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

A Journey into $x^k+y^k=a^k$

Welcome, future engineers. Today, we are not just solving an equation; we are exploring the architecture of a family of curves. When you see $x^k+y^k=a^k$, do not just see variables and exponents. See a shape.

For $k=2$, this is a circle. For $k=1$, it is a straight line. As $k$ changes, the curve morphs, breathing and shifting. Our task is to find the specific value of $k$ that satisfies a very particular condition regarding its slope.

The Art of Implicit Differentiation

The problem gives us a condition involving the derivative: $\frac{dy}{dx}+(\frac{y}{x}{)}^{\frac{1}{3}}=0$. To use this, we need to find the derivative of our curve.

Since $y$ is trapped inside the power $k$, we cannot easily isolate it. This is where we summon the power of Implicit Differentiation. We treat $y$ as a hidden function of $x$, denoted as $y(x)$, and differentiate both sides with respect to $x$.

When we apply the operator $\frac{d}{dx}$ to $x^k+y^k=a^k$, we are essentially asking: "How does the sum of these powers change as $x$ changes?"

On the left, we have two terms. The derivative of $x^k$ is straightforward: $k{x}^{k-1}$. But the second term, $y^k$, requires the Chain Rule.

We differentiate the outer function $y^k$ to get $k{y}^{k-1}$, and then multiply by the derivative of the inner function, $\frac{dy}{dx}$. Thus, $\frac{d}{dx}(y^k)=k{y}^{k-1}\frac{dy}{dx}$.

On the right side, $a^k$ is a constant, so its rate of change is zero. Our equation becomes:

Isolating the Slope

Now, we isolate $\frac{dy}{dx}$. We subtract $k{x}^{k-1}$ from both sides and divide by $k{y}^{k-1}$. The $k$ cancels out beautifully, leaving us with:

We can rewrite this as:

This is the general expression for the slope of our curve at any point $(x,y)$. It is elegant, symmetric, and powerful.

The Bridge of Equality

The problem provides us with a specific condition: $\frac{dy}{dx}+(\frac{y}{x}{)}^{\frac{1}{3}}=0$. Rearranging this, we get:

Now, we have two expressions for the same slope. We equate them:

The negative signs cancel out, leaving us with $(\frac{x}{y}{)}^{k-1}=(\frac{y}{x}{)}^{\frac{1}{3}}$. To compare the exponents, we need the bases to match.

We know that $\frac{y}{x}=(\frac{x}{y}{)}^{-1}$. Substituting this into the right side, we get:

Using the laws of exponents, this simplifies to:

The Final Unveiling

Since the bases are identical, the exponents must be equal. We set:

Solving for $k$, we add $1$ to both sides:

And there it is. The value of $k$ is $\frac{2}{3}$. You have successfully navigated the implicit landscape, avoided the chain rule trap, and unified the algebraic expressions. This is the essence of JEE Advanced mathematics—not just calculating, but connecting.