, $\sin y \to 1$. Numerator $\approx (0)(2x) + (1)(1) = 1$.

Denominator: $\cos^3 y$ As $x \to 0$, $\cos y \approx \sqrt{2x}$. Denominator $\approx (\sqrt{2x})^3 = (2x)^{3/2} = 2\sqrt{2} x^{3/2}$.

So, as $x \to 0$, $\frac{d^2y}{dx^2} \approx \frac{1}{2\sqrt{2} x^{3/2}}$.

Now, consider the limit: $\lim_{x \to 0} x^{\alpha} \frac{d^2y}{dx^2} = \lim_{x \to 0} x^{\alpha} \left(\frac{1}{2\sqrt{2} x^{3/2}}\right) = \lim_{x \to 0} \frac{1}{2\sqrt{2}} x^{\alpha - 3/2}$.

For this limit to be a non-zero finite value $L$, the power of $x$ must be zero. $\alpha - 3/2 = 0 \implies \alpha = 3/2$.

With $\alpha = 3/2$, the limit becomes: $L = \lim_{x \to 0} \frac{1}{2\sqrt{2}} x^{3/2 - 3/2} = \lim_{x \to 0} \frac{1}{2\sqrt{2}} x^0 = \frac{1}{2\sqrt{2}}$.

The value of $L$ is $\frac{1}{2\sqrt{2}}$.