Problem 16
Question
Which of the following does not represent the orthogonal trajectory of the system of curves \(\left(\frac{d y}{d x}\right)^{2}\) \(=\frac{a}{x}\) (A) \(9 a(y+c)^{2}=4 x^{3}\) (B) \(y+c=\frac{-2}{3 \sqrt{a}} x^{3 / 2}\) (C) \(y+c=\frac{2}{3 \sqrt{a}} x^{3 / 2}\) (D) All are orthogonal trajectories
Step-by-Step Solution
Verified Answer
Option C is not an orthogonal trajectory.
1Step 1: Understand Orthogonal Trajectories
Orthogonal trajectories are curves that intersect a given family of curves at right angles. If you have a family of curves with a differential equation in the form of \(\frac{dy}{dx}\), the orthogonal trajectory will have a differential equation that is the negative reciprocal of the original.
2Step 2: Derive the Differential Equation of Orthogonal Trajectories
The given system of curves is \( \left(\frac{dy}{dx}\right)^2 = \frac{a}{x} \). Let's find the typical \( \frac{dy}{dx} = \pm\sqrt{\frac{a}{x}} \). This means orthogonal trajectories should have \( \frac{dy}{dx} = \mp \sqrt{\frac{x}{a}} \).
3Step 3: Solve for the Orthogonal Trajectory
For the orthogonal trajectory, we solve the differential equation \( \frac{dy}{dx} = -\sqrt{\frac{x}{a}} \). Integrating, we get \( \int dy = -\int \sqrt{\frac{x}{a}} dx \). This leads to \( y = -\frac{2}{3\sqrt{a}} x^{3/2} + c \).
4Step 4: Compare Solutions with Given Options
The result from Step 3 gives the equation \( y + c = -\frac{2}{3\sqrt{a}} x^{3/2} \). Comparing this with the options:- Option (A) is in a different form, so we check calculations.- Option (B) matches exactly with \( y+c=\frac{-2}{3 \sqrt{a}} x^{3 / 2} \).- Option (C) contradicts, showing the opposite sign, \( y+c=\frac{2}{3 \sqrt{a}} x^{3 / 2} \).
5Step 5: Identify the Incorrect Orthogonal Trajectory
Based on Step 4's comparison, Option (C) \( y+c=\frac{2}{3 \sqrt{a}} x^{3 / 2} \) does not fit the derived orthogonal trajectory's equation. This means it's not an orthogonal trajectory as it has the incorrect sign.
Key Concepts
Differential EquationsCurve OrthogonalitySystem of Curves
Differential Equations
Differential equations are mathematical statements that relate a function with its derivatives. They play a crucial role in describing the behavior of dynamic systems in fields like physics, biology, and engineering. In simple terms, they help us understand how changing one quantity affects another.
A "differential equation" usually involves expressions like \( \frac{dy}{dx} \), which represents the rate of change of the variable \( y \) with respect to \( x \). Solving these equations allows us to find the function \( y(x) \) that satisfies the relationship.
For example, let's consider the differential equation from the original problem: \( \left(\frac{dy}{dx}\right)^2 = \frac{a}{x} \). This equation tells us that the square of the rate of change of \( y \) with respect to \( x \) is equal to \( \frac{a}{x} \). Solutions to this equation would represent a family of curves in a coordinate plane.
A "differential equation" usually involves expressions like \( \frac{dy}{dx} \), which represents the rate of change of the variable \( y \) with respect to \( x \). Solving these equations allows us to find the function \( y(x) \) that satisfies the relationship.
For example, let's consider the differential equation from the original problem: \( \left(\frac{dy}{dx}\right)^2 = \frac{a}{x} \). This equation tells us that the square of the rate of change of \( y \) with respect to \( x \) is equal to \( \frac{a}{x} \). Solutions to this equation would represent a family of curves in a coordinate plane.
Curve Orthogonality
Curve orthogonality means that two curves intersect at right angles (90 degrees). This concept becomes important when solving problems involving families of curves and looking for their orthogonal trajectories.
An orthogonal trajectory is a curve that crosses each curve in a given family of curves perpendicularly. To find the orthogonal trajectory of a family of curves represented by a differential equation \( \frac{dy}{dx} = f(x, y) \), we use the negative reciprocal slope, \( \frac{dy}{dx} = -\frac{1}{f(x, y)} \).
In our problem, the orthogonal trajectories are found using \( \frac{dy}{dx} = \mp \sqrt{\frac{x}{a}} \), which is the negative reciprocal of the original slope \( \pm\sqrt{\frac{a}{x}} \). This adjustment ensures the new curves intersect their counterparts at 90-degree angles.
An orthogonal trajectory is a curve that crosses each curve in a given family of curves perpendicularly. To find the orthogonal trajectory of a family of curves represented by a differential equation \( \frac{dy}{dx} = f(x, y) \), we use the negative reciprocal slope, \( \frac{dy}{dx} = -\frac{1}{f(x, y)} \).
In our problem, the orthogonal trajectories are found using \( \frac{dy}{dx} = \mp \sqrt{\frac{x}{a}} \), which is the negative reciprocal of the original slope \( \pm\sqrt{\frac{a}{x}} \). This adjustment ensures the new curves intersect their counterparts at 90-degree angles.
System of Curves
A system of curves refers to a family of curves that can be expressed through a common mathematical formula. These curves share some geometric properties and can often be described by a differential equation.
For the system of curves in our exercise example, the equation \( \left(\frac{dy}{dx}\right)^2 = \frac{a}{x} \) describes such a system. Each specific curve within this family has different characteristics and may be visualized as several similar shapes that vary in position, size, or orientation.
Understanding and working with systems of curves is highly valuable in various applications, including physics and engineering. It allows for the modeling of phenomena where multiple similar patterns or behaviors are observed, helping to predict interactions between elements of the system.
For the system of curves in our exercise example, the equation \( \left(\frac{dy}{dx}\right)^2 = \frac{a}{x} \) describes such a system. Each specific curve within this family has different characteristics and may be visualized as several similar shapes that vary in position, size, or orientation.
Understanding and working with systems of curves is highly valuable in various applications, including physics and engineering. It allows for the modeling of phenomena where multiple similar patterns or behaviors are observed, helping to predict interactions between elements of the system.
Other exercises in this chapter
Problem 14
The solution of the differential equation \((x-y)(2 d y-d x)=3 d x-5 d y\) is (A) \(2 x-y=\log (x-y+z)+c\) (B) \(2 x+y=\log (x-y+z)+c\) (C) \(2 y-x=\log (x-y+z)
View solution Problem 15
Solution of \(\frac{x d x+y d y}{x d y-y d x}=\frac{\sqrt{1-\left(x^{2}+y^{2}\right)}}{\sqrt{x^{2}+y^{2}}}\) (A) \(\sin ^{-1} \sqrt{x^{2}+y^{2}}=c\) (B) \(\tan
View solution Problem 17
Solution of \(\left(\frac{x+y-1}{x+y-2}\right) \frac{d y}{d x}=\left(\frac{x+y+1}{x+y+2}\right)\), given that \(y=1\) when \(x=1\), is (A) \(\log \left|\frac{(x
View solution Problem 18
Solution of \((y+x \sqrt{x y}(x+y)+y) d x+(y \sqrt{x y}(x+y)-x) d y=0\) is (A) \(x^{2}+y^{2}=2 \tan ^{-1} \sqrt{\frac{y}{x}}+c\) (B) \(x^{2}+y^{2}=4 \tan ^{-1}
View solution