Let a curve y = y(x) pass through the point (3, 3) and the area of the region under this curve, abov

Q: Wait, why do we use the Leibniz Rule here? Can't we just integrate directly?

Great question! We don't know the function y(x) yet, so we cannot perform the integration. The Leibniz Rule allows us to 'unlock' the function trapped inside the integral by differentiating both sides, effectively turning an integral equation into a differential equation that we can solve.

Q: Why did we substitute t = y^2? Is there a specific reason?

Look closely at the differential equation after differentiation: it contains terms of y and y'. This is a Bernoulli-type equation. By setting t = y^2, we simplify the non-linear term into a linear one, which makes the equation much easier to handle using the Integrating Factor method.

Q: Why do we reject the negative root for alpha?

The problem explicitly states that the point (alpha, 6*sqrt(10)) lies in the first quadrant. In the first quadrant, both x and y coordinates must be positive. Since alpha represents the x-coordinate, we must choose the positive root.

Q: How do I know if my Integrating Factor is correct?

A quick check is to see if multiplying the entire differential equation by your I.F. makes the left-hand side a perfect derivative of the product (I.F. * t). If it does, you've found the right one!

Q: Is it possible to solve this without the substitution t = y^2?

Technically, you could try other methods, but the substitution t = y^2 is the most elegant path. It reduces the complexity of the equation immediately, saving you time and reducing the chance of algebraic errors—a crucial skill for JEE Advanced.

Question

Let a curve $y = y(x)$ pass through the point $(3, 3)$ and the area of the region under this curve, above the x-axis and between the abscissae 3 and $x(>3)$ be $(\frac{y}{x})^3$. If this curve also passes through the point $(\alpha, 6\sqrt{10})$ in the first quadrant, then $\alpha$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You see a curve $y = y(x)$ snaking through the point $(3, 3)$.

You are told that the area trapped under this curve, bounded by the $x$-axis and the vertical lines at $x=3$ and some arbitrary $x$, is exactly equal to $(\frac{y}{x})^3$. This curve is defined by its own accumulation of area.

### The Calculus of Accumulation

We start by writing the area as an integral. The area under the curve from $3$ to $x$ is given by $\int_3^x y(t) dt$. According to the problem, this integral is equal to $(\frac{y}{x})^3$.

So, we have our starting equation:

$$\int_3^x y(t) dt = \left( \frac{y}{x} ight)^3$$

To isolate $y$, we use the Leibniz Rule. By differentiating both sides with respect to $x$, we invoke the Fundamental Theorem of Calculus on the left side, which returns the integrand evaluated at $x$.

On the right side, we apply the chain rule and the quotient rule:

$$\frac{d}{dx} \left[ \int_3^x y(t) dt ight] = \frac{d}{dx} \left[ \left( \frac{y}{x} ight)^3 ight]$$

This yields $y = 3(\frac{y}{x})^2 \cdot \frac{d}{dx}(\frac{y}{x})$. Expanding the derivative of the quotient, we get:

$$y = \frac{3y^2}{x^2} \cdot \left( \frac{x y' - y}{x^2} ight)$$

### The Transformation

Now, we simplify the expression. By rearranging the terms, we arrive at the differential equation:

$$x^4 y = 3xy^2 y' - 3y^3$$

Dividing by $y$ (assuming $y 
eq 0$), we get $3y y' - \frac{3}{x} y^2 = x^3$. We use the substitution $t = y^2$, which implies $\frac{dt}{dx} = 2y y'$, or $y y' = \frac{1}{2} \frac{dt}{dx}$.

Substituting this into our equation transforms the non-linear expression into a linear differential equation:

$$\frac{3}{2} \frac{dt}{dx} - \frac{3}{x} t = x^3$$

After normalizing by multiplying by $\frac{2}{3}$, we obtain:

$$\frac{dt}{dx} - \frac{2}{x} t = \frac{2}{3} x^3$$

### The Final Descent

The Integrating Factor (I.F.) is calculated as follows:

$$I.F. = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2}$$

Multiplying our linear equation by this factor allows us to integrate both sides. The general solution becomes:

$$\frac{t}{x^2} = \int \frac{2}{3} x dx + C \Rightarrow \frac{t}{x^2} = \frac{x^2}{3} + C$$

Substituting $t = y^2$ back in, we find the general equation of our curve:

$$y^2 = \frac{x^4}{3} + Cx^2$$

Using the point $(3, 3)$, we solve for $C$: $9 = \frac{81}{3} + 9C \Rightarrow 9 = 27 + 9C \Rightarrow 9C = -18 \Rightarrow C = -2$. Thus, the curve is defined by **$y^2 = \frac{x^4}{3} - 2x^2$**.

### The Victory Lap

Finally, we look for the point $(\alpha, 6\sqrt{10})$. Plugging these coordinates into our curve equation:

$$(6\sqrt{10})^2 = \frac{\alpha^4}{3} - 2\alpha^2$$
$$360 = \frac{\alpha^4}{3} - 2\alpha^2$$

This leads to the quadratic equation $\alpha^4 - 6\alpha^2 - 1080 = 0$. Solving for $\alpha^2$ using the quadratic formula:

$$\alpha^2 = \frac{6 \pm \sqrt{36 - 4(1)(-1080)}}{2} = \frac{6 \pm \sqrt{4356}}{2} = \frac{6 \pm 66}{2}$$

Since $\alpha^2$ must be positive, $\alpha^2 = \frac{72}{2} = 36$. Therefore, **$\alpha = 6$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Problem

The Integral Equation

Applying Leibniz Rule

Expanding the Derivative

Simplifying to a Differential Equation

The Substitution Trick

Forming the Linear DE

Finding the Integrating Factor

Solving the Linear DE

Finding the Constant of Integration

The Final Curve Equation

Finding the Target Point

Solving for Alpha

The Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Calculus of Accumulation

The Transformation

The Final Descent

The Victory Lap

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