Problem 30 - A Complete Resource Book in Mathematics for JEE Main

Question

The points on the curve \(x y^{2}=1\) which are nearest to the origin are (A) \(\left[\left(\frac{1}{2}\right)^{1 / 3}, \pm\left(\frac{1}{2}\right)^{-1 / 6}\right]\) (B) \(\left[\left(\frac{1}{2}\right)^{1 / 3}, 2^{-1 / 6}\right]\) (C) \(\left(2^{1 / 3}, \pm\left(\frac{1}{2}\right)^{-1 / 6}\right)\) (D) None of these

Step-by-Step Solution

Answer

Option (A) \(\left[\left(\frac{1}{2}\right)^{1/3}, \pm\left(\frac{1}{2}\right)^{-1/6}\right]\) is nearest to the origin.

Step 1: Express the Distance Function

We want to find the distance from any point \((x, y)\) on the curve \(xy^2 = 1\) to the origin \((0,0)\). The distance function is given by \( d = \sqrt{x^2 + y^2} \).

Step 2: Set Up the Optimization Problem

Since the square root function is increasing, minimizing \(d\) is equivalent to minimizing \(d^2 = x^2 + y^2\). Substitute \(y^2 = \frac{1}{x}\) (from the curve equation) into \(d^2\). Thus, \(d^2 = x^2 + \frac{1}{x}\).

Step 3: Differentiate the Function

Differentiate \(d^2 = x^2 + \frac{1}{x}\) with respect to \(x\). The derivative is \(\frac{d}{dx}(x^2 + \frac{1}{x}) = 2x - \frac{1}{x^2}\).

Step 4: Find the Critical Points

Set the derivative equal to zero to find the critical points. Solve \(2x - \frac{1}{x^2} = 0\). Rearrange to get \(2x^3 = 1\). Thus, \(x = \left(\frac{1}{2}\right)^{1/3}\).

Step 5: Solve for y

Knowing \(x = \left(\frac{1}{2}\right)^{1/3}\), use the curve equation \(xy^2 = 1\) to find \(y\). Substitute \(x\) to get \(y^2 = \frac{1}{\left(\frac{1}{2}\right)^{1/3}} = 2^{1/3}\), so \(y = \pm\left(2^{1/3}\right)^{1/2} = \pm \left(\frac{1}{2}\right)^{-1/6}\).

Step 6: Conclusion

Thus, the points on the curve closest to the origin are \(\left(\left(\frac{1}{2}\right)^{1/3}, \pm\left(\frac{1}{2}\right)^{-1/6}\right)\).

Key Concepts

Distance Minimization

When we talk about distance minimization in calculus, it's all about finding the point on a given curve closest to a specific location. In this scenario, we aim to discover the point on the curve \( xy^2 = 1 \) that is closest to the origin, which is the point \((0, 0)\).

The principle of minimizing distance often involves working with the distance formula. The distance between any point \( (x, y) \) and the origin is given by \( d = \sqrt{x^2 + y^2} \). Although this gives us the actual distance, optimizing becomes more manageable when we work with the square of the distance instead, which removes the square root and simplifies our calculations. Consequently, we focus on minimizing \( d^2 = x^2 + y^2 \), which retains the properties necessary to find the closest point.

This strategy can often save time because squaring is a monotonic transformation, meaning if \( d_1 < d_2 \), then \( d_1^2 < d_2^2 \) as well. By minimizing \( d^2 \), we're indirectly minimizing \( d \), helping to find the optimal solution effectively.

Differentiation

Differentiation is the cornerstone of calculus that allows us to analyze the behavior of functions in detail. In the context of optimization problems like distance minimization, differentiation helps identify how functions change and reveals where these changes indicate a minimum or maximum value.

In our case, we are tasked with differentiating the function \( d^2 = x^2 + \frac{1}{x} \). By taking the derivative with respect to \( x \), we gain insight into how the function \( d^2 \) behaves as \( x \) varies. Through differentiation, we obtain \( \frac{d}{dx}(x^2 + \frac{1}{x}) = 2x - \frac{1}{x^2} \). This derivative tells us about the slope of the distance squared function. At points where the derivative equals zero, the slope is zero, indicating potential locations for minima or maxima.

Differentiation is critical because it provides the necessary tools to solve practical problems by unraveling the hidden characteristics within mathematical relationships. Whether maximizing profit in economics or minimizing distance in geometry, differentiation simplifies these complex tasks.

Critical Points

Critical points are a major focus in optimization problems as they reveal where a function might attain its minima or maxima. A critical point occurs where the derivative of a function is zero or undefined. In terms of finding the closest point on a curve, it could correspond to the location where the distance is minimized.

Once we have the derivative \( 2x - \frac{1}{x^2} \) from the previous section, we set it equal to zero to find the critical points. Solving the equation \( 2x - \frac{1}{x^2} = 0 \) leads us to \( 2x^3 = 1 \). Therefore, \( x = \left(\frac{1}{2}\right)^{1/3} \). Finding \( x \) is just one part of the puzzle. It provides a promising location where the distance squared function doesn't increase or decrease—indicative of a potential minimum.

Next, to determine the corresponding \( y \) values, we use the equation of the curve \( xy^2 = 1 \), substituting \( x = \left(\frac{1}{2}\right)^{1/3} \). This yields \( y = \pm\left(\frac{1}{2}\right)^{-1/6} \). These \((x, y)\) pairs are the points on the curve nearest the origin. Understanding and locating critical points drive the process of discovering such optimal solutions in calculus.