Problem 110
Question
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) \((x-2)^{2} y^{\prime 2}=25-(y-2)^{2}\)
Step-by-Step Solution
Verified Answer
The correct answer is (C):
\((y-2)^2 y'^2 = 25-(y-2)^2\).
1Step 1: 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 \((x-a)^2 + (y-2)^2 = 25\).
2Step 2: 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 \(y' = \frac{dy}{dx}\).
3Step 3: Express 'a' in terms of x, y, and y'
From \((x-a) + (y-2)y' = 0\), solve for \(a\): \(a = x + (y-2)y'\).
4Step 4: 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 \(((y-2)y')^2 + (y-2)^2 = 25\).
5Step 5: 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).
Key Concepts
Implicit DifferentiationCircle GeometryDifferential Calculus
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.
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.
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.
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.
Other exercises in this chapter
Problem 108
The differential equation of all circles passing through the origin and having their centres on the \(x\)-axis is (A) \(x^{2}=y^{2}+x y \frac{d y}{d x}\) [2007]
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The solution of the differential equation \(\frac{d y}{d x}=\frac{x+y}{x}\) satisfying the condition \(y(1)=1\) is \([2008]\) (A) \(y=\ln x+x\) (B) \(y=x \ln x+
View solution Problem 111
The differential equation which represents the family of curves \(y=c_{1} e^{c_{x} x}\), where \(c_{1}\) and \(c_{2}\) are arbitrary constants is (A) \(y^{\prim
View solution Problem 112
Solution to the differential equation \(\cos x d y=y(\sin\) \(x-y) d x, 0
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