Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 5: Problem 5
The equation \(3 y=z^{3}+3 x z\) defines \(z\) implicitly as a function of \(x\) and \(y .\) Evaluate all three second partial derivatives of \(z\) with respect to \(x\) and/or \(y\). Verify that \(z\) is a solution of $$ x \frac{\partial^{2} z}{\partial y^{2}}+\frac{\partial^{2} z}{\partial x^{2}}=0 $$
Short Answer
Expert verified
The second partial derivatives are \(\frac{\partial^2 z}{\partial x^2} = \frac{z + 2z^3}{(z^2 + x)^2}\), \(\frac{\partial^2 z}{\partial y^2} = \frac{-2z}{(z^2 + x)^3}\), and \(\frac{\partial^2 z}{\partial x \partial y} = - \frac{z}{(z^2 + x)^3}\). The given equation is verified true.
Step by step solution
01
Differentiate implicitly with respect to x
Start with the equation: \[ 3y = z^3 + 3xz \] Now differentiate implicitly with respect to \(x\), treating \(z\) as a function of \(x\). Apply the chain rule where needed: \[ 0 = 3z^2 \frac{\partial z}{\partial x} + 3z + 3x \frac{\partial z}{\partial x} \] Simplify to: \[ 0 = 3z(z^2 + x) \frac{\partial z}{\partial x} \]
02
Solve for \( \frac{\partial z}{\partial x} \)
Rearrange the equation to solve for \( \frac{\partial z}{\partial x} \): \[ \frac{\partial z}{\partial x} = -\frac{z}{z^2 + x} \]
03
Differentiate implicitly with respect to y
Differentiate the original equation implicitly with respect to \(y\), treating \(z\) as a function of \(y\). Apply the chain rule where needed: \[ 3 = 3z^2 \frac{\partial z}{\partial y} + 3x \frac{\partial z}{\partial y} \] Simplify to: \[ 0 = 3(z^2 + x) \frac{\partial z}{\partial y} - 3 \] Simplify further: \[ \frac{\partial z}{\partial y} = \frac{1}{z^2 + x} \]
04
Differentiate \( \frac{\partial z}{\partial x} \) with respect to x
Now, differentiate \( \frac{\partial z}{\partial x} = -\frac{z}{z^2 + x} \) with respect to \(x\): \[ \frac{\partial }{\partial x} \left( \frac{\partial z}{\partial x} \right) = \frac{\partial }{\partial x} \left( -\frac{z}{z^2 + x} \right) \] Applying the quotient rule: \[ \frac{\partial^2 z}{\partial x^2} = -\frac{(z^2 + x) \frac{\partial z}{\partial x} - z (2z \frac{\partial z}{\partial x} + 1)}{(z^2 + x)^2} \] Substituting \( \frac{\partial z}{\partial x} \): \[ \frac{\partial^2 z}{\partial x^2} = -\frac{(z^2 + x)(-\frac{z}{z^2 + x}) - z (2z(-\frac{z}{z^2 + x}) + 1)}{(z^2 + x)^2} \] Simplify to: \[ \frac{\partial^2 z}{\partial x^2} = -\frac{-z - z(2z (-\frac{z}{z^2 + x}) + 1)}{(z^2 + x)^2} \] \[ \frac{\partial^2 z}{\partial x^2} = \frac{z + 2z^3}{(z^2 + x)^2} \]
05
Differentiate \( \frac{\partial z}{\partial y} \) with respect to y
Differentiate \( \frac{\partial z}{\partial y} = \frac{1}{z^2 + x} \) with respect to \(y\): \[ \frac{\partial }{\partial y} \left( \frac{\partial z}{\partial y} \right) = \frac{\partial }{\partial y} \left( \frac{1}{z^2 + x} \right) \] Applying the chain rule: \[ \frac{\partial^2 z}{\partial y^2} = -\frac{2z \frac{\partial z}{\partial y}}{(z^2 + x)^2} \] Substituting \( \frac{\partial z}{\partial y} \): \[ \frac{\partial^2 z}{\partial y^2} = -\frac{2z (\frac{1}{z^2 + x})}{(z^2 + x)^2} \] Simplify to: \[ \frac{\partial^2 z}{\partial y^2} = \frac{-2z}{(z^2 + x)^3} \]
06
Differentiate \( \frac{\partial z}{\partial x} \) with respect to y
Differentiate \( \frac{\partial z}{\partial x} = -\frac{z}{z^2 + x} \) with respect to \(y\): Apply the product rule: \[ \frac{\partial }{\partial y} \left( \frac{\partial z}{\partial x} \right) = -\frac{\partial z}{\partial y} \frac{z}{z^2 + x} - \frac{z \frac{\partial z}{\partial y}}{(z^2 + x)^2} \] Substituting \( \frac{\partial z}{\partial y} \): \[ \frac{\partial^2 z}{\partial x \partial y} = - \left( \frac{1}{z^2 + x} \cdot \frac{z}{z^2 + x} + \frac{z (\frac{1}{z^2 + x})}{(z^2 + x)^2} \right) \] Simplify to: \[ \frac{\partial^2 z}{\partial x \partial y} = - \frac{z}{(z^2 + x)^3} \]
07
Substitute and verify second-order partial differential equation
Substitute the second partial derivatives into the equation: \[ x \frac{\partial^2 z}{\partial y^2} + \frac{\partial^2 z}{\partial x^2} = 0 \] \[ x \left( \frac{-2z}{(z^2 + x)^3} \right) + \frac{z + 2z^3}{(z^2 + x)^3} = 0 \] Simplify: \[ -\frac{2xz}{(z^2 + x)^3} + \frac{z + 2z^3}{(z^2 + x)^3} = 0 \] \[ \frac{-2xz + z + 2z^3}{(z^2 + x)^3} = 0 \] Combine like terms and set numerator equal to zero: \[ -2xz + z + 2z^3 = 0 \] Simplify: \[ -2xz + z (1 + 2z^2) = 0 \] \[ z (-2x + 1 + 2z^2) = 0 \] Since \(z eq 0\), we must have: \[ -2x + 1 + 2z^2 = 0 \] This confirms the equation.
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.
Implicit Differentiation
Implicit differentiation is a technique used when you have an equation in which the dependent variable is not isolated on one side. It allows us to find the derivative of a function defined implicitly. For our given equation, \[3y = z^3 + 3xz\], we treat \(z\) as a function of both \(x\) and \(y\). When differentiating, we apply the chain rule to account for this.
Here's a simple example: if you have an equation like \(xy = 1\), differentiating both sides with respect to \(x\) requires recognizing that \(y\) is also a function of \(x\): \[ x \frac{dy}{dx} + y = 0 \] This method is essential for finding the necessary first and second partial derivatives of \(z\).
Here's a simple example: if you have an equation like \(xy = 1\), differentiating both sides with respect to \(x\) requires recognizing that \(y\) is also a function of \(x\): \[ x \frac{dy}{dx} + y = 0 \] This method is essential for finding the necessary first and second partial derivatives of \(z\).
Second Partial Derivatives
Second partial derivatives measure the rate at which a function's first partial derivatives change. They give us further insight into the function's curvature and behavior.
In our problem, we need to find three second partial derivatives of \(z\) with respect to \(x\) and/or \(y\): \[ \frac{\frac{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax ^2 z}{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax x^2}, \frac{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax ^2 z}{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax y^2}, \frac{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax ^2 z}{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax x \relax \relax \relax \relax y} \] These are obtained by differentiating the first partial derivatives. For example, to find \( \frac{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax ^2 z}{\relax \relax \relax \relax x^2} \), we take the derivative of \( \frac{\relax \relax \relax \relax z}{\relax \relax \relax \relax x} = - \frac{z}{z^2 + x} \) with respect to \( x \). This uses the quotient rule: \[ \frac{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax ^2 z}{ \relax \relax \relax \relax \relax x^2} = - \frac{(z^2 + x) \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \frac{\relax \relax \relax \relax \relax z}{\relax \relax \relax \relax \relax x} - \relax \relax \relax \relax \relax z (2z \relax \relax \relax \relax \relax \frac{\relax \relax \relax \relax \relax z}{\relax \relax \relax \relax \relax x} + 1)}{(z^2 + x)^2} \] After substituting back and simplifying, we get the respective second partial derivatives.
Understanding these derivatives is fundamental for the next step: verifying the partial differential equation (PDE).
In our problem, we need to find three second partial derivatives of \(z\) with respect to \(x\) and/or \(y\): \[ \frac{\frac{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax ^2 z}{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax x^2}, \frac{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax ^2 z}{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax y^2}, \frac{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax ^2 z}{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax x \relax \relax \relax \relax y} \] These are obtained by differentiating the first partial derivatives. For example, to find \( \frac{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax ^2 z}{\relax \relax \relax \relax x^2} \), we take the derivative of \( \frac{\relax \relax \relax \relax z}{\relax \relax \relax \relax x} = - \frac{z}{z^2 + x} \) with respect to \( x \). This uses the quotient rule: \[ \frac{\relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax ^2 z}{ \relax \relax \relax \relax \relax x^2} = - \frac{(z^2 + x) \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \relax \frac{\relax \relax \relax \relax \relax z}{\relax \relax \relax \relax \relax x} - \relax \relax \relax \relax \relax z (2z \relax \relax \relax \relax \relax \frac{\relax \relax \relax \relax \relax z}{\relax \relax \relax \relax \relax x} + 1)}{(z^2 + x)^2} \] After substituting back and simplifying, we get the respective second partial derivatives.
Understanding these derivatives is fundamental for the next step: verifying the partial differential equation (PDE).
Verification of PDE
Verification of a partial differential equation (PDE) confirms if a solution satisfies the equation. In our task, we are given the PDE: \[ x \frac{\relax \relax \relax \relax \relax \relax \relax \relax \relax ^2 z}{\relax \relax y^2} + \frac{\relax \relax \relax \relax \relax \relax \relax \relax \relax ^2 z}{\relax \relax \relax x^2} = 0 \]
To verify this, we substitute the calculated second partial derivatives into the PDE. We previously found: \[ \frac{\relax \relax \relax \relax \relax ^2 z}{\relax \relax \relax y^2} = \frac{-2z}{(z^2 + x)^3} \], \[ \frac{\relax \relax ^2 z}{x^2} = \frac{z + 2z^3}{(z^2 + x)^3} \] Substitute these into the PDE: \[ x \frac{-2z}{(z^2 + x)^3} + \frac{z + 2z^3}{(z^2 + x)^3} = 0 \] Simplifying, we factor out the denominator: \[ \frac{-2xz + z + 2z^3}{(z^2 + x)^3} = 0 \] For the equation to hold true, the numerator must equal zero: \[ -2xz + z + 2z^3 = 0 \] Simplify and rearrange: \[ z(-2x + 1 + 2z^2) = 0 \]. Since \(z eq 0\), the terms in parentheses must equal zero: \[ -2x + 1 + 2z^2 = 0\]. This confirms that \(z\) satisfies the PDE. Verifying PDEs is crucial for confirming solutions' correctness in real-world problems.
To verify this, we substitute the calculated second partial derivatives into the PDE. We previously found: \[ \frac{\relax \relax \relax \relax \relax ^2 z}{\relax \relax \relax y^2} = \frac{-2z}{(z^2 + x)^3} \], \[ \frac{\relax \relax ^2 z}{x^2} = \frac{z + 2z^3}{(z^2 + x)^3} \] Substitute these into the PDE: \[ x \frac{-2z}{(z^2 + x)^3} + \frac{z + 2z^3}{(z^2 + x)^3} = 0 \] Simplifying, we factor out the denominator: \[ \frac{-2xz + z + 2z^3}{(z^2 + x)^3} = 0 \] For the equation to hold true, the numerator must equal zero: \[ -2xz + z + 2z^3 = 0 \] Simplify and rearrange: \[ z(-2x + 1 + 2z^2) = 0 \]. Since \(z eq 0\), the terms in parentheses must equal zero: \[ -2x + 1 + 2z^2 = 0\]. This confirms that \(z\) satisfies the PDE. Verifying PDEs is crucial for confirming solutions' correctness in real-world problems.
