Question

Assume \(F(x, y, z(x, y))=0\) implicitly defines \(z\) as a differentiable function of \(x\) and \(y .\) Extend Theorem 15.9 to show that $$\frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}} \text { and } \frac{\partial z}{\partial y}=-\frac{F_{y}}{F_{z}}$$.

Short answer

Question: Given a function F(x, y, z(x, y)) = 0, which implicitly defines z as a differentiable function of x and y, find the partial derivatives of z with respect to x and y, by extending Theorem 15.9. Answer: The partial derivatives of z with respect to x and y are given by: $$\frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}} \text { and } \frac{\partial z}{\partial y}=-\frac{F_{y}}{F_{z}}$$

Step-by-step solution

Implicit differentiation with respect to x

Given the equation F(x, y, z(x, y)) = 0, we will first find the partial derivative with respect to x. In order to do this, we will implicitly differentiate both sides of the equation with respect to x, treating y as a constant and using the chain rule for z(x, y). Now, $$\frac{\partial F}{\partial x} = \frac{\partial F}{\partial x} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial x} = 0$$

Solving for ∂z/∂x

To find the partial derivative of z with respect to x, we can now solve the above equation for ∂z/∂x. $$\frac{\partial z}{\partial x} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}} = -\frac{F_{x}}{F_{z}}$$

Implicit differentiation with respect to y

Now, we will repeat the previous steps but with respect to y. We'll implicitly differentiate both sides of the equation F(x, y, z(x, y)) = 0 with respect to y, treating x as a constant and using the chain rule again for z(x, y): $$\frac{\partial F}{\partial y} = \frac{\partial F}{\partial y} + \frac{\partial F}{\partial z}\frac{\partial z}{\partial y} = 0$$

Solving for ∂z/∂y

To find the partial derivative of z with respect to y, we can now solve the above equation for ∂z/∂y. $$\frac{\partial z}{\partial y} = -\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}} = -\frac{F_{y}}{F_{z}}$$

Conclusion

Finally, we have found the partial derivatives of z with respect to x and y: $$\frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}} \text { and } \frac{\partial z}{\partial y}=-\frac{F_{y}}{F_{z}}$$

Key concepts

Partial Derivatives

Partial derivatives are a fundamental tool in calculus when dealing with functions of multiple variables. They allow us to understand how a function changes with respect to one variable while keeping the others constant. In the context of the given problem, the partial derivatives \(\frac{\partial F}{\partial x}\) and \(\frac{\partial F}{\partial y}\) denote how the function \(F\) changes as \(x\) or \(y\) varies, holding other variables constant.

Partial derivatives are especially useful in problems like the one here, where \(z\) is defined implicitly by an equation \(F(x, y, z(x, y)) = 0\). Instead of solving for \(z\) explicitly, partial derivatives provide a way to calculate how \(z\) changes by computing gradients and slopes. Understanding this concept is vital to differentiate implicitly and find the rate of change of \(z\) with respect to both \(x\) and \(y\).

• For \(\frac{\partial z}{\partial x}\), treat \(y\) as constant, and similarly, for \(\frac{\partial z}{\partial y}\), treat \(x\) as constant.
• Focus on the implications of each variable individually while others are constant.

Chain Rule

The chain rule is an essential concept that helps in differentiating composite functions. It explains how to find the derivative of a function that is the composition of multiple other functions. When dealing with multiple variables and implicit differentiation, the chain rule becomes incredibly powerful.

In this exercise, the chain rule plays a vital role in computing the partial derivatives of \(z\) with respect to \(x\) and \(y\). Since \(z\) itself is a function of both \(x\) and \(y\), we use the chain rule to differentiate \(F(x, y, z(x, y)) = 0\) with respect to \(x\) and \(y\).

• For differentiating with respect to \(x\), the rule applied as \(\frac{\partial F}{\partial x} + \frac{\partial F}{\partial z} \cdot \frac{\partial z}{\partial x} = 0\).
• For differentiating with respect to \(y\), the application is similar, i.e., \(\frac{\partial F}{\partial y} + \frac{\partial F}{\partial z} \cdot \frac{\partial z}{\partial y} = 0\).

Thus, the chain rule helps break down these composite functions, allowing us to find the individual impacts of each variable's change on \(z\).

Function of Multiple Variables

In mathematics, functions of multiple variables are a way to describe relationships between multiple input values and an output. They are central to understanding complex systems in many scientific fields. In this exercise, the function \(F(x, y, z(x, y))\) depends on three variables, where \(z\) is implicitly defined as a function of \(x\) and \(y\).

Working with multivariable functions requires considering each variable independently as well as how they interact collectively. This expands the possibility of observing dynamics that cannot be captured by single-variable functions. For our problem:

• Recognize that each variable might influence \(z\) directly and indirectly through other variables.
• Understand how changes in one input could potentially modify the others, leading to variations in the output, \(z\).

By analyzing functions of multiple variables, mathematical models can provide detailed insights into phenomena involving several interdependent factors.

Implicit Function Theorem

The Implicit Function Theorem is a highly useful result in calculus, providing conditions under which a system of equations implies the existence of a function. It states that, if a function \(F\) and its derivatives satisfy certain conditions, we can solve for one variable as a function of others. This theorem simplifies the process of finding derivatives in a complex system of equations.

In the given problem setting, the Implicit Function Theorem is used to derive the formulas for the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). Since \(F(x, y, z(x, y)) = 0\), and \(z\) is a differentiable function of \(x\) and \(y\), the theorem predicts:

• If \(\frac{\partial F}{\partial z} eq 0\), then \(z\) can be expressed as a function of \(x\) and \(y\).
• The derivatives with respect to \(x\) and \(y\) can then be calculated as \(\frac{\partial z}{\partial x}=-\frac{F_{x}}{F_{z}}\) and \(\frac{\partial z}{\partial y}=-\frac{F_{y}}{F_{z}}\).

This theorem elegantly provides a way to analyze implicit functions without needing to solve them explicitly.