Question

Differential equation of the curves having the subnormal with \((7 / 2)\) units and passes through \((0,0)\) is: (A) \(x^{2}=7 y\) (B) \(\mathrm{y}^{2}=7 \mathrm{x}+\mathrm{c}, \mathrm{c} \neq 0\) (C) \(y^{2}=7 x\) (D) None of these

Short answer

The short answer to the problem is: (D) None of these The differential equation of the curves having a subnormal of \((7 / 2)\) units and passing through the point \((0, 0)\) is given by the equation: \(x^2 = e^{\frac{4y}{7}} + C\) None of the given options match the derived equation.

Step-by-step solution

Noting the Given Information

We are given: 1. The length of the subnormal is \((7 / 2)\) units. 2. The curves pass through the point \((0, 0)\).

Write the Expression for Subnormal Length

Let \(y = f(x)\) be the curve. Its differential is: \[ dy = f'(x) dx \] If \(P(x, y)\) is any point on the curve and its subnormal length is \(p\), we have: \[ p = x \tan \theta \] where \(\theta\) is the angle between the tangent at \(P\) and the x-axis. Note that \(\tan \theta = \frac{dy}{dx} = f'(x)\). From the given information, we have \(p = \frac{7}{2}\).

Write the Expression for the Derivative and Integrate

By substituting the value of \(p\) in the equation for subnormal, we get: \[ x f'(x) = \frac{7}{2} \] Now, we need to find \(f(x)\) by integrating with respect to \(x\). So, we rewrite the equation as: \[ f'(x) = \frac{7}{2x} \] Now integrate on both sides: \[ \int f'(x) dx = \int \frac{7}{2x} dx \] Using integration, we get: \[ f(x) = \frac{7}{2} \ln |x| + C \] Here, \(C\) is an arbitrary constant.

Applying the Given Condition (\(0, 0\))

The curve passes through the point \((0, 0)\). So, we can use \((x, y) = (0, 0)\) to find the value of \(C\). However, since ln(x) is undefined at x = 0, we must use L'Hôpital's rule to find the limit as x approaches 0 for f(x). Consider the limit: \[ \lim_{x \to 0} \frac{y}{\ln x} = \lim_{x \to 0} \frac{f(x)}{\ln x} \] Applying L'Hôpital's rule: \[ \lim_{x \to 0} \frac{f'(x)}{1/x} = \lim_{x \to 0} \frac{7/2x}{1/x} \] Evaluating the limit, we get: \[ \lim_{x \to 0} \frac{7/2x}{1/x} = \frac{7}{2} \] So, the equation of the curve becomes: \[ y = \frac{7}{2} \ln |x| \]

Rewrite Equation in Implicit Form using Exponentials

Finally, we use exponentials to rewrite the equation in an implicit form: First, let \(2y = 7 \ln |x|\), then: \[ \frac{2y}{7} = \ln |x| \] Now, using exponentials: \[ |x| = e^{\frac{2y}{7}} \] In implicit form, we have: \[ x^2 = e^{\frac{4y}{7}} + C \] Now, we have to compare our derived equation with the given options: (A) \(x^{2}=7 y\) (B) \(\mathrm{y}^{2}=7 \mathrm{x}+\mathrm{c}, \mathrm{c} \neq 0\) (C) \(y^{2}=7 x\) (D) None of these Since no given options match our derived equation, the correct answer is (D) None of these.

Key concepts

Subnormal Length

The subnormal length of a curve at a given point is a geometrical concept that relates to the tangent of the curve. Imagine a curve described by a function \(y = f(x)\). When you draw a tangent to this curve at a point \(P(x, y)\), it makes an angle \(\theta\) with the x-axis. The subnormal length (denoted as \(p\)) is the horizontal projection of the part of the normal line, which is perpendicular to the tangent, between the point \(P\) and the x-axis. This is given by the expression:

• \(p = x \tan \theta\)

Substituting \(\tan \theta\) with the slope of the tangent \(\frac{dy}{dx}\), we can express the subnormal length in terms of the derivative of the function. In cases like the original exercise, knowing the subnormal length enables us to find the derivative and allows further exploration of the curve by integration.

Integration

Integration is a fundamental tool in calculus that helps us find a function when its derivative is known. In the scenario of the original exercise, we needed to find \(f(x)\), for which the derivative \(f'(x)\) was given. By applying integration, we essentially reverse the differentiation process. Here’s how it works:

• Start with the equation for the derivative: \(f'(x) = \frac{7}{2x}\).
• Integrate both sides: \(\int f'(x) \, dx = \int \frac{7}{2x} \, dx\).
• Complete the integration to find \(f(x)\): \(f(x) = \frac{7}{2} \ln |x| + C\), where \(C\) is the constant of integration.

In this example, finding \(C\) depends on the conditions given—such as the curve passing through a specific point. Integration transforms derivatives into usable functions, unlocking further analysis possibilities for curves.

Curve Analysis

Curve analysis encompasses the study of various properties and features of curves, often represented as functions or equations. In the context of differential equations, understanding the behavior of a curve includes examining elements like tangents, normals, and special points like intercepts or vertices.

Using known values (such as a specified subnormal length and passing through a known point), analysts can derive specific curve equations. Once the equation is known, you can run a comprehensive analysis of the curve’s properties:

• Determine whether it's open or closed.
• Identify symmetry.
• Explore curvature and inflection points.

In our exercise, the analysis also involves comparing the derived equation to provided options, which helps ascertain which, if any, match the calculated conditions of the problem.

Implicit Function

Implicit functions are equations that define a relationship between variables indirectly, without solving explicitly for one in terms of the others. In the original problem, after deriving the curve equation, rewriting it in implicit form provides a broader yet concise representation.

Here’s an example of translating our derived explicit function into an implicit one:

• Start with \(2y = 7 \ln |x|\) derived from \(y = \frac{7}{2} \ln |x|\).
• Incorporate exponentials: \(x = e^{(\frac{2y}{7})}\).
• Square both sides to frame it implicitly: \(x^2 = (e^{(\frac{2y}{7})})^2\).

The implicit function form \(x^2 = e^{\frac{4y}{7}} + C\) simplifies and generalizes the expression. These forms are often vital in differential equations because they allow curve representations without isolating specific variables, which can simplify complex problem-solving.