Problem 58
Question
Suppose that the equation \(F(x, y, z)=0\) implicitly defines each of the three variables \(x, y,\) and \(z\) as functions of the other two: \(z=f(x, y), y=g(x, z), x=h(y, z) .\) If \(F\) is differentiable and \(F_{x}, F_{y},\) and \(F_{z}\) are all nonzero, show that $$\frac{\partial z}{\partial x} \frac{\partial x}{\partial y} \frac{\partial y}{\partial z}=-1$$
Step-by-Step Solution
Verified Answer
The product of the partial derivatives is -1 using implicit differentiation.
1Step 1: Understand the problem
We need to show that the product of three partial derivatives \( \frac{\partial z}{\partial x} \), \( \frac{\partial x}{\partial y} \), and \( \frac{\partial y}{\partial z} \) equals \(-1\). Each variable \(x\), \(y\), and \(z\) is implicitly defined as a function of the other two via a differentiable function \(F\).
2Step 2: Use the Implicit Function Theorem
According to the implicit function theorem, if \(F(x, y, z) = 0\) defines \(z = f(x, y)\), then the partial derivative of \(z\) with respect to \(x\) can be expressed as \(\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}\). Similarly, apply it to find \(\frac{\partial y}{\partial z}\) and \(\frac{\partial x}{\partial y}\).
3Step 3: Calculate \(\frac{\partial y}{\partial z}\) and \(\frac{\partial x}{\partial y}\)
For \(y = g(x, z)\), we have \(\frac{\partial y}{\partial z} = -\frac{F_z}{F_y}\). For \(x = h(y, z)\), \(\frac{\partial x}{\partial y} = -\frac{F_y}{F_x}\). These results are derived using the implicit function theorem similarly as in step 2.
4Step 4: Multiply the Partial Derivatives
Now, multiply the partial derivatives obtained: \(\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \frac{\partial x}{\partial y} = -\frac{F_y}{F_x}, \frac{\partial y}{\partial z} = -\frac{F_z}{F_y}\).
5Step 5: Simplify the Product
Multiply the expressions: \(\left( -\frac{F_x}{F_z} \right)\left( -\frac{F_y}{F_x} \right)\left( -\frac{F_z}{F_y} \right)\). The negatives cancel each other out, resulting in \(-1\). The terms \(F_x, F_y\), and \(F_z\) cancel each other, verifying the desired identity.
Key Concepts
Partial DerivativesImplicit DifferentiationCalculus
Partial Derivatives
Partial derivatives are a fundamental part of calculus, especially when dealing with functions of multiple variables. They represent the rate at which a function changes as one of the variables changes, while all other variables are held constant. For example, in a function of two variables like \( f(x, y) \), the partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), indicates how \( f \) changes as \( x \) changes, keeping \( y \) constant.
In the context of the problem, we work with a function \( F(x, y, z) \) where every variable can be considered as a function of the other two. Each partial derivative such as \( \frac{\partial z}{\partial x} \) shows how the implicit function \( z = f(x, y) \) changes the value of \( z \) in response to \( x \). Hence, understanding partial derivatives helps unravel the behavior of implicit functions by focusing on one perspective of change at a time. It's a way of breaking down complex systems into more manageable parts.
In the context of the problem, we work with a function \( F(x, y, z) \) where every variable can be considered as a function of the other two. Each partial derivative such as \( \frac{\partial z}{\partial x} \) shows how the implicit function \( z = f(x, y) \) changes the value of \( z \) in response to \( x \). Hence, understanding partial derivatives helps unravel the behavior of implicit functions by focusing on one perspective of change at a time. It's a way of breaking down complex systems into more manageable parts.
- Partial derivatives focus on changes in one variable
- Useful in functions with multiple variables
- Key in understanding implicit relationships between variables
Implicit Differentiation
Implicit differentiation is a technique used when the relationship between variables is given in a form that does not explicitly express one variable in terms of another, such as \( F(x, y, z) = 0 \). In these cases, finding derivatives directly is not possible, but implicit differentiation allows us to differentiate both sides of an equation and solve for the derivative of the desired variable.
This is crucial when functions are defined implicitly rather than explicitly. The Implicit Function Theorem plays a pivotal role here, as it allows us to express one variable as a function of the others. Given \( F(x, y, z) = 0 \) and assuming \( F \) is differentiable with non-zero derivatives \( F_x \), \( F_y \), and \( F_z \), implicit differentiation derives expressions like \( \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} \), indicating how changes in \( x \) can lead to changes in \( z \). This technique is instrumental in advancing calculus solutions to more complex scenarios.
This is crucial when functions are defined implicitly rather than explicitly. The Implicit Function Theorem plays a pivotal role here, as it allows us to express one variable as a function of the others. Given \( F(x, y, z) = 0 \) and assuming \( F \) is differentiable with non-zero derivatives \( F_x \), \( F_y \), and \( F_z \), implicit differentiation derives expressions like \( \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} \), indicating how changes in \( x \) can lead to changes in \( z \). This technique is instrumental in advancing calculus solutions to more complex scenarios.
- Used when variables are intertwined
- Generates derivatives without explicit functions
- Relies on the Implicit Function Theorem
Calculus
Calculus is a branch of mathematics dealing with rates of change (differential calculus) and accumulation of quantities (integral calculus). Fundamental to calculus is the concept of a derivative, which signifies how a quantity changes in response to a change in another quantity.
When it comes to solving equations such as \( F(x, y, z) = 0 \), calculus provides us with robust tools like derivative computation and integration techniques. In the given problem, we employ partial derivatives and implicit differentiation—both crucial elements of calculus—to demonstrate the equation \( \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z} = -1 \). These operations form the backbone of solving such mathematical problems.
When it comes to solving equations such as \( F(x, y, z) = 0 \), calculus provides us with robust tools like derivative computation and integration techniques. In the given problem, we employ partial derivatives and implicit differentiation—both crucial elements of calculus—to demonstrate the equation \( \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z} = -1 \). These operations form the backbone of solving such mathematical problems.
- Calculus handles change and rates of change
- Includes differential and integral calculus
- Essential for understanding functions and their behaviors
Other exercises in this chapter
Problem 57
Verify that the conclusion of Clairaut's Theorem holds, that is, \(u_{x y}=u_{y x}\) $$u=x \sin (x+2 y)$$
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Show that the sum of the \(x^{-}, y_{-}\) and \(z\) -intercepts of any tangent plane to the surface \(\sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{c}\) is a constant.
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Verify that the conclusion of Clairaut's Theorem holds, that is, \(u_{x y}=u_{y x}\) $$u=x^{4} y^{2}-2 x y^{5}$$
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Show that the pyramids cut off from the first octant by any tangent planes to the surface \(x y z=1\) at points in the first octant must all have the same volum
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