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The differential equation of the family of circles with fixed radius 5 units and centre on the line \(y=2\) is [2008] (A) \((x-2) y^{\prime 2}=25-(y-2)^{2}\) (B) \((y-2) y^{\prime 2}=25-(y-2)^{2}\) (C) \((y-2)^{2} y^{\prime 2}=25-(y-2)^{2}\) (D) \((x-2)^{2} y^{\prime 2}=25-(y-2)^{2}\)

Short Answer

Expert verified
The correct answer is (C): \((y-2)^2 y'^2 = 25-(y-2)^2\).

Step by step solution

01

Understand the Problem

We need to find the differential equation of circles with radius 5 units with centers on the line \(y=2\). This means all centers have the form \((a, 2)\). The equation of such a circle is \((x-a)^2 + (y-2)^2 = 25\).
02

Differentiate the Circle Equation

Differentiate the equation \((x-a)^2 + (y-2)^2 = 25\) with respect to \(x\). Using implicit differentiation, we get \(2(x-a) + 2(y-2)\cdot\frac{dy}{dx} = 0\). Simplifying gives \((x-a) + (y-2)y' = 0\), where \(y' = \frac{dy}{dx}\).
03

Express 'a' in terms of x, y, and y'

From \((x-a) + (y-2)y' = 0\), solve for \(a\): \(a = x + (y-2)y'\).
04

Substitution into Circle Equation

Substitute \(a = x + (y-2)y'\) back into the circle equation \((x-a)^2 + (y-2)^2 = 25\). Replacing \(a\), we have:\((x - (x + (y-2)y'))^2 + (y-2)^2 = 25\).This simplifies to \(((y-2)y')^2 + (y-2)^2 = 25\).
05

Identify the Correct Option

Compare the equation \(((y-2)y')^2 + (y-2)^2 = 25\) with the given options. Simplifying further, we get:\((y-2)^2 \cdot y'^2 + (y-2)^2 = 25\), which can be written as\((y-2)^2(y'^2 + 1) = 25\).Hence \((y-2)^2 y'^2 = 25 - (y-2)^2\).So the correct option is (C).

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Key Concepts

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

Implicit Differentiation
Implicit differentiation is a technique used to find derivatives when a function is not explicitly solved for one variable in terms of another. Instead of solving for one variable and then differentiating, we apply the differentiation directly to the equation involving both variables. This is particularly useful for complex equations like those involving circles or other geometric shapes.
In the context of the circle equation \[(x-a)^2 + (y-2)^2 = 25,\]we differentiate both sides with respect to \(x\). Since \(y\) is implicitly a function of \(x\), when we differentiate terms involving \(y\), we use the chain rule. This means multiplying by \(\frac{dy}{dx}\), often denoted as \(y'\). This approach helps manage equations where both variables are intertwined.
Following the differentiation, we obtain:\[2(x-a) + 2(y-2)\cdot\frac{dy}{dx} = 0.\]This equation contains both \(x\) and \(y\), reflecting their implicit relationship, and allows us to isolate terms involving derivatives.
Circle Geometry
Circle geometry is an essential area of mathematics focused on properties and equations related to circles. In problems like the one at hand, we consider circles having a specific radius and center. A circle's equation in standard form is given by\[(x-h)^2 + (y-k)^2 = r^2,\]where \((h, k)\) is the center, and \(r\) is the radius. This problem takes it further by restricting centers to a particular line, \(y=2\), and maintaining a fixed radius of 5 units.
This means the centers of all possible circles lie on the line \(y=2\), giving centers structured as \((a, 2)\). Consequently, each circle in this family has the equation:\[(x-a)^2 + (y-2)^2 = 25.\]The constant value of 25 is derived from \(r^2\), since the radius \(r = 5\).
Circles' geometric properties become critical when analyzing or converting these equations into differential forms, as is required in this task.
Differential Calculus
Differential calculus plays a crucial role in analyzing rates of change and slopes of curves described by equations. In this exercise, it's about transforming the geometric circle equation into a differential form that describes how the circle's geometry behaves as variables change.
Starting with the basic circle equation \[(x-a)^2 + (y-2)^2 = 25,\]we derive its differential equation using implicit differentiation. This leads us to\[(x-a) + (y-2)y' = 0,\]where \(y' = \frac{dy}{dx}\). Solving for \(a\) gives us a way to express it in terms of \(x\), \(y\), and \(y'\).
The substitution of this back into the original circle equation transforms it into the differential equation:\[(y-2)^2 y'^2 = 25 - (y-2)^2.\]This expression not only encapsulates the continuous nature of the circle but also highlights how differential calculus can break down static geometry into fluid mathematical relationships. Understanding this helps in grasping how motion or changes can be modeled mathematically.

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Most popular questions from this chapter

Passage 2 Some equations which are not exact can be made exact on multiplication by some suitable function known as an integrating factor. The equation $$ x d y-y d x=0 $$ which is not exact becomes so on multiplication by \(1 / y^{2}\), for then we $$ \frac{x}{y^{2}} d y-\frac{1}{y} d x=0 $$ which is easily seen to be exact. We can solve it either by re-arranging the terms and making them exact differential or by the method of exact equations. We now give some rules for finding integrating factors of differential equation $$ M d x+N d y=0 $$ to make it exact. I. If \(M x+N y \neq 0\) and the equation is homogeneous, then \(\frac{1}{M x+N y}\) is an I.F. II. If the equation \(M d x+N d y=0\) is not exact but is of the form \(f_{1}(x y) y d x+f_{2}(x y) x d y=0\), then \(\frac{1}{M x-N y}\) is an I.F., provided \(M x-N y \neq 0\) III. When \(\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\) is a function of \(x\) alone, say \(f(x)\), then I.F. \(=e^{\int f(x) d x}\) IV. When \(\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\) is a function of \(y\) alone, say \(f(y)\), then I.F. \(=e^{\int f(y) d y}\) The solution of the differential equation \(\left(x^{2} y-2 x y^{2}\right) d x-\left(x^{3}-3 x^{2} y\right) d y=0\) is (A) \(\frac{x}{y}-2 \log x+3 \log y=c\) (B) \(\frac{x}{y}+2 \log x+3 \log y=c\) (C) \(\frac{x}{y}-2 \log x-3 \log y=c\) (D) None of these

If \(y=y(x)\) and \(\frac{2+\sin x}{y+1}\left(\frac{d y}{d x}\right)=-\cos x, y(0)=1\), then \(y\left(\frac{\pi}{2}\right)=\) (A) \(\frac{1}{3}\) (B) \(\frac{2}{3}\) (C) \(-\frac{1}{3}\) (D) 1

The solution of the differential equation \(x\left(y^{2} e^{x y}+e^{x / y}\right) d y=y\left(e^{x / y}-y^{2} e^{x y}\right) d x\) is (A) \(x y=\ln \left(e^{y_{x}}+c\right)\) (B) \(x y=\ln \left(e^{x / y}+c\right)\) (C) \(\frac{y}{x}=\ln \left(e^{x y}+c\right)\) (D) \(\frac{x}{y}=\ln \left(e^{x y}+c\right)\)

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

In the following questions an Assertion \((A)\) is given followed by a Reason \((R) .\) Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is acorrect explanation forAssertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: The order of the differential equation, of which \(x y=c e^{x}+b e^{-x}+x^{2}\) is a solution, is 2 . Reason: The differential equation is \(x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2=0\)

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