/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 If the solution of \(\frac{d y}{... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 22

If the solution of \(\frac{d y}{d x}=\frac{a x+3}{2 y+f}\) represents a circle, then the value of \(a\) is (A) 2 \(\begin{array}{lll}\text { (B) }-2 & \text { (C) } 3 & \text { (D) }-4\end{array}\)

Short Answer

Expert verified
The value of \(a\) is -2.

Step by step solution

01

Recognize the Equation of the Circle

We know that the general equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where \((h,k)\) is the center of the circle and \(r\) is the radius. For such a solution, the derivative \(\frac{dy}{dx}\) must be of a specific form.
02

Differentiate and Compare

Start by considering the form of the differential equation \(\frac{dy}{dx} = \frac{ax + 3}{2y + f}\), which must be derived from a circle's equation. Recognize that for \(\frac{dy}{dx}\) to be a consistent circular derivative, the numerator and denominator must imply a constant.
03

Simplify and Examine Constants

The circle equation, when parameterized into derivatives, often translates to terms involving symmetries in \(x\) and \(y\). For this problem, equate the coefficients in such a way as to eliminate dependencies on \(x\), suggesting that \(a\) produces opposing symmetry across \(x\)-intersecting terms.
04

Solve for 'a'

By setting \(a\) such that the transformation recapitulates a known circular symmetry, solve through attempts or patterning to find suitable constancy: this symmetry emerges when \(a = -2\).
05

Confirm and Validate the Calculation

Ensure correctness by substituting \(a = -2\) into the original differential equation, verifying it corresponds correctly to a circle. This ensures uniform interpretation of derivatives yielding resisting terms.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Circle Equation
Understanding the equation of a circle is fundamental when dealing with differential equations that represent circles. A circle's typical equation in the Cartesian coordinate system is given by \[ (x - h)^2 + (y - k)^2 = r^2 \] where:
  • \((h, k)\) is the center of the circle.
  • \(r\) is the radius of the circle.
This equation is derived from the distance formula, ensuring that every point \( (x, y) \) on the circle is equidistant from the center. To have a differential equation represent a circle, it must align with this basic form or structure. This alignment ensures that when differentiating, the terms reflect the circle's symmetry and constant radius.
Derivative
The derivative, in this context, signifies how the function changes as the variables alter. For the differential equation given \[ \frac{dy}{dx} = \frac{ax + 3}{2y + f} \] we want it to reflect characteristics of a circle. The term \(\frac{dy}{dx}\) represents the slope of the tangent to the curve at any point \( (x, y)\). For a function to represent a circle, the resultant derivative must reflect a balance that mirrors the geometric symmetry of a circle. This involves:
  • Ensuring that the slope remains uniform around the circle.
  • Equal contributions from both x and y directions, implying a lack of dependency solely on one variable.
Interpreting these derivatives correctly helps ensure that the equation remains consistent with circular motion or symmetry.
Solution Verification
Verification of a proposed solution is crucial in mathematics to ensure correctness. After hypothesizing that \(a = -2\) yields a circle through the given differential equation, this must be checked through substitution. When substituting \(a = -2\) back into the differential equation:
  • Recompute the slope \(\frac{dy}{dx}\) and ensure it logically binds to the known geometry of a circle.
  • Check that the slopes derived reflect equidistance and symmetry around a central point, consistent with the expected behavior of a circle.
  • Confirm that the resultant equation aligns with the original circle equation's parameters.
If everything holds true, the choice of 'a' aligns, confirming the solution's validity and adequacy.
Symmetry in Mathematics
Symmetry plays a pivotal role in understanding mathematical functions and equations. For circles, symmetry is their most distinguishing feature, spanning evenly in every direction from a central point. The notion of symmetry is crucial when assessing whether an equation can depict a circle, as was the case with the exercise.
  • Symmetry in a circle means the equation must exhibit equal dependencies and geometric balance in all axes involved.
  • For the differential equation provided, symmetry would dictate how changes in \(x\) and \(y\) counterbalance.
  • This balance ensures consistent radial distance, aligning with the derivative requirements.
Recognizing symmetry assures that the mathematical structure reflects the intended geometric form, solidifying the equation's claim to circular representation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The solution of the differential equation \((x \cos x-\sin\) \(\left.x+y x^{2}\right) d x+x^{3} d y=0\) is equal to (A) \(\frac{\sin x}{x}+x y=c\) (B) \(\frac{\sin x}{x}+x=c\) (C) \(\frac{\sin x}{x}+y=c\) (D) None of these

The solution of the differential equation \(\frac{d y}{d x}=\frac{x+y}{x}\) satisfying the condition \(y(1)=1\) is (A) \(y=\ln x+x\) (B) \(y=x \ln x+x^{2}\) (C) \(y=x e^{(x-1)}\) (D) \(y=x \ln x+x\)

Let \(l\) be the purchase value of an equipment and \(V(t)\) be the value of equipment after it has been used for \(t\) years. The value \(V(t)\) depreciates at a rate given by the differential equation \(\frac{d V(t)}{d t}=k(T-t)\), where \(k>0\) is a constant and \(T\) is the total life in years of the equipment. Then, the scrap value \(V(T)\) of the equipment is (A) \(l-\frac{k T^{2}}{2}\) (B) \(l-\frac{k(T-t)^{2}}{2}\) (C) \(e^{-k T}\) (D) \(T^{2}-\frac{l}{k}\)

The population \(p(t)\) at time \(t\) of a certain mouse species satisfies the differential equation \(\frac{d p(t)}{d t}\) \(=0.5 p(t)-450 .\) If \(p(0)=850\), then the time at which the population become zero is (A) \(2 \ln 18\) (B) \(\ln 9\) (C) \(1 / 2 \ln 18\) (D) \(\ln 18\)

The differential equation of all non-vertical lines in a plane is (A) \(\frac{d^{2} y}{d x^{2}}=0\) (B) \(\frac{d^{2} x}{d y^{2}}=0\) (C) \(\frac{d y}{d x}=0\) (D) \(\frac{d x}{d y}=0\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.