Animated Solution for Mathematics - Differentiation: Let $x^k+y^k=a^k,(a,k>0)$ and $\frac{dy}{dx}+(\frac{y}{x}{)}^{1/3}=0$, then $k$ is

Unveiling the Astroid

Welcome, fellow traveler in the world of calculus! Today, we are not just solving a problem; we are embarking on a journey to uncover the hidden nature of a curve.

We are given the equation $x^k+y^k=a^k$ and a mysterious differential equation $\frac{dy}{dx}+(\frac{y}{x}{)}^{1/3}=0$. Our mission is to find the value of $k$ that binds these two mathematical truths together.

The Power of Implicit Differentiation

When you look at $x^k+y^k=a^k$, you might feel the urge to isolate $y$. Resist that urge! In the realm of JEE Advanced, the most elegant path is often the one that respects the structure of the equation.

We are looking for the slope, $\frac{dy}{dx}$. Since $y$ is implicitly defined as a function of $x$, we use Implicit Differentiation. We apply the derivative operator $\frac{d}{dx}$ to both sides of our equation:

The Chain Rule Trap

Here is where the magic—and the danger—lies. Differentiating $x^k$ is simple: it is $k{x}^{k-1}$.

But when we differentiate $y^k$, we must remember that $y$ is a function of $x$. This is the Chain Rule in action. The derivative of $y^k$ with respect to $x$ is $k{y}^{k-1}\cdot \frac{dy}{dx}$.

If you forget that $\frac{dy}{dx}$ term, the entire solution will collapse. On the right side, $a^k$ is a constant, so its derivative is simply $0$. Our equation now stands as:

The Algebraic Dance

Now, we isolate our target, $\frac{dy}{dx}$. We subtract $k{x}^{k-1}$ from both sides and divide by $k{y}^{k-1}$:

The $k$ terms cancel out beautifully, leaving us with $\frac{dy}{dx}=-(\frac{x}{y}{)}^{k-1}$. But wait! The problem gives us the slope in terms of $(\frac{y}{x}{)}^{1/3}$.

We need to flip our fraction. Using the rule $(\frac{x}{y}{)}^{k-1}=(\frac{y}{x}{)}^{-(k-1)}$, we rewrite our expression as:

The Final Comparison

We have arrived at the climax. The problem states that $\frac{dy}{dx}+(\frac{y}{x}{)}^{1/3}=0$, which rearranges to $\frac{dy}{dx}=-(\frac{y}{x}{)}^{1/3}$.

Since both expressions represent the same slope, their exponents must be equal:

Solving this simple linear equation, we find:

The Astroid

We have found it! The value of $k$ is $\frac{2}{3}$.

When $k=\frac{2}{3}$, the curve $x^{2/3}+y^{2/3}=a^{2/3}$ is known as an Astroid. It is a beautiful, star-shaped curve that appears frequently in physics and geometry.

You didn't just solve for a variable; you identified a fundamental geometric structure. Keep this curiosity alive, and remember: every equation has a story to tell. You just have to listen.