Problem 250
Question
For the following exercises, use this information: A function \(f(x, y)\) is said to be homogeneous of degree \(n\) if \(f(t x, t y)=t^{n} f(x, y) .\) For all homogeneous functions of degree \(n, \quad\) the following equation is true: \(x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=n f(x, y)\) . Show that the given function is homogeneous and verify that \(x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=n f(x, y)\) $$ f(x, y)=x^{2} y-2 y^{3} $$
Step-by-Step Solution
Verified Answer
The function is homogeneous of degree 3, and the equation is verified for this degree.
1Step 1: Verify Homogeneity of the Function
To assess whether the function \( f(x, y) = x^2y - 2y^3 \) is homogeneous, we apply the definition: a function is homogeneous of degree \( n \) if \( f(tx, ty) = t^n f(x, y) \). We substitute \( tx \) and \( ty \) in place of \( x \) and \( y \) in the function:\[ f(tx, ty) = (tx)^2(ty) - 2(ty)^3 = t^2 \cdot x^2 \cdot t \cdot y - 2t^3y^3 = t^3x^2y - 2t^3y^3 \]Simplifying this gives:\[ t^3(x^2y - 2y^3) = t^3f(x, y) \]Therefore, \( n = 3 \). The function is homogeneous of degree 3.
2Step 2: Calculate Partial Derivatives
To verify the given equation, find the partial derivatives of \( f(x, y) = x^2y - 2y^3 \). First, calculate \( \frac{\partial f}{\partial x} \):\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2y - 2y^3) = 2xy \]Next, calculate \( \frac{\partial f}{\partial y} \):\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2y - 2y^3) = x^2 - 6y^2 \]
3Step 3: Verify Given Equation
The given equation to verify is:\[ x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = nf(x, y) \]Substitute the partial derivatives:\[ x(2xy) + y(x^2 - 6y^2) = 2x^2y + x^2y - 6y^3 = 3x^2y - 6y^3 \]Since \( n = 3 \), the right side is:\[ 3(x^2y - 2y^3) = 3x^2y - 6y^3 \]Both sides are equal: \( 3x^2y - 6y^3 = 3x^2y - 6y^3 \).Therefore, the equation is satisfied.
Key Concepts
Partial DerivativesDegree of HomogeneityVerification of Homogeneous Property
Partial Derivatives
Partial derivatives are essential when dealing with functions that have multiple variables. Consider a function of two variables, like in our exercise, where we have \( f(x, y) = x^2y - 2y^3 \). Here, the function depends on both \( x \) and \( y \), and a partial derivative provides the rate at which the function changes as one variable changes while keeping the other constant.
To compute the partial derivative of \( f \) with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant. Therefore:\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2y - 2y^3) = 2xy \] For the partial derivative with respect to \( y \), represented as \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant:\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2y - 2y^3) = x^2 - 6y^2 \]
To compute the partial derivative of \( f \) with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant. Therefore:\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^2y - 2y^3) = 2xy \] For the partial derivative with respect to \( y \), represented as \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant:\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^2y - 2y^3) = x^2 - 6y^2 \]
- Use \( \frac{\partial f}{\partial x} \) when assessing changes due to \( x \) variation.
- Use \( \frac{\partial f}{\partial y} \) when assessing changes due to \( y \) variation.
Degree of Homogeneity
The degree of homogeneity is a measure that indicates how a function scales when its variables are multiplied by a constant factor. A function \( f(x, y) \) is said to be homogeneous of degree \( n \) if when each variable is replaced by its corresponding scaled version, it satisfies:\[ f(tx, ty) = t^n f(x, y) \]This tells us that scaling the input variables by \( t \) results in the output being scaled by \( t^n \).
Let's verify this for the function given in our exercise. If we substitute \( tx \) and \( ty \) for \( x \) and \( y \), respectively, we get:\[ f(tx, ty) = (tx)^2 (ty) - 2 (ty)^3 = t^3 (x^2y - 2y^3) = t^3 f(x, y) \]From this calculation, we can see that the function is homogeneous of degree \( 3 \). This property is particularly useful in simplifying complex equations and analyzing systems where proportional scaling occurs, like in economics or physics.
Let's verify this for the function given in our exercise. If we substitute \( tx \) and \( ty \) for \( x \) and \( y \), respectively, we get:\[ f(tx, ty) = (tx)^2 (ty) - 2 (ty)^3 = t^3 (x^2y - 2y^3) = t^3 f(x, y) \]From this calculation, we can see that the function is homogeneous of degree \( 3 \). This property is particularly useful in simplifying complex equations and analyzing systems where proportional scaling occurs, like in economics or physics.
Verification of Homogeneous Property
Verification involves ensuring that a homogeneous function respects the well-known identity that connects its degree of homogeneity with partial derivatives. For a homogeneous function of degree \( n \), the expression\[ x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = n f(x, y) \]should hold true. Using this identity helps confirm the consistency of the function's defined scaling behavior.
To verify, substitute the partial derivatives from our function: \( \frac{\partial f}{\partial x} = 2xy \) and \( \frac{\partial f}{\partial y} = x^2 - 6y^2 \).Replacing in the identity, we have:\[ x(2xy) + y(x^2 - 6y^2) = 2x^2y + x^2y - 6y^3 = 3x^2y - 6y^3 \]Verify this matches \( n f(x, y) \) where \( n = 3 \):\[ 3(x^2y - 2y^3) = 3x^2y - 6y^3 \]Both sides match, confirming the function satisfies the homogeneous property. Such verification is crucial in mathematical modeling to ensure equations adhere to theoretical constructs.
To verify, substitute the partial derivatives from our function: \( \frac{\partial f}{\partial x} = 2xy \) and \( \frac{\partial f}{\partial y} = x^2 - 6y^2 \).Replacing in the identity, we have:\[ x(2xy) + y(x^2 - 6y^2) = 2x^2y + x^2y - 6y^3 = 3x^2y - 6y^3 \]Verify this matches \( n f(x, y) \) where \( n = 3 \):\[ 3(x^2y - 2y^3) = 3x^2y - 6y^3 \]Both sides match, confirming the function satisfies the homogeneous property. Such verification is crucial in mathematical modeling to ensure equations adhere to theoretical constructs.
Other exercises in this chapter
Problem 248
For the following exercises, use this information: A function \(f(x, y)\) is said to be homogeneous of degree \(n\) if \(f(t x, t y)=t^{n} f(x, y) .\) For all h
View solution Problem 249
For the following exercises, use this information: A function \(f(x, y)\) is said to be homogeneous of degree \(n\) if \(f(t x, t y)=t^{n} f(x, y) .\) For all h
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